YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Trafo_06/dup16.srs-torpacyc2out-split.srs

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 QTRSRRRProof (⇔, 367 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 141 ms)
↳4 QTRS
↳5 Overlay + Local Confluence (⇔, 535 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 159 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 AND
    ↳11 QDP
        ↳12 UsableRulesProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QReductionProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QReductionProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 QReductionProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QReductionProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 QDPSizeChangeProof (⇔, 0 ms)
        ↳38 YES
    ↳39 QDP
        ↳40 UsableRulesProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 QReductionProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 QDPSizeChangeProof (⇔, 0 ms)
        ↳45 YES
    ↳46 QDP
        ↳47 UsableRulesProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 QReductionProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 QDPSizeChangeProof (⇔, 0 ms)
        ↳52 YES
    ↳53 QDP
        ↳54 UsableRulesProof (⇔, 0 ms)
        ↳55 QDP
        ↳56 QReductionProof (⇔, 0 ms)
        ↳57 QDP
        ↳58 QDPSizeChangeProof (⇔, 0 ms)
        ↳59 YES
    ↳60 QDP
        ↳61 UsableRulesProof (⇔, 0 ms)
        ↳62 QDP
        ↳63 QReductionProof (⇔, 0 ms)
        ↳64 QDP
        ↳65 QDPSizeChangeProof (⇔, 0 ms)
        ↳66 YES
    ↳67 QDP
        ↳68 UsableRulesProof (⇔, 0 ms)
        ↳69 QDP
        ↳70 QReductionProof (⇔, 0 ms)
        ↳71 QDP
        ↳72 QDPSizeChangeProof (⇔, 0 ms)
        ↳73 YES
    ↳74 QDP
        ↳75 UsableRulesProof (⇔, 0 ms)
        ↳76 QDP
        ↳77 QReductionProof (⇔, 0 ms)
        ↳78 QDP
        ↳79 QDPSizeChangeProof (⇔, 0 ms)
        ↳80 YES
    ↳81 QDP
        ↳82 UsableRulesProof (⇔, 0 ms)
        ↳83 QDP
        ↳84 QReductionProof (⇔, 0 ms)
        ↳85 QDP
        ↳86 QDPSizeChangeProof (⇔, 0 ms)
        ↳87 YES
    ↳88 QDP
        ↳89 UsableRulesProof (⇔, 0 ms)
        ↳90 QDP
        ↳91 QReductionProof (⇔, 0 ms)
        ↳92 QDP
        ↳93 QDPSizeChangeProof (⇔, 0 ms)
        ↳94 YES
    ↳95 QDP
        ↳96 UsableRulesProof (⇔, 0 ms)
        ↳97 QDP
        ↳98 QReductionProof (⇔, 0 ms)
        ↳99 QDP
        ↳100 QDPSizeChangeProof (⇔, 0 ms)
        ↳101 YES
    ↳102 QDP
        ↳103 UsableRulesProof (⇔, 0 ms)
        ↳104 QDP
        ↳105 QReductionProof (⇔, 0 ms)
        ↳106 QDP
        ↳107 QDPSizeChangeProof (⇔, 0 ms)
        ↳108 YES
    ↳109 QDP
        ↳110 UsableRulesProof (⇔, 0 ms)
        ↳111 QDP
        ↳112 QReductionProof (⇔, 0 ms)
        ↳113 QDP
        ↳114 QDPSizeChangeProof (⇔, 0 ms)
        ↳115 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(b(b(b(x)))) → Wait(Right4(x))
Begin(b(b(x))) → Wait(Right5(x))
Begin(b(x)) → Wait(Right6(x))
Begin(c(c(c(x)))) → Wait(Right7(x))
Begin(c(c(x))) → Wait(Right8(x))
Begin(c(x)) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(c(d(d(d(d(x)))))) → Wait(Right11(x))
Begin(d(d(d(d(x))))) → Wait(Right12(x))
Begin(d(d(d(x)))) → Wait(Right13(x))
Begin(d(d(x))) → Wait(Right14(x))
Begin(d(x)) → Wait(Right15(x))
Right1(a(End(x))) → Left(b(b(b(b(b(b(End(x))))))))
Right2(a(a(End(x)))) → Left(b(b(b(b(b(b(End(x))))))))
Right3(a(a(a(End(x))))) → Left(b(b(b(b(b(b(End(x))))))))
Right4(b(End(x))) → Left(c(c(c(c(c(c(End(x))))))))
Right5(b(b(End(x)))) → Left(c(c(c(c(c(c(End(x))))))))
Right6(b(b(b(End(x))))) → Left(c(c(c(c(c(c(End(x))))))))
Right7(c(End(x))) → Left(d(d(d(d(d(d(End(x))))))))
Right8(c(c(End(x)))) → Left(d(d(d(d(d(d(End(x))))))))
Right9(c(c(c(End(x))))) → Left(d(d(d(d(d(d(End(x))))))))
Right10(b(End(x))) → Left(d(d(d(d(End(x))))))
Right11(c(End(x))) → Left(a(a(End(x))))
Right12(c(c(End(x)))) → Left(a(a(End(x))))
Right13(c(c(d(End(x))))) → Left(a(a(End(x))))
Right14(c(c(d(d(End(x)))))) → Left(a(a(End(x))))
Right15(c(c(d(d(d(End(x))))))) → Left(a(a(End(x))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
a(a(a(a(x)))) → b(b(b(b(b(b(x))))))
b(b(b(b(x)))) → c(c(c(c(c(c(x))))))
c(c(c(c(x)))) → d(d(d(d(d(d(x))))))
b(b(x)) → d(d(d(d(x))))
c(c(d(d(d(d(x)))))) → a(a(x))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 323 + x1   
POL(Ab(x1)) = 215 + x1   
POL(Ac(x1)) = 143 + x1   
POL(Ad(x1)) = 95 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 968 + x1   
POL(Right10(x1)) = 212 + x1   
POL(Right11(x1)) = 515 + x1   
POL(Right12(x1)) = 376 + x1   
POL(Right13(x1)) = 284 + x1   
POL(Right14(x1)) = 188 + x1   
POL(Right15(x1)) = 82 + x1   
POL(Right2(x1)) = 645 + x1   
POL(Right3(x1)) = 322 + x1   
POL(Right4(x1)) = 644 + x1   
POL(Right5(x1)) = 429 + x1   
POL(Right6(x1)) = 214 + x1   
POL(Right7(x1)) = 428 + x1   
POL(Right8(x1)) = 285 + x1   
POL(Right9(x1)) = 142 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 323 + x1   
POL(b(x1)) = 215 + x1   
POL(c(x1)) = 143 + x1   
POL(d(x1)) = 95 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Begin(a(a(a(x)))) → Wait(Right1(x))
Begin(a(a(x))) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Begin(b(b(b(x)))) → Wait(Right4(x))
Begin(b(b(x))) → Wait(Right5(x))
Begin(b(x)) → Wait(Right6(x))
Begin(c(c(c(x)))) → Wait(Right7(x))
Begin(c(c(x))) → Wait(Right8(x))
Begin(c(x)) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(c(d(d(d(d(x)))))) → Wait(Right11(x))
Begin(d(d(d(d(x))))) → Wait(Right12(x))
Begin(d(d(d(x)))) → Wait(Right13(x))
Begin(d(d(x))) → Wait(Right14(x))
Begin(d(x)) → Wait(Right15(x))
Right1(a(End(x))) → Left(b(b(b(b(b(b(End(x))))))))
Right2(a(a(End(x)))) → Left(b(b(b(b(b(b(End(x))))))))
Right3(a(a(a(End(x))))) → Left(b(b(b(b(b(b(End(x))))))))
Right4(b(End(x))) → Left(c(c(c(c(c(c(End(x))))))))
Right5(b(b(End(x)))) → Left(c(c(c(c(c(c(End(x))))))))
Right6(b(b(b(End(x))))) → Left(c(c(c(c(c(c(End(x))))))))
Right7(c(End(x))) → Left(d(d(d(d(d(d(End(x))))))))
Right8(c(c(End(x)))) → Left(d(d(d(d(d(d(End(x))))))))
Right9(c(c(c(End(x))))) → Left(d(d(d(d(d(d(End(x))))))))
Right10(b(End(x))) → Left(d(d(d(d(End(x))))))
Right11(c(End(x))) → Left(a(a(End(x))))
Right12(c(c(End(x)))) → Left(a(a(End(x))))
Right13(c(c(d(End(x))))) → Left(a(a(End(x))))
Right14(c(c(d(d(End(x)))))) → Left(a(a(End(x))))
Right15(c(c(d(d(d(End(x))))))) → Left(a(a(End(x))))
a(a(a(a(x)))) → b(b(b(b(b(b(x))))))
b(b(b(b(x)))) → c(c(c(c(c(c(x))))))
c(c(c(c(x)))) → d(d(d(d(d(d(x))))))
b(b(x)) → d(d(d(d(x))))
c(c(d(d(d(d(x)))))) → a(a(x))


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(Ac(x1)) = x1   
POL(Ad(x1)) = x1   
POL(Begin(x1)) = x1   
POL(Left(x1)) = 1 + x1   
POL(Right1(x1)) = x1   
POL(Right10(x1)) = x1   
POL(Right11(x1)) = x1   
POL(Right12(x1)) = x1   
POL(Right13(x1)) = x1   
POL(Right14(x1)) = x1   
POL(Right15(x1)) = x1   
POL(Right2(x1)) = x1   
POL(Right3(x1)) = x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = x1   
POL(Right6(x1)) = x1   
POL(Right7(x1)) = x1   
POL(Right8(x1)) = x1   
POL(Right9(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

Wait(Left(x)) → Begin(x)


(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

Q is empty.

(5) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(6) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(7) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
RIGHT4(a(x)) → AA(Right4(x))
RIGHT4(a(x)) → RIGHT4(x)
RIGHT5(a(x)) → AA(Right5(x))
RIGHT5(a(x)) → RIGHT5(x)
RIGHT6(a(x)) → AA(Right6(x))
RIGHT6(a(x)) → RIGHT6(x)
RIGHT7(a(x)) → AA(Right7(x))
RIGHT7(a(x)) → RIGHT7(x)
RIGHT8(a(x)) → AA(Right8(x))
RIGHT8(a(x)) → RIGHT8(x)
RIGHT9(a(x)) → AA(Right9(x))
RIGHT9(a(x)) → RIGHT9(x)
RIGHT10(a(x)) → AA(Right10(x))
RIGHT10(a(x)) → RIGHT10(x)
RIGHT11(a(x)) → AA(Right11(x))
RIGHT11(a(x)) → RIGHT11(x)
RIGHT12(a(x)) → AA(Right12(x))
RIGHT12(a(x)) → RIGHT12(x)
RIGHT13(a(x)) → AA(Right13(x))
RIGHT13(a(x)) → RIGHT13(x)
RIGHT14(a(x)) → AA(Right14(x))
RIGHT14(a(x)) → RIGHT14(x)
RIGHT15(a(x)) → AA(Right15(x))
RIGHT15(a(x)) → RIGHT15(x)
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT4(b(x)) → AB(Right4(x))
RIGHT4(b(x)) → RIGHT4(x)
RIGHT5(b(x)) → AB(Right5(x))
RIGHT5(b(x)) → RIGHT5(x)
RIGHT6(b(x)) → AB(Right6(x))
RIGHT6(b(x)) → RIGHT6(x)
RIGHT7(b(x)) → AB(Right7(x))
RIGHT7(b(x)) → RIGHT7(x)
RIGHT8(b(x)) → AB(Right8(x))
RIGHT8(b(x)) → RIGHT8(x)
RIGHT9(b(x)) → AB(Right9(x))
RIGHT9(b(x)) → RIGHT9(x)
RIGHT10(b(x)) → AB(Right10(x))
RIGHT10(b(x)) → RIGHT10(x)
RIGHT11(b(x)) → AB(Right11(x))
RIGHT11(b(x)) → RIGHT11(x)
RIGHT12(b(x)) → AB(Right12(x))
RIGHT12(b(x)) → RIGHT12(x)
RIGHT13(b(x)) → AB(Right13(x))
RIGHT13(b(x)) → RIGHT13(x)
RIGHT14(b(x)) → AB(Right14(x))
RIGHT14(b(x)) → RIGHT14(x)
RIGHT15(b(x)) → AB(Right15(x))
RIGHT15(b(x)) → RIGHT15(x)
RIGHT1(c(x)) → AC(Right1(x))
RIGHT1(c(x)) → RIGHT1(x)
RIGHT2(c(x)) → AC(Right2(x))
RIGHT2(c(x)) → RIGHT2(x)
RIGHT3(c(x)) → AC(Right3(x))
RIGHT3(c(x)) → RIGHT3(x)
RIGHT4(c(x)) → AC(Right4(x))
RIGHT4(c(x)) → RIGHT4(x)
RIGHT5(c(x)) → AC(Right5(x))
RIGHT5(c(x)) → RIGHT5(x)
RIGHT6(c(x)) → AC(Right6(x))
RIGHT6(c(x)) → RIGHT6(x)
RIGHT7(c(x)) → AC(Right7(x))
RIGHT7(c(x)) → RIGHT7(x)
RIGHT8(c(x)) → AC(Right8(x))
RIGHT8(c(x)) → RIGHT8(x)
RIGHT9(c(x)) → AC(Right9(x))
RIGHT9(c(x)) → RIGHT9(x)
RIGHT10(c(x)) → AC(Right10(x))
RIGHT10(c(x)) → RIGHT10(x)
RIGHT11(c(x)) → AC(Right11(x))
RIGHT11(c(x)) → RIGHT11(x)
RIGHT12(c(x)) → AC(Right12(x))
RIGHT12(c(x)) → RIGHT12(x)
RIGHT13(c(x)) → AC(Right13(x))
RIGHT13(c(x)) → RIGHT13(x)
RIGHT14(c(x)) → AC(Right14(x))
RIGHT14(c(x)) → RIGHT14(x)
RIGHT15(c(x)) → AC(Right15(x))
RIGHT15(c(x)) → RIGHT15(x)
RIGHT1(d(x)) → AD(Right1(x))
RIGHT1(d(x)) → RIGHT1(x)
RIGHT2(d(x)) → AD(Right2(x))
RIGHT2(d(x)) → RIGHT2(x)
RIGHT3(d(x)) → AD(Right3(x))
RIGHT3(d(x)) → RIGHT3(x)
RIGHT4(d(x)) → AD(Right4(x))
RIGHT4(d(x)) → RIGHT4(x)
RIGHT5(d(x)) → AD(Right5(x))
RIGHT5(d(x)) → RIGHT5(x)
RIGHT6(d(x)) → AD(Right6(x))
RIGHT6(d(x)) → RIGHT6(x)
RIGHT7(d(x)) → AD(Right7(x))
RIGHT7(d(x)) → RIGHT7(x)
RIGHT8(d(x)) → AD(Right8(x))
RIGHT8(d(x)) → RIGHT8(x)
RIGHT9(d(x)) → AD(Right9(x))
RIGHT9(d(x)) → RIGHT9(x)
RIGHT10(d(x)) → AD(Right10(x))
RIGHT10(d(x)) → RIGHT10(x)
RIGHT11(d(x)) → AD(Right11(x))
RIGHT11(d(x)) → RIGHT11(x)
RIGHT12(d(x)) → AD(Right12(x))
RIGHT12(d(x)) → RIGHT12(x)
RIGHT13(d(x)) → AD(Right13(x))
RIGHT13(d(x)) → RIGHT13(x)
RIGHT14(d(x)) → AD(Right14(x))
RIGHT14(d(x)) → RIGHT14(x)
RIGHT15(d(x)) → AD(Right15(x))
RIGHT15(d(x)) → RIGHT15(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(9) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 15 SCCs with 60 less nodes.

(10) Complex Obligation (AND)

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT15(b(x)) → RIGHT15(x)
RIGHT15(a(x)) → RIGHT15(x)
RIGHT15(c(x)) → RIGHT15(x)
RIGHT15(d(x)) → RIGHT15(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(12) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(13) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT15(b(x)) → RIGHT15(x)
RIGHT15(a(x)) → RIGHT15(x)
RIGHT15(c(x)) → RIGHT15(x)
RIGHT15(d(x)) → RIGHT15(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(14) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT15(b(x)) → RIGHT15(x)
RIGHT15(a(x)) → RIGHT15(x)
RIGHT15(c(x)) → RIGHT15(x)
RIGHT15(d(x)) → RIGHT15(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT15(b(x)) → RIGHT15(x)
    The graph contains the following edges 1 > 1

  • RIGHT15(a(x)) → RIGHT15(x)
    The graph contains the following edges 1 > 1

  • RIGHT15(c(x)) → RIGHT15(x)
    The graph contains the following edges 1 > 1

  • RIGHT15(d(x)) → RIGHT15(x)
    The graph contains the following edges 1 > 1

(17) YES

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT14(b(x)) → RIGHT14(x)
RIGHT14(a(x)) → RIGHT14(x)
RIGHT14(c(x)) → RIGHT14(x)
RIGHT14(d(x)) → RIGHT14(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(19) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT14(b(x)) → RIGHT14(x)
RIGHT14(a(x)) → RIGHT14(x)
RIGHT14(c(x)) → RIGHT14(x)
RIGHT14(d(x)) → RIGHT14(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(21) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(22) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT14(b(x)) → RIGHT14(x)
RIGHT14(a(x)) → RIGHT14(x)
RIGHT14(c(x)) → RIGHT14(x)
RIGHT14(d(x)) → RIGHT14(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT14(b(x)) → RIGHT14(x)
    The graph contains the following edges 1 > 1

  • RIGHT14(a(x)) → RIGHT14(x)
    The graph contains the following edges 1 > 1

  • RIGHT14(c(x)) → RIGHT14(x)
    The graph contains the following edges 1 > 1

  • RIGHT14(d(x)) → RIGHT14(x)
    The graph contains the following edges 1 > 1

(24) YES

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT13(b(x)) → RIGHT13(x)
RIGHT13(a(x)) → RIGHT13(x)
RIGHT13(c(x)) → RIGHT13(x)
RIGHT13(d(x)) → RIGHT13(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(26) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT13(b(x)) → RIGHT13(x)
RIGHT13(a(x)) → RIGHT13(x)
RIGHT13(c(x)) → RIGHT13(x)
RIGHT13(d(x)) → RIGHT13(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(28) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT13(b(x)) → RIGHT13(x)
RIGHT13(a(x)) → RIGHT13(x)
RIGHT13(c(x)) → RIGHT13(x)
RIGHT13(d(x)) → RIGHT13(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(30) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT13(b(x)) → RIGHT13(x)
    The graph contains the following edges 1 > 1

  • RIGHT13(a(x)) → RIGHT13(x)
    The graph contains the following edges 1 > 1

  • RIGHT13(c(x)) → RIGHT13(x)
    The graph contains the following edges 1 > 1

  • RIGHT13(d(x)) → RIGHT13(x)
    The graph contains the following edges 1 > 1

(31) YES

(32) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT12(b(x)) → RIGHT12(x)
RIGHT12(a(x)) → RIGHT12(x)
RIGHT12(c(x)) → RIGHT12(x)
RIGHT12(d(x)) → RIGHT12(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(33) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(34) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT12(b(x)) → RIGHT12(x)
RIGHT12(a(x)) → RIGHT12(x)
RIGHT12(c(x)) → RIGHT12(x)
RIGHT12(d(x)) → RIGHT12(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(35) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(36) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT12(b(x)) → RIGHT12(x)
RIGHT12(a(x)) → RIGHT12(x)
RIGHT12(c(x)) → RIGHT12(x)
RIGHT12(d(x)) → RIGHT12(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(37) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT12(b(x)) → RIGHT12(x)
    The graph contains the following edges 1 > 1

  • RIGHT12(a(x)) → RIGHT12(x)
    The graph contains the following edges 1 > 1

  • RIGHT12(c(x)) → RIGHT12(x)
    The graph contains the following edges 1 > 1

  • RIGHT12(d(x)) → RIGHT12(x)
    The graph contains the following edges 1 > 1

(38) YES

(39) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT11(b(x)) → RIGHT11(x)
RIGHT11(a(x)) → RIGHT11(x)
RIGHT11(c(x)) → RIGHT11(x)
RIGHT11(d(x)) → RIGHT11(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(40) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT11(b(x)) → RIGHT11(x)
RIGHT11(a(x)) → RIGHT11(x)
RIGHT11(c(x)) → RIGHT11(x)
RIGHT11(d(x)) → RIGHT11(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(42) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT11(b(x)) → RIGHT11(x)
RIGHT11(a(x)) → RIGHT11(x)
RIGHT11(c(x)) → RIGHT11(x)
RIGHT11(d(x)) → RIGHT11(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT11(b(x)) → RIGHT11(x)
    The graph contains the following edges 1 > 1

  • RIGHT11(a(x)) → RIGHT11(x)
    The graph contains the following edges 1 > 1

  • RIGHT11(c(x)) → RIGHT11(x)
    The graph contains the following edges 1 > 1

  • RIGHT11(d(x)) → RIGHT11(x)
    The graph contains the following edges 1 > 1

(45) YES

(46) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT10(b(x)) → RIGHT10(x)
RIGHT10(a(x)) → RIGHT10(x)
RIGHT10(c(x)) → RIGHT10(x)
RIGHT10(d(x)) → RIGHT10(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(47) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(48) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT10(b(x)) → RIGHT10(x)
RIGHT10(a(x)) → RIGHT10(x)
RIGHT10(c(x)) → RIGHT10(x)
RIGHT10(d(x)) → RIGHT10(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(49) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(50) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT10(b(x)) → RIGHT10(x)
RIGHT10(a(x)) → RIGHT10(x)
RIGHT10(c(x)) → RIGHT10(x)
RIGHT10(d(x)) → RIGHT10(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(51) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT10(b(x)) → RIGHT10(x)
    The graph contains the following edges 1 > 1

  • RIGHT10(a(x)) → RIGHT10(x)
    The graph contains the following edges 1 > 1

  • RIGHT10(c(x)) → RIGHT10(x)
    The graph contains the following edges 1 > 1

  • RIGHT10(d(x)) → RIGHT10(x)
    The graph contains the following edges 1 > 1

(52) YES

(53) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT9(b(x)) → RIGHT9(x)
RIGHT9(a(x)) → RIGHT9(x)
RIGHT9(c(x)) → RIGHT9(x)
RIGHT9(d(x)) → RIGHT9(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(54) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(55) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT9(b(x)) → RIGHT9(x)
RIGHT9(a(x)) → RIGHT9(x)
RIGHT9(c(x)) → RIGHT9(x)
RIGHT9(d(x)) → RIGHT9(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(56) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(57) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT9(b(x)) → RIGHT9(x)
RIGHT9(a(x)) → RIGHT9(x)
RIGHT9(c(x)) → RIGHT9(x)
RIGHT9(d(x)) → RIGHT9(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(58) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT9(b(x)) → RIGHT9(x)
    The graph contains the following edges 1 > 1

  • RIGHT9(a(x)) → RIGHT9(x)
    The graph contains the following edges 1 > 1

  • RIGHT9(c(x)) → RIGHT9(x)
    The graph contains the following edges 1 > 1

  • RIGHT9(d(x)) → RIGHT9(x)
    The graph contains the following edges 1 > 1

(59) YES

(60) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT8(b(x)) → RIGHT8(x)
RIGHT8(a(x)) → RIGHT8(x)
RIGHT8(c(x)) → RIGHT8(x)
RIGHT8(d(x)) → RIGHT8(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(61) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(62) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT8(b(x)) → RIGHT8(x)
RIGHT8(a(x)) → RIGHT8(x)
RIGHT8(c(x)) → RIGHT8(x)
RIGHT8(d(x)) → RIGHT8(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(63) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(64) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT8(b(x)) → RIGHT8(x)
RIGHT8(a(x)) → RIGHT8(x)
RIGHT8(c(x)) → RIGHT8(x)
RIGHT8(d(x)) → RIGHT8(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(65) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT8(b(x)) → RIGHT8(x)
    The graph contains the following edges 1 > 1

  • RIGHT8(a(x)) → RIGHT8(x)
    The graph contains the following edges 1 > 1

  • RIGHT8(c(x)) → RIGHT8(x)
    The graph contains the following edges 1 > 1

  • RIGHT8(d(x)) → RIGHT8(x)
    The graph contains the following edges 1 > 1

(66) YES

(67) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT7(b(x)) → RIGHT7(x)
RIGHT7(a(x)) → RIGHT7(x)
RIGHT7(c(x)) → RIGHT7(x)
RIGHT7(d(x)) → RIGHT7(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(68) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(69) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT7(b(x)) → RIGHT7(x)
RIGHT7(a(x)) → RIGHT7(x)
RIGHT7(c(x)) → RIGHT7(x)
RIGHT7(d(x)) → RIGHT7(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(70) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(71) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT7(b(x)) → RIGHT7(x)
RIGHT7(a(x)) → RIGHT7(x)
RIGHT7(c(x)) → RIGHT7(x)
RIGHT7(d(x)) → RIGHT7(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(72) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT7(b(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(a(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(c(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(d(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

(73) YES

(74) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT6(b(x)) → RIGHT6(x)
RIGHT6(a(x)) → RIGHT6(x)
RIGHT6(c(x)) → RIGHT6(x)
RIGHT6(d(x)) → RIGHT6(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(75) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(76) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT6(b(x)) → RIGHT6(x)
RIGHT6(a(x)) → RIGHT6(x)
RIGHT6(c(x)) → RIGHT6(x)
RIGHT6(d(x)) → RIGHT6(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(77) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(78) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT6(b(x)) → RIGHT6(x)
RIGHT6(a(x)) → RIGHT6(x)
RIGHT6(c(x)) → RIGHT6(x)
RIGHT6(d(x)) → RIGHT6(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(79) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT6(b(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(a(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(c(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(d(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

(80) YES

(81) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)
RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(d(x)) → RIGHT5(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(82) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(83) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)
RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(d(x)) → RIGHT5(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(84) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(85) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)
RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(d(x)) → RIGHT5(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(86) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT5(b(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(a(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(c(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(d(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

(87) YES

(88) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(89) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(90) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(91) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(92) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(d(x)) → RIGHT4(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(93) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT4(b(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(a(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(c(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(d(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

(94) YES

(95) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(96) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(97) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(98) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(99) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(d(x)) → RIGHT3(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(100) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT3(b(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(a(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(c(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(d(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

(101) YES

(102) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(103) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(104) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(105) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(106) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(d(x)) → RIGHT2(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(107) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT2(b(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(a(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(c(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(d(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

(108) YES

(109) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)

The TRS R consists of the following rules:

Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right6(a(x)) → Aa(Right6(x))
Right7(a(x)) → Aa(Right7(x))
Right8(a(x)) → Aa(Right8(x))
Right9(a(x)) → Aa(Right9(x))
Right10(a(x)) → Aa(Right10(x))
Right11(a(x)) → Aa(Right11(x))
Right12(a(x)) → Aa(Right12(x))
Right13(a(x)) → Aa(Right13(x))
Right14(a(x)) → Aa(Right14(x))
Right15(a(x)) → Aa(Right15(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right6(b(x)) → Ab(Right6(x))
Right7(b(x)) → Ab(Right7(x))
Right8(b(x)) → Ab(Right8(x))
Right9(b(x)) → Ab(Right9(x))
Right10(b(x)) → Ab(Right10(x))
Right11(b(x)) → Ab(Right11(x))
Right12(b(x)) → Ab(Right12(x))
Right13(b(x)) → Ab(Right13(x))
Right14(b(x)) → Ab(Right14(x))
Right15(b(x)) → Ab(Right15(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Right6(c(x)) → Ac(Right6(x))
Right7(c(x)) → Ac(Right7(x))
Right8(c(x)) → Ac(Right8(x))
Right9(c(x)) → Ac(Right9(x))
Right10(c(x)) → Ac(Right10(x))
Right11(c(x)) → Ac(Right11(x))
Right12(c(x)) → Ac(Right12(x))
Right13(c(x)) → Ac(Right13(x))
Right14(c(x)) → Ac(Right14(x))
Right15(c(x)) → Ac(Right15(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right6(d(x)) → Ad(Right6(x))
Right7(d(x)) → Ad(Right7(x))
Right8(d(x)) → Ad(Right8(x))
Right9(d(x)) → Ad(Right9(x))
Right10(d(x)) → Ad(Right10(x))
Right11(d(x)) → Ad(Right11(x))
Right12(d(x)) → Ad(Right12(x))
Right13(d(x)) → Ad(Right13(x))
Right14(d(x)) → Ad(Right14(x))
Right15(d(x)) → Ad(Right15(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Ad(Left(x)) → Left(d(x))

The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(110) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(111) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)

R is empty.
The set Q consists of the following terms:

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

We have to consider all minimal (P,Q,R)-chains.

(112) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

Right1(a(x0))
Right2(a(x0))
Right3(a(x0))
Right4(a(x0))
Right5(a(x0))
Right6(a(x0))
Right7(a(x0))
Right8(a(x0))
Right9(a(x0))
Right10(a(x0))
Right11(a(x0))
Right12(a(x0))
Right13(a(x0))
Right14(a(x0))
Right15(a(x0))
Right1(b(x0))
Right2(b(x0))
Right3(b(x0))
Right4(b(x0))
Right5(b(x0))
Right6(b(x0))
Right7(b(x0))
Right8(b(x0))
Right9(b(x0))
Right10(b(x0))
Right11(b(x0))
Right12(b(x0))
Right13(b(x0))
Right14(b(x0))
Right15(b(x0))
Right1(c(x0))
Right2(c(x0))
Right3(c(x0))
Right4(c(x0))
Right5(c(x0))
Right6(c(x0))
Right7(c(x0))
Right8(c(x0))
Right9(c(x0))
Right10(c(x0))
Right11(c(x0))
Right12(c(x0))
Right13(c(x0))
Right14(c(x0))
Right15(c(x0))
Right1(d(x0))
Right2(d(x0))
Right3(d(x0))
Right4(d(x0))
Right5(d(x0))
Right6(d(x0))
Right7(d(x0))
Right8(d(x0))
Right9(d(x0))
Right10(d(x0))
Right11(d(x0))
Right12(d(x0))
Right13(d(x0))
Right14(d(x0))
Right15(d(x0))
Aa(Left(x0))
Ab(Left(x0))
Ac(Left(x0))
Ad(Left(x0))

(113) Obligation:

Q DP problem:
The TRS P consists of the following rules:

RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(d(x)) → RIGHT1(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(114) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • RIGHT1(b(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(a(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(c(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(d(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

(115) YES