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

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 508 ms)
↳4 QTRS
↳5 Overlay + Local Confluence (⇔, 76 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 0 ms)
↳8 QDP
↳9 DependencyGraphProof (⇔, 0 ms)
↳10 QDP
↳11 UsableRulesProof (⇔, 0 ms)
↳12 QDP
↳13 QReductionProof (⇔, 0 ms)
↳14 QDP
↳15 QDPSizeChangeProof (⇔, 0 ms)
↳16 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) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 403 + x1   
POL(Ab(x1)) = 268 + x1   
POL(Ac(x1)) = 178 + x1   
POL(Ad(x1)) = 118 + x1   
POL(Begin(x1)) = 19 + x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 1206 + x1   
POL(Right10(x1)) = 205 + x1   
POL(Right11(x1)) = 646 + x1   
POL(Right12(x1)) = 469 + x1   
POL(Right13(x1)) = 351 + x1   
POL(Right14(x1)) = 233 + x1   
POL(Right15(x1)) = 115 + x1   
POL(Right2(x1)) = 803 + x1   
POL(Right3(x1)) = 400 + x1   
POL(Right4(x1)) = 801 + x1   
POL(Right5(x1)) = 533 + x1   
POL(Right6(x1)) = 265 + x1   
POL(Right7(x1)) = 531 + x1   
POL(Right8(x1)) = 353 + x1   
POL(Right9(x1)) = 175 + x1   
POL(Wait(x1)) = 21 + x1   
POL(a(x1)) = 403 + x1   
POL(b(x1)) = 268 + x1   
POL(c(x1)) = 178 + x1   
POL(d(x1)) = 118 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(4) Obligation:

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

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

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

The set Q consists of the following terms:

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

LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → D(Left(x))
LEFT(Ad(x)) → LEFT(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

a(Right1(x0))
a(Right2(x0))
a(Right3(x0))
a(Right4(x0))
a(Right5(x0))
a(Right6(x0))
a(Right7(x0))
a(Right8(x0))
a(Right9(x0))
a(Right10(x0))
a(Right11(x0))
a(Right12(x0))
a(Right13(x0))
a(Right14(x0))
a(Right15(x0))
b(Right1(x0))
b(Right2(x0))
b(Right3(x0))
b(Right4(x0))
b(Right5(x0))
b(Right6(x0))
b(Right7(x0))
b(Right8(x0))
b(Right9(x0))
b(Right10(x0))
b(Right11(x0))
b(Right12(x0))
b(Right13(x0))
b(Right14(x0))
b(Right15(x0))
c(Right1(x0))
c(Right2(x0))
c(Right3(x0))
c(Right4(x0))
c(Right5(x0))
c(Right6(x0))
c(Right7(x0))
c(Right8(x0))
c(Right9(x0))
c(Right10(x0))
c(Right11(x0))
c(Right12(x0))
c(Right13(x0))
c(Right14(x0))
c(Right15(x0))
d(Right1(x0))
d(Right2(x0))
d(Right3(x0))
d(Right4(x0))
d(Right5(x0))
d(Right6(x0))
d(Right7(x0))
d(Right8(x0))
d(Right9(x0))
d(Right10(x0))
d(Right11(x0))
d(Right12(x0))
d(Right13(x0))
d(Right14(x0))
d(Right15(x0))
Left(Aa(x0))
Left(Ab(x0))
Left(Ac(x0))
Left(Ad(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 1 SCC with 4 less nodes.

(10) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(11) 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.

(12) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)

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

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

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

(13) 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].

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

(14) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)

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

(15) 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:

  • LEFT(Ab(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Aa(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Ac(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Ad(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

(16) YES