YES Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/ICFP_2010/54622.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 QTRSRRRProof (⇔, 540 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 116 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPSizeChangeProof (⇔, 0 ms)
        ↳11 YES
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES
    ↳17 QDP
        ↳18 UsableRulesProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 UsableRulesProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QDPSizeChangeProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QDPSizeChangeProof (⇔, 0 ms)
        ↳41 YES
    ↳42 QDP
        ↳43 UsableRulesProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 QDPOrderProof (⇔, 505 ms)
        ↳46 QDP
        ↳47 DependencyGraphProof (⇔, 0 ms)
        ↳48 TRUE

(0) Obligation:

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

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(x) → 1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
2(x) → 3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(1(x1)) = x1   
POL(2(x1)) = 1 + x1   
POL(3(x1)) = x1   
POL(4(x1)) = x1   
POL(5(x1)) = x1   
POL(A0(x1)) = 1 + x1   
POL(A1(x1)) = x1   
POL(A2(x1)) = 1 + x1   
POL(A3(x1)) = x1   
POL(A4(x1)) = x1   
POL(A5(x1)) = x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 1 + x1   
POL(Right2(x1)) = 1 + x1   
POL(Right3(x1)) = x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = x1   
POL(Right6(x1)) = x1   
POL(Right7(x1)) = x1   
POL(Wait(x1)) = x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

0(x) → 1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(1(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
2(x) → 3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(3(x))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))


(2) Obligation:

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

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

BEGIN(1(2(3(4(5(4(5(x)))))))) → WAIT(Right1(x))
BEGIN(1(2(3(4(5(4(5(x)))))))) → RIGHT1(x)
BEGIN(2(3(4(5(4(5(x))))))) → WAIT(Right2(x))
BEGIN(2(3(4(5(4(5(x))))))) → RIGHT2(x)
BEGIN(3(4(5(4(5(x)))))) → WAIT(Right3(x))
BEGIN(3(4(5(4(5(x)))))) → RIGHT3(x)
BEGIN(4(5(4(5(x))))) → WAIT(Right4(x))
BEGIN(4(5(4(5(x))))) → RIGHT4(x)
BEGIN(5(4(5(x)))) → WAIT(Right5(x))
BEGIN(5(4(5(x)))) → RIGHT5(x)
BEGIN(4(5(x))) → WAIT(Right6(x))
BEGIN(4(5(x))) → RIGHT6(x)
BEGIN(5(x)) → WAIT(Right7(x))
BEGIN(5(x)) → RIGHT7(x)
RIGHT1(0(End(x))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT2(0(1(End(x)))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT3(0(1(2(End(x))))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT4(0(1(2(3(End(x)))))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT5(0(1(2(3(4(End(x))))))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT6(0(1(2(3(4(5(End(x)))))))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT7(0(1(2(3(4(5(4(End(x))))))))) → 01(1(2(3(4(4(5(5(End(x)))))))))
RIGHT1(0(x)) → A01(Right1(x))
RIGHT1(0(x)) → RIGHT1(x)
RIGHT2(0(x)) → A01(Right2(x))
RIGHT2(0(x)) → RIGHT2(x)
RIGHT3(0(x)) → A01(Right3(x))
RIGHT3(0(x)) → RIGHT3(x)
RIGHT4(0(x)) → A01(Right4(x))
RIGHT4(0(x)) → RIGHT4(x)
RIGHT5(0(x)) → A01(Right5(x))
RIGHT5(0(x)) → RIGHT5(x)
RIGHT6(0(x)) → A01(Right6(x))
RIGHT6(0(x)) → RIGHT6(x)
RIGHT7(0(x)) → A01(Right7(x))
RIGHT7(0(x)) → RIGHT7(x)
RIGHT1(1(x)) → A11(Right1(x))
RIGHT1(1(x)) → RIGHT1(x)
RIGHT2(1(x)) → A11(Right2(x))
RIGHT2(1(x)) → RIGHT2(x)
RIGHT3(1(x)) → A11(Right3(x))
RIGHT3(1(x)) → RIGHT3(x)
RIGHT4(1(x)) → A11(Right4(x))
RIGHT4(1(x)) → RIGHT4(x)
RIGHT5(1(x)) → A11(Right5(x))
RIGHT5(1(x)) → RIGHT5(x)
RIGHT6(1(x)) → A11(Right6(x))
RIGHT6(1(x)) → RIGHT6(x)
RIGHT7(1(x)) → A11(Right7(x))
RIGHT7(1(x)) → RIGHT7(x)
RIGHT1(2(x)) → A21(Right1(x))
RIGHT1(2(x)) → RIGHT1(x)
RIGHT2(2(x)) → A21(Right2(x))
RIGHT2(2(x)) → RIGHT2(x)
RIGHT3(2(x)) → A21(Right3(x))
RIGHT3(2(x)) → RIGHT3(x)
RIGHT4(2(x)) → A21(Right4(x))
RIGHT4(2(x)) → RIGHT4(x)
RIGHT5(2(x)) → A21(Right5(x))
RIGHT5(2(x)) → RIGHT5(x)
RIGHT6(2(x)) → A21(Right6(x))
RIGHT6(2(x)) → RIGHT6(x)
RIGHT7(2(x)) → A21(Right7(x))
RIGHT7(2(x)) → RIGHT7(x)
RIGHT1(3(x)) → A31(Right1(x))
RIGHT1(3(x)) → RIGHT1(x)
RIGHT2(3(x)) → A31(Right2(x))
RIGHT2(3(x)) → RIGHT2(x)
RIGHT3(3(x)) → A31(Right3(x))
RIGHT3(3(x)) → RIGHT3(x)
RIGHT4(3(x)) → A31(Right4(x))
RIGHT4(3(x)) → RIGHT4(x)
RIGHT5(3(x)) → A31(Right5(x))
RIGHT5(3(x)) → RIGHT5(x)
RIGHT6(3(x)) → A31(Right6(x))
RIGHT6(3(x)) → RIGHT6(x)
RIGHT7(3(x)) → A31(Right7(x))
RIGHT7(3(x)) → RIGHT7(x)
RIGHT1(4(x)) → A41(Right1(x))
RIGHT1(4(x)) → RIGHT1(x)
RIGHT2(4(x)) → A41(Right2(x))
RIGHT2(4(x)) → RIGHT2(x)
RIGHT3(4(x)) → A41(Right3(x))
RIGHT3(4(x)) → RIGHT3(x)
RIGHT4(4(x)) → A41(Right4(x))
RIGHT4(4(x)) → RIGHT4(x)
RIGHT5(4(x)) → A41(Right5(x))
RIGHT5(4(x)) → RIGHT5(x)
RIGHT6(4(x)) → A41(Right6(x))
RIGHT6(4(x)) → RIGHT6(x)
RIGHT7(4(x)) → A41(Right7(x))
RIGHT7(4(x)) → RIGHT7(x)
RIGHT1(5(x)) → A51(Right1(x))
RIGHT1(5(x)) → RIGHT1(x)
RIGHT2(5(x)) → A51(Right2(x))
RIGHT2(5(x)) → RIGHT2(x)
RIGHT3(5(x)) → A51(Right3(x))
RIGHT3(5(x)) → RIGHT3(x)
RIGHT4(5(x)) → A51(Right4(x))
RIGHT4(5(x)) → RIGHT4(x)
RIGHT5(5(x)) → A51(Right5(x))
RIGHT5(5(x)) → RIGHT5(x)
RIGHT6(5(x)) → A51(Right6(x))
RIGHT6(5(x)) → RIGHT6(x)
RIGHT7(5(x)) → A51(Right7(x))
RIGHT7(5(x)) → RIGHT7(x)
A01(Left(x)) → 01(x)
WAIT(Left(x)) → BEGIN(x)
01(1(2(3(4(5(4(5(x)))))))) → 01(1(2(3(4(4(5(5(x))))))))

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 58 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

RIGHT7(1(x)) → RIGHT7(x)
RIGHT7(0(x)) → RIGHT7(x)
RIGHT7(2(x)) → RIGHT7(x)
RIGHT7(3(x)) → RIGHT7(x)
RIGHT7(4(x)) → RIGHT7(x)
RIGHT7(5(x)) → RIGHT7(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(8) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

RIGHT7(1(x)) → RIGHT7(x)
RIGHT7(0(x)) → RIGHT7(x)
RIGHT7(2(x)) → RIGHT7(x)
RIGHT7(3(x)) → RIGHT7(x)
RIGHT7(4(x)) → RIGHT7(x)
RIGHT7(5(x)) → RIGHT7(x)

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

(10) 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(1(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(0(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(2(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(3(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(4(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

  • RIGHT7(5(x)) → RIGHT7(x)
    The graph contains the following edges 1 > 1

(11) YES

(12) Obligation:

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

RIGHT6(1(x)) → RIGHT6(x)
RIGHT6(0(x)) → RIGHT6(x)
RIGHT6(2(x)) → RIGHT6(x)
RIGHT6(3(x)) → RIGHT6(x)
RIGHT6(4(x)) → RIGHT6(x)
RIGHT6(5(x)) → RIGHT6(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(13) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

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

RIGHT6(1(x)) → RIGHT6(x)
RIGHT6(0(x)) → RIGHT6(x)
RIGHT6(2(x)) → RIGHT6(x)
RIGHT6(3(x)) → RIGHT6(x)
RIGHT6(4(x)) → RIGHT6(x)
RIGHT6(5(x)) → RIGHT6(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:

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

  • RIGHT6(0(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(2(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(3(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(4(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(5(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

(16) YES

(17) Obligation:

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

RIGHT5(1(x)) → RIGHT5(x)
RIGHT5(0(x)) → RIGHT5(x)
RIGHT5(2(x)) → RIGHT5(x)
RIGHT5(3(x)) → RIGHT5(x)
RIGHT5(4(x)) → RIGHT5(x)
RIGHT5(5(x)) → RIGHT5(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(18) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(19) Obligation:

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

RIGHT5(1(x)) → RIGHT5(x)
RIGHT5(0(x)) → RIGHT5(x)
RIGHT5(2(x)) → RIGHT5(x)
RIGHT5(3(x)) → RIGHT5(x)
RIGHT5(4(x)) → RIGHT5(x)
RIGHT5(5(x)) → RIGHT5(x)

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

(20) 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(1(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(0(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(2(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(3(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(4(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(5(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

(21) YES

(22) Obligation:

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

RIGHT4(1(x)) → RIGHT4(x)
RIGHT4(0(x)) → RIGHT4(x)
RIGHT4(2(x)) → RIGHT4(x)
RIGHT4(3(x)) → RIGHT4(x)
RIGHT4(4(x)) → RIGHT4(x)
RIGHT4(5(x)) → RIGHT4(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(23) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(24) Obligation:

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

RIGHT4(1(x)) → RIGHT4(x)
RIGHT4(0(x)) → RIGHT4(x)
RIGHT4(2(x)) → RIGHT4(x)
RIGHT4(3(x)) → RIGHT4(x)
RIGHT4(4(x)) → RIGHT4(x)
RIGHT4(5(x)) → RIGHT4(x)

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

(25) 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(1(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(0(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(2(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(3(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(4(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(5(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

(26) YES

(27) Obligation:

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

RIGHT3(1(x)) → RIGHT3(x)
RIGHT3(0(x)) → RIGHT3(x)
RIGHT3(2(x)) → RIGHT3(x)
RIGHT3(3(x)) → RIGHT3(x)
RIGHT3(4(x)) → RIGHT3(x)
RIGHT3(5(x)) → RIGHT3(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(28) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(29) Obligation:

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

RIGHT3(1(x)) → RIGHT3(x)
RIGHT3(0(x)) → RIGHT3(x)
RIGHT3(2(x)) → RIGHT3(x)
RIGHT3(3(x)) → RIGHT3(x)
RIGHT3(4(x)) → RIGHT3(x)
RIGHT3(5(x)) → RIGHT3(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:

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

  • RIGHT3(0(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(2(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(3(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(4(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(5(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

(31) YES

(32) Obligation:

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

RIGHT2(1(x)) → RIGHT2(x)
RIGHT2(0(x)) → RIGHT2(x)
RIGHT2(2(x)) → RIGHT2(x)
RIGHT2(3(x)) → RIGHT2(x)
RIGHT2(4(x)) → RIGHT2(x)
RIGHT2(5(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(33) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(34) Obligation:

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

RIGHT2(1(x)) → RIGHT2(x)
RIGHT2(0(x)) → RIGHT2(x)
RIGHT2(2(x)) → RIGHT2(x)
RIGHT2(3(x)) → RIGHT2(x)
RIGHT2(4(x)) → RIGHT2(x)
RIGHT2(5(x)) → RIGHT2(x)

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

(35) 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(1(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(0(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(2(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(3(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(4(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(5(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

(36) YES

(37) Obligation:

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

RIGHT1(1(x)) → RIGHT1(x)
RIGHT1(0(x)) → RIGHT1(x)
RIGHT1(2(x)) → RIGHT1(x)
RIGHT1(3(x)) → RIGHT1(x)
RIGHT1(4(x)) → RIGHT1(x)
RIGHT1(5(x)) → RIGHT1(x)

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(38) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(39) Obligation:

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

RIGHT1(1(x)) → RIGHT1(x)
RIGHT1(0(x)) → RIGHT1(x)
RIGHT1(2(x)) → RIGHT1(x)
RIGHT1(3(x)) → RIGHT1(x)
RIGHT1(4(x)) → RIGHT1(x)
RIGHT1(5(x)) → RIGHT1(x)

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

(40) 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(1(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(0(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(2(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(3(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(4(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(5(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

(41) YES

(42) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(1(2(3(4(5(4(5(x)))))))) → WAIT(Right1(x))
BEGIN(2(3(4(5(4(5(x))))))) → WAIT(Right2(x))
BEGIN(3(4(5(4(5(x)))))) → WAIT(Right3(x))
BEGIN(4(5(4(5(x))))) → WAIT(Right4(x))
BEGIN(5(4(5(x)))) → WAIT(Right5(x))
BEGIN(4(5(x))) → WAIT(Right6(x))
BEGIN(5(x)) → WAIT(Right7(x))

The TRS R consists of the following rules:

Begin(1(2(3(4(5(4(5(x)))))))) → Wait(Right1(x))
Begin(2(3(4(5(4(5(x))))))) → Wait(Right2(x))
Begin(3(4(5(4(5(x)))))) → Wait(Right3(x))
Begin(4(5(4(5(x))))) → Wait(Right4(x))
Begin(5(4(5(x)))) → Wait(Right5(x))
Begin(4(5(x))) → Wait(Right6(x))
Begin(5(x)) → Wait(Right7(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right1(3(x)) → A3(Right1(x))
Right2(3(x)) → A3(Right2(x))
Right3(3(x)) → A3(Right3(x))
Right4(3(x)) → A3(Right4(x))
Right5(3(x)) → A3(Right5(x))
Right6(3(x)) → A3(Right6(x))
Right7(3(x)) → A3(Right7(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
Wait(Left(x)) → Begin(x)
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

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

(43) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(44) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(1(2(3(4(5(4(5(x)))))))) → WAIT(Right1(x))
BEGIN(2(3(4(5(4(5(x))))))) → WAIT(Right2(x))
BEGIN(3(4(5(4(5(x)))))) → WAIT(Right3(x))
BEGIN(4(5(4(5(x))))) → WAIT(Right4(x))
BEGIN(5(4(5(x)))) → WAIT(Right5(x))
BEGIN(4(5(x))) → WAIT(Right6(x))
BEGIN(5(x)) → WAIT(Right7(x))

The TRS R consists of the following rules:

Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(x)) → A0(Right7(x))
Right7(1(x)) → A1(Right7(x))
Right7(2(x)) → A2(Right7(x))
Right7(3(x)) → A3(Right7(x))
Right7(4(x)) → A4(Right7(x))
Right7(5(x)) → A5(Right7(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
A3(Left(x)) → Left(3(x))
A2(Left(x)) → Left(2(x))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(x)) → A0(Right6(x))
Right6(1(x)) → A1(Right6(x))
Right6(2(x)) → A2(Right6(x))
Right6(3(x)) → A3(Right6(x))
Right6(4(x)) → A4(Right6(x))
Right6(5(x)) → A5(Right6(x))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(x)) → A0(Right5(x))
Right5(1(x)) → A1(Right5(x))
Right5(2(x)) → A2(Right5(x))
Right5(3(x)) → A3(Right5(x))
Right5(4(x)) → A4(Right5(x))
Right5(5(x)) → A5(Right5(x))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(x)) → A0(Right4(x))
Right4(1(x)) → A1(Right4(x))
Right4(2(x)) → A2(Right4(x))
Right4(3(x)) → A3(Right4(x))
Right4(4(x)) → A4(Right4(x))
Right4(5(x)) → A5(Right4(x))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(2(x)) → A2(Right3(x))
Right3(3(x)) → A3(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(x)) → A0(Right2(x))
Right2(1(x)) → A1(Right2(x))
Right2(2(x)) → A2(Right2(x))
Right2(3(x)) → A3(Right2(x))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right1(1(x)) → A1(Right1(x))
Right1(2(x)) → A2(Right1(x))
Right1(3(x)) → A3(Right1(x))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


WAIT(Left(x)) → BEGIN(x)
BEGIN(1(2(3(4(5(4(5(x)))))))) → WAIT(Right1(x))
BEGIN(2(3(4(5(4(5(x))))))) → WAIT(Right2(x))
BEGIN(3(4(5(4(5(x)))))) → WAIT(Right3(x))
BEGIN(4(5(4(5(x))))) → WAIT(Right4(x))
BEGIN(5(4(5(x)))) → WAIT(Right5(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 0   
POL(1(x1)) = x1   
POL(2(x1)) = x1   
POL(3(x1)) = x1   
POL(4(x1)) = x1   
POL(5(x1)) = 1 + x1   
POL(A0(x1)) = 1   
POL(A1(x1)) = x1   
POL(A2(x1)) = x1   
POL(A3(x1)) = x1   
POL(A4(x1)) = x1   
POL(A5(x1)) = 1 + x1   
POL(BEGIN(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = 1 + x1   
POL(Right1(x1)) = 1 + x1   
POL(Right2(x1)) = 1 + x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = 1 + x1   
POL(Right5(x1)) = 1 + x1   
POL(Right6(x1)) = 1 + x1   
POL(Right7(x1)) = 1 + x1   
POL(WAIT(x1)) = x1   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right1(1(x)) → A1(Right1(x))
Right1(2(x)) → A2(Right1(x))
Right1(3(x)) → A3(Right1(x))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(x)) → A0(Right2(x))
Right2(1(x)) → A1(Right2(x))
Right2(2(x)) → A2(Right2(x))
Right2(3(x)) → A3(Right2(x))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(2(x)) → A2(Right3(x))
Right3(3(x)) → A3(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(x)) → A0(Right4(x))
Right4(1(x)) → A1(Right4(x))
Right4(2(x)) → A2(Right4(x))
Right4(3(x)) → A3(Right4(x))
Right4(4(x)) → A4(Right4(x))
Right4(5(x)) → A5(Right4(x))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(x)) → A0(Right5(x))
Right5(1(x)) → A1(Right5(x))
Right5(2(x)) → A2(Right5(x))
Right5(3(x)) → A3(Right5(x))
Right5(4(x)) → A4(Right5(x))
Right5(5(x)) → A5(Right5(x))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(x)) → A0(Right6(x))
Right6(1(x)) → A1(Right6(x))
Right6(2(x)) → A2(Right6(x))
Right6(3(x)) → A3(Right6(x))
Right6(4(x)) → A4(Right6(x))
Right6(5(x)) → A5(Right6(x))
Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(x)) → A0(Right7(x))
Right7(1(x)) → A1(Right7(x))
Right7(2(x)) → A2(Right7(x))
Right7(3(x)) → A3(Right7(x))
Right7(4(x)) → A4(Right7(x))
Right7(5(x)) → A5(Right7(x))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
A2(Left(x)) → Left(2(x))
A3(Left(x)) → Left(3(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))

(46) Obligation:

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

BEGIN(4(5(x))) → WAIT(Right6(x))
BEGIN(5(x)) → WAIT(Right7(x))

The TRS R consists of the following rules:

Right7(0(1(2(3(4(5(4(End(x))))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right7(0(x)) → A0(Right7(x))
Right7(1(x)) → A1(Right7(x))
Right7(2(x)) → A2(Right7(x))
Right7(3(x)) → A3(Right7(x))
Right7(4(x)) → A4(Right7(x))
Right7(5(x)) → A5(Right7(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
A3(Left(x)) → Left(3(x))
A2(Left(x)) → Left(2(x))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
0(1(2(3(4(5(4(5(x)))))))) → 0(1(2(3(4(4(5(5(x))))))))
Right6(0(1(2(3(4(5(End(x)))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right6(0(x)) → A0(Right6(x))
Right6(1(x)) → A1(Right6(x))
Right6(2(x)) → A2(Right6(x))
Right6(3(x)) → A3(Right6(x))
Right6(4(x)) → A4(Right6(x))
Right6(5(x)) → A5(Right6(x))
Right5(0(1(2(3(4(End(x))))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right5(0(x)) → A0(Right5(x))
Right5(1(x)) → A1(Right5(x))
Right5(2(x)) → A2(Right5(x))
Right5(3(x)) → A3(Right5(x))
Right5(4(x)) → A4(Right5(x))
Right5(5(x)) → A5(Right5(x))
Right4(0(1(2(3(End(x)))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right4(0(x)) → A0(Right4(x))
Right4(1(x)) → A1(Right4(x))
Right4(2(x)) → A2(Right4(x))
Right4(3(x)) → A3(Right4(x))
Right4(4(x)) → A4(Right4(x))
Right4(5(x)) → A5(Right4(x))
Right3(0(1(2(End(x))))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(2(x)) → A2(Right3(x))
Right3(3(x)) → A3(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right2(0(1(End(x)))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right2(0(x)) → A0(Right2(x))
Right2(1(x)) → A1(Right2(x))
Right2(2(x)) → A2(Right2(x))
Right2(3(x)) → A3(Right2(x))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right1(0(End(x))) → Left(0(1(2(3(4(4(5(5(End(x))))))))))
Right1(0(x)) → A0(Right1(x))
Right1(1(x)) → A1(Right1(x))
Right1(2(x)) → A2(Right1(x))
Right1(3(x)) → A3(Right1(x))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))

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

(47) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(48) TRUE