NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z066.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 140 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 68 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 204 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 154 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 QDPOrderProof (⇔, 109 ms)
        ↳15 QDP
        ↳16 DependencyGraphProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPOrderProof (⇔, 103 ms)
        ↳19 QDP
        ↳20 DependencyGraphProof (⇔, 0 ms)
        ↳21 TRUE
    ↳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 QDPSizeChangeProof (⇔, 0 ms)
        ↳46 YES
    ↳47 QDP
        ↳48 UsableRulesProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 QDPSizeChangeProof (⇔, 0 ms)
        ↳51 YES
    ↳52 QDP
        ↳53 UsableRulesProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 QDPSizeChangeProof (⇔, 0 ms)
        ↳56 YES
    ↳57 QDP
        ↳58 UsableRulesProof (⇔, 0 ms)
        ↳59 QDP
        ↳60 QDPSizeChangeProof (⇔, 0 ms)
        ↳61 YES
    ↳62 QDP
        ↳63 UsableRulesProof (⇔, 0 ms)
        ↳64 QDP
        ↳65 QDPSizeChangeProof (⇔, 0 ms)
        ↳66 YES
    ↳67 QDP
        ↳68 UsableRulesProof (⇔, 0 ms)
        ↳69 QDP
        ↳70 QDPSizeChangeProof (⇔, 0 ms)
        ↳71 YES
    ↳72 QDP
        ↳73 UsableRulesProof (⇔, 0 ms)
        ↳74 QDP
        ↳75 QDPSizeChangeProof (⇔, 0 ms)
        ↳76 YES
    ↳77 QDP
        ↳78 UsableRulesProof (⇔, 0 ms)
        ↳79 QDP
        ↳80 QDPSizeChangeProof (⇔, 0 ms)
        ↳81 YES
    ↳82 QDP
        ↳83 UsableRulesProof (⇔, 0 ms)
        ↳84 QDP
        ↳85 NonTerminationLoopProof (⇒, 1415 ms)
        ↳86 NO

(0) Obligation:

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

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

BEGIN(b(c(x))) → WAIT(Right1(x))
BEGIN(b(c(x))) → RIGHT1(x)
BEGIN(c(x)) → WAIT(Right2(x))
BEGIN(c(x)) → RIGHT2(x)
BEGIN(B(A(x))) → WAIT(Right3(x))
BEGIN(B(A(x))) → RIGHT3(x)
BEGIN(A(x)) → WAIT(Right4(x))
BEGIN(A(x)) → RIGHT4(x)
BEGIN(a(C(x))) → WAIT(Right5(x))
BEGIN(a(C(x))) → RIGHT5(x)
BEGIN(C(x)) → WAIT(Right6(x))
BEGIN(C(x)) → RIGHT6(x)
BEGIN(A(B(x))) → WAIT(Right7(x))
BEGIN(A(B(x))) → RIGHT7(x)
BEGIN(B(x)) → WAIT(Right8(x))
BEGIN(B(x)) → RIGHT8(x)
BEGIN(c(b(x))) → WAIT(Right9(x))
BEGIN(c(b(x))) → RIGHT9(x)
BEGIN(b(x)) → WAIT(Right10(x))
BEGIN(b(x)) → RIGHT10(x)
BEGIN(C(a(x))) → WAIT(Right11(x))
BEGIN(C(a(x))) → RIGHT11(x)
BEGIN(a(x)) → WAIT(Right12(x))
BEGIN(a(x)) → RIGHT12(x)
RIGHT1(a(End(x))) → C1(b(a(End(x))))
RIGHT1(a(End(x))) → B1(a(End(x)))
RIGHT2(a(b(End(x)))) → C1(b(a(End(x))))
RIGHT2(a(b(End(x)))) → B1(a(End(x)))
RIGHT2(a(b(End(x)))) → A1(End(x))
RIGHT3(C(End(x))) → A2(B(C(End(x))))
RIGHT3(C(End(x))) → B2(C(End(x)))
RIGHT4(C(B(End(x)))) → A2(B(C(End(x))))
RIGHT4(C(B(End(x)))) → B2(C(End(x)))
RIGHT4(C(B(End(x)))) → C2(End(x))
RIGHT5(b(End(x))) → C2(a(b(End(x))))
RIGHT5(b(End(x))) → A1(b(End(x)))
RIGHT6(b(a(End(x)))) → C2(a(b(End(x))))
RIGHT6(b(a(End(x)))) → A1(b(End(x)))
RIGHT6(b(a(End(x)))) → B1(End(x))
RIGHT7(c(End(x))) → B2(A(c(End(x))))
RIGHT7(c(End(x))) → A2(c(End(x)))
RIGHT8(c(A(End(x)))) → B2(A(c(End(x))))
RIGHT8(c(A(End(x)))) → A2(c(End(x)))
RIGHT8(c(A(End(x)))) → C1(End(x))
RIGHT9(A(End(x))) → B1(c(A(End(x))))
RIGHT9(A(End(x))) → C1(A(End(x)))
RIGHT10(A(c(End(x)))) → B1(c(A(End(x))))
RIGHT10(A(c(End(x)))) → C1(A(End(x)))
RIGHT10(A(c(End(x)))) → A2(End(x))
RIGHT11(B(End(x))) → A1(C(B(End(x))))
RIGHT11(B(End(x))) → C2(B(End(x)))
RIGHT12(B(C(End(x)))) → A1(C(B(End(x))))
RIGHT12(B(C(End(x)))) → C2(B(End(x)))
RIGHT12(B(C(End(x)))) → B2(End(x))
RIGHT1(a(x)) → AA1(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA1(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA1(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
RIGHT4(a(x)) → AA1(Right4(x))
RIGHT4(a(x)) → RIGHT4(x)
RIGHT5(a(x)) → AA1(Right5(x))
RIGHT5(a(x)) → RIGHT5(x)
RIGHT6(a(x)) → AA1(Right6(x))
RIGHT6(a(x)) → RIGHT6(x)
RIGHT7(a(x)) → AA1(Right7(x))
RIGHT7(a(x)) → RIGHT7(x)
RIGHT8(a(x)) → AA1(Right8(x))
RIGHT8(a(x)) → RIGHT8(x)
RIGHT9(a(x)) → AA1(Right9(x))
RIGHT9(a(x)) → RIGHT9(x)
RIGHT10(a(x)) → AA1(Right10(x))
RIGHT10(a(x)) → RIGHT10(x)
RIGHT11(a(x)) → AA1(Right11(x))
RIGHT11(a(x)) → RIGHT11(x)
RIGHT12(a(x)) → AA1(Right12(x))
RIGHT12(a(x)) → RIGHT12(x)
RIGHT1(b(x)) → AB1(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB1(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB1(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT4(b(x)) → AB1(Right4(x))
RIGHT4(b(x)) → RIGHT4(x)
RIGHT5(b(x)) → AB1(Right5(x))
RIGHT5(b(x)) → RIGHT5(x)
RIGHT6(b(x)) → AB1(Right6(x))
RIGHT6(b(x)) → RIGHT6(x)
RIGHT7(b(x)) → AB1(Right7(x))
RIGHT7(b(x)) → RIGHT7(x)
RIGHT8(b(x)) → AB1(Right8(x))
RIGHT8(b(x)) → RIGHT8(x)
RIGHT9(b(x)) → AB1(Right9(x))
RIGHT9(b(x)) → RIGHT9(x)
RIGHT10(b(x)) → AB1(Right10(x))
RIGHT10(b(x)) → RIGHT10(x)
RIGHT11(b(x)) → AB1(Right11(x))
RIGHT11(b(x)) → RIGHT11(x)
RIGHT12(b(x)) → AB1(Right12(x))
RIGHT12(b(x)) → RIGHT12(x)
RIGHT1(c(x)) → AC1(Right1(x))
RIGHT1(c(x)) → RIGHT1(x)
RIGHT2(c(x)) → AC1(Right2(x))
RIGHT2(c(x)) → RIGHT2(x)
RIGHT3(c(x)) → AC1(Right3(x))
RIGHT3(c(x)) → RIGHT3(x)
RIGHT4(c(x)) → AC1(Right4(x))
RIGHT4(c(x)) → RIGHT4(x)
RIGHT5(c(x)) → AC1(Right5(x))
RIGHT5(c(x)) → RIGHT5(x)
RIGHT6(c(x)) → AC1(Right6(x))
RIGHT6(c(x)) → RIGHT6(x)
RIGHT7(c(x)) → AC1(Right7(x))
RIGHT7(c(x)) → RIGHT7(x)
RIGHT8(c(x)) → AC1(Right8(x))
RIGHT8(c(x)) → RIGHT8(x)
RIGHT9(c(x)) → AC1(Right9(x))
RIGHT9(c(x)) → RIGHT9(x)
RIGHT10(c(x)) → AC1(Right10(x))
RIGHT10(c(x)) → RIGHT10(x)
RIGHT11(c(x)) → AC1(Right11(x))
RIGHT11(c(x)) → RIGHT11(x)
RIGHT12(c(x)) → AC1(Right12(x))
RIGHT12(c(x)) → RIGHT12(x)
RIGHT1(C(x)) → AC2(Right1(x))
RIGHT1(C(x)) → RIGHT1(x)
RIGHT2(C(x)) → AC2(Right2(x))
RIGHT2(C(x)) → RIGHT2(x)
RIGHT3(C(x)) → AC2(Right3(x))
RIGHT3(C(x)) → RIGHT3(x)
RIGHT4(C(x)) → AC2(Right4(x))
RIGHT4(C(x)) → RIGHT4(x)
RIGHT5(C(x)) → AC2(Right5(x))
RIGHT5(C(x)) → RIGHT5(x)
RIGHT6(C(x)) → AC2(Right6(x))
RIGHT6(C(x)) → RIGHT6(x)
RIGHT7(C(x)) → AC2(Right7(x))
RIGHT7(C(x)) → RIGHT7(x)
RIGHT8(C(x)) → AC2(Right8(x))
RIGHT8(C(x)) → RIGHT8(x)
RIGHT9(C(x)) → AC2(Right9(x))
RIGHT9(C(x)) → RIGHT9(x)
RIGHT10(C(x)) → AC2(Right10(x))
RIGHT10(C(x)) → RIGHT10(x)
RIGHT11(C(x)) → AC2(Right11(x))
RIGHT11(C(x)) → RIGHT11(x)
RIGHT12(C(x)) → AC2(Right12(x))
RIGHT12(C(x)) → RIGHT12(x)
RIGHT1(B(x)) → AB2(Right1(x))
RIGHT1(B(x)) → RIGHT1(x)
RIGHT2(B(x)) → AB2(Right2(x))
RIGHT2(B(x)) → RIGHT2(x)
RIGHT3(B(x)) → AB2(Right3(x))
RIGHT3(B(x)) → RIGHT3(x)
RIGHT4(B(x)) → AB2(Right4(x))
RIGHT4(B(x)) → RIGHT4(x)
RIGHT5(B(x)) → AB2(Right5(x))
RIGHT5(B(x)) → RIGHT5(x)
RIGHT6(B(x)) → AB2(Right6(x))
RIGHT6(B(x)) → RIGHT6(x)
RIGHT7(B(x)) → AB2(Right7(x))
RIGHT7(B(x)) → RIGHT7(x)
RIGHT8(B(x)) → AB2(Right8(x))
RIGHT8(B(x)) → RIGHT8(x)
RIGHT9(B(x)) → AB2(Right9(x))
RIGHT9(B(x)) → RIGHT9(x)
RIGHT10(B(x)) → AB2(Right10(x))
RIGHT10(B(x)) → RIGHT10(x)
RIGHT11(B(x)) → AB2(Right11(x))
RIGHT11(B(x)) → RIGHT11(x)
RIGHT12(B(x)) → AB2(Right12(x))
RIGHT12(B(x)) → RIGHT12(x)
RIGHT1(A(x)) → AA2(Right1(x))
RIGHT1(A(x)) → RIGHT1(x)
RIGHT2(A(x)) → AA2(Right2(x))
RIGHT2(A(x)) → RIGHT2(x)
RIGHT3(A(x)) → AA2(Right3(x))
RIGHT3(A(x)) → RIGHT3(x)
RIGHT4(A(x)) → AA2(Right4(x))
RIGHT4(A(x)) → RIGHT4(x)
RIGHT5(A(x)) → AA2(Right5(x))
RIGHT5(A(x)) → RIGHT5(x)
RIGHT6(A(x)) → AA2(Right6(x))
RIGHT6(A(x)) → RIGHT6(x)
RIGHT7(A(x)) → AA2(Right7(x))
RIGHT7(A(x)) → RIGHT7(x)
RIGHT8(A(x)) → AA2(Right8(x))
RIGHT8(A(x)) → RIGHT8(x)
RIGHT9(A(x)) → AA2(Right9(x))
RIGHT9(A(x)) → RIGHT9(x)
RIGHT10(A(x)) → AA2(Right10(x))
RIGHT10(A(x)) → RIGHT10(x)
RIGHT11(A(x)) → AA2(Right11(x))
RIGHT11(A(x)) → RIGHT11(x)
RIGHT12(A(x)) → AA2(Right12(x))
RIGHT12(A(x)) → RIGHT12(x)
AA1(Left(x)) → A1(x)
AB1(Left(x)) → B1(x)
AC1(Left(x)) → C1(x)
AC2(Left(x)) → C2(x)
AB2(Left(x)) → B2(x)
AA2(Left(x)) → A2(x)
WAIT(Left(x)) → BEGIN(x)
A1(b(c(x))) → C1(b(a(x)))
A1(b(c(x))) → B1(a(x))
A1(b(c(x))) → A1(x)
C2(B(A(x))) → A2(B(C(x)))
C2(B(A(x))) → B2(C(x))
C2(B(A(x))) → C2(x)
B1(a(C(x))) → C2(a(b(x)))
B1(a(C(x))) → A1(b(x))
B1(a(C(x))) → B1(x)
C1(A(B(x))) → B2(A(c(x)))
C1(A(B(x))) → A2(c(x))
C1(A(B(x))) → C1(x)
A2(c(b(x))) → B1(c(A(x)))
A2(c(b(x))) → C1(A(x))
A2(c(b(x))) → A2(x)
B2(C(a(x))) → A1(C(B(x)))
B2(C(a(x))) → C2(B(x))
B2(C(a(x))) → B2(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 14 SCCs with 120 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

C1(A(B(x))) → B2(A(c(x)))
B2(C(a(x))) → A1(C(B(x)))
A1(b(c(x))) → C1(b(a(x)))
C1(A(B(x))) → A2(c(x))
A2(c(b(x))) → B1(c(A(x)))
B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → A2(B(C(x)))
A2(c(b(x))) → C1(A(x))
C1(A(B(x))) → C1(x)
A2(c(b(x))) → A2(x)
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)
B1(a(C(x))) → A1(b(x))
A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → B1(x)
A1(b(c(x))) → A1(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(6) 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.

(7) Obligation:

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

C1(A(B(x))) → B2(A(c(x)))
B2(C(a(x))) → A1(C(B(x)))
A1(b(c(x))) → C1(b(a(x)))
C1(A(B(x))) → A2(c(x))
A2(c(b(x))) → B1(c(A(x)))
B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → A2(B(C(x)))
A2(c(b(x))) → C1(A(x))
C1(A(B(x))) → C1(x)
A2(c(b(x))) → A2(x)
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)
B1(a(C(x))) → A1(b(x))
A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → B1(x)
A1(b(c(x))) → A1(x)

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(8) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


A2(c(b(x))) → C1(A(x))
A2(c(b(x))) → A2(x)
A1(b(c(x))) → A1(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(A1(x1)) = x1   
POL(A2(x1)) = x1   
POL(B(x1)) = x1   
POL(B1(x1)) = 1 + x1   
POL(B2(x1)) = x1   
POL(C(x1)) = x1   
POL(C1(x1)) = x1   
POL(C2(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = x1   

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

B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
c(A(B(x))) → B(A(c(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))
A(c(b(x))) → b(c(A(x)))

(9) Obligation:

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

C1(A(B(x))) → B2(A(c(x)))
B2(C(a(x))) → A1(C(B(x)))
A1(b(c(x))) → C1(b(a(x)))
C1(A(B(x))) → A2(c(x))
A2(c(b(x))) → B1(c(A(x)))
B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → A2(B(C(x)))
C1(A(B(x))) → C1(x)
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)
B1(a(C(x))) → A1(b(x))
A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → B1(x)

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


C1(A(B(x))) → B2(A(c(x)))
C1(A(B(x))) → A2(c(x))
C1(A(B(x))) → C1(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(A1(x1)) = 0   
POL(A2(x1)) = 0   
POL(B(x1)) = 1 + x1   
POL(B1(x1)) = 0   
POL(B2(x1)) = 0   
POL(C(x1)) = x1   
POL(C1(x1)) = x1   
POL(C2(x1)) = 0   
POL(a(x1)) = 0   
POL(b(x1)) = 0   
POL(c(x1)) = x1   

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

B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
c(A(B(x))) → B(A(c(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))
A(c(b(x))) → b(c(A(x)))

(11) Obligation:

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

B2(C(a(x))) → A1(C(B(x)))
A1(b(c(x))) → C1(b(a(x)))
A2(c(b(x))) → B1(c(A(x)))
B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → A2(B(C(x)))
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)
B1(a(C(x))) → A1(b(x))
A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → B1(x)

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(13) Obligation:

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

A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → A2(B(C(x)))
A2(c(b(x))) → B1(c(A(x)))
B1(a(C(x))) → A1(b(x))
B1(a(C(x))) → B1(x)
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → A1(C(B(x)))
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(14) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


A2(c(b(x))) → B1(c(A(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(A1(x1)) = 0   
POL(A2(x1)) = x1   
POL(B(x1)) = 0   
POL(B1(x1)) = 0   
POL(B2(x1)) = 0   
POL(C(x1)) = 0   
POL(C2(x1)) = 0   
POL(a(x1)) = x1   
POL(b(x1)) = 1   
POL(c(x1)) = 1   

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

B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
c(A(B(x))) → B(A(c(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))
A(c(b(x))) → b(c(A(x)))

(15) Obligation:

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

A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → A2(B(C(x)))
B1(a(C(x))) → A1(b(x))
B1(a(C(x))) → B1(x)
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → A1(C(B(x)))
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(16) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(17) Obligation:

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

B1(a(C(x))) → C2(a(b(x)))
C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → A1(C(B(x)))
A1(b(c(x))) → B1(a(x))
B1(a(C(x))) → A1(b(x))
B1(a(C(x))) → B1(x)
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(18) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


C2(B(A(x))) → B2(C(x))
B2(C(a(x))) → A1(C(B(x)))
B1(a(C(x))) → A1(b(x))
B1(a(C(x))) → B1(x)
B2(C(a(x))) → C2(B(x))
C2(B(A(x))) → C2(x)
B2(C(a(x))) → B2(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 1 + x1   
POL(A1(x1)) = 1 + x1   
POL(B(x1)) = x1   
POL(B1(x1)) = 1 + x1   
POL(B2(x1)) = 1 + x1   
POL(C(x1)) = x1   
POL(C2(x1)) = 1 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = x1   
POL(c(x1)) = 1 + x1   

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

b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
c(A(B(x))) → B(A(c(x)))

(19) Obligation:

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

B1(a(C(x))) → C2(a(b(x)))
A1(b(c(x))) → B1(a(x))

The TRS R consists of the following rules:

c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))

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

(20) DependencyGraphProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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(C(x)) → RIGHT12(x)
RIGHT12(B(x)) → RIGHT12(x)
RIGHT12(A(x)) → RIGHT12(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(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:

RIGHT12(b(x)) → RIGHT12(x)
RIGHT12(a(x)) → RIGHT12(x)
RIGHT12(c(x)) → RIGHT12(x)
RIGHT12(C(x)) → RIGHT12(x)
RIGHT12(B(x)) → RIGHT12(x)
RIGHT12(A(x)) → RIGHT12(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:

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

  • 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

(26) YES

(27) 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(C(x)) → RIGHT11(x)
RIGHT11(B(x)) → RIGHT11(x)
RIGHT11(A(x)) → RIGHT11(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(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:

RIGHT11(b(x)) → RIGHT11(x)
RIGHT11(a(x)) → RIGHT11(x)
RIGHT11(c(x)) → RIGHT11(x)
RIGHT11(C(x)) → RIGHT11(x)
RIGHT11(B(x)) → RIGHT11(x)
RIGHT11(A(x)) → RIGHT11(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:

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

  • 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

(31) YES

(32) 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(C(x)) → RIGHT10(x)
RIGHT10(B(x)) → RIGHT10(x)
RIGHT10(A(x)) → RIGHT10(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(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:

RIGHT10(b(x)) → RIGHT10(x)
RIGHT10(a(x)) → RIGHT10(x)
RIGHT10(c(x)) → RIGHT10(x)
RIGHT10(C(x)) → RIGHT10(x)
RIGHT10(B(x)) → RIGHT10(x)
RIGHT10(A(x)) → RIGHT10(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:

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

  • 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

(36) YES

(37) 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(C(x)) → RIGHT9(x)
RIGHT9(B(x)) → RIGHT9(x)
RIGHT9(A(x)) → RIGHT9(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(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:

RIGHT9(b(x)) → RIGHT9(x)
RIGHT9(a(x)) → RIGHT9(x)
RIGHT9(c(x)) → RIGHT9(x)
RIGHT9(C(x)) → RIGHT9(x)
RIGHT9(B(x)) → RIGHT9(x)
RIGHT9(A(x)) → RIGHT9(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:

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

  • 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

(41) YES

(42) 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(C(x)) → RIGHT8(x)
RIGHT8(B(x)) → RIGHT8(x)
RIGHT8(A(x)) → RIGHT8(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(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:

RIGHT8(b(x)) → RIGHT8(x)
RIGHT8(a(x)) → RIGHT8(x)
RIGHT8(c(x)) → RIGHT8(x)
RIGHT8(C(x)) → RIGHT8(x)
RIGHT8(B(x)) → RIGHT8(x)
RIGHT8(A(x)) → RIGHT8(x)

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

(45) 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(C(x)) → RIGHT8(x)
    The graph contains the following edges 1 > 1

  • 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

(46) YES

(47) 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(C(x)) → RIGHT7(x)
RIGHT7(B(x)) → RIGHT7(x)
RIGHT7(A(x)) → RIGHT7(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(48) 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.

(49) 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(C(x)) → RIGHT7(x)
RIGHT7(B(x)) → RIGHT7(x)
RIGHT7(A(x)) → RIGHT7(x)

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

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

  • 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

(51) YES

(52) 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(C(x)) → RIGHT6(x)
RIGHT6(B(x)) → RIGHT6(x)
RIGHT6(A(x)) → RIGHT6(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(53) 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.

(54) 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(C(x)) → RIGHT6(x)
RIGHT6(B(x)) → RIGHT6(x)
RIGHT6(A(x)) → RIGHT6(x)

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

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

  • 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

(56) YES

(57) 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(C(x)) → RIGHT5(x)
RIGHT5(B(x)) → RIGHT5(x)
RIGHT5(A(x)) → RIGHT5(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(58) 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.

(59) 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(C(x)) → RIGHT5(x)
RIGHT5(B(x)) → RIGHT5(x)
RIGHT5(A(x)) → RIGHT5(x)

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

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

  • 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

(61) YES

(62) 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(C(x)) → RIGHT4(x)
RIGHT4(B(x)) → RIGHT4(x)
RIGHT4(A(x)) → RIGHT4(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(63) 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.

(64) 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(C(x)) → RIGHT4(x)
RIGHT4(B(x)) → RIGHT4(x)
RIGHT4(A(x)) → RIGHT4(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:

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

  • 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

(66) YES

(67) 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(C(x)) → RIGHT3(x)
RIGHT3(B(x)) → RIGHT3(x)
RIGHT3(A(x)) → RIGHT3(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(68) 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.

(69) 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(C(x)) → RIGHT3(x)
RIGHT3(B(x)) → RIGHT3(x)
RIGHT3(A(x)) → RIGHT3(x)

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

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

  • 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

(71) YES

(72) 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(C(x)) → RIGHT2(x)
RIGHT2(B(x)) → RIGHT2(x)
RIGHT2(A(x)) → RIGHT2(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(73) 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.

(74) 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(C(x)) → RIGHT2(x)
RIGHT2(B(x)) → RIGHT2(x)
RIGHT2(A(x)) → RIGHT2(x)

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

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

  • 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

(76) YES

(77) 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(C(x)) → RIGHT1(x)
RIGHT1(B(x)) → RIGHT1(x)
RIGHT1(A(x)) → RIGHT1(x)

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(78) 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.

(79) 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(C(x)) → RIGHT1(x)
RIGHT1(B(x)) → RIGHT1(x)
RIGHT1(A(x)) → RIGHT1(x)

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

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

  • 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

(81) YES

(82) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(c(x))) → WAIT(Right1(x))
BEGIN(c(x)) → WAIT(Right2(x))
BEGIN(B(A(x))) → WAIT(Right3(x))
BEGIN(A(x)) → WAIT(Right4(x))
BEGIN(a(C(x))) → WAIT(Right5(x))
BEGIN(C(x)) → WAIT(Right6(x))
BEGIN(A(B(x))) → WAIT(Right7(x))
BEGIN(B(x)) → WAIT(Right8(x))
BEGIN(c(b(x))) → WAIT(Right9(x))
BEGIN(b(x)) → WAIT(Right10(x))
BEGIN(C(a(x))) → WAIT(Right11(x))
BEGIN(a(x)) → WAIT(Right12(x))

The TRS R consists of the following rules:

Begin(b(c(x))) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(B(A(x))) → Wait(Right3(x))
Begin(A(x)) → Wait(Right4(x))
Begin(a(C(x))) → Wait(Right5(x))
Begin(C(x)) → Wait(Right6(x))
Begin(A(B(x))) → Wait(Right7(x))
Begin(B(x)) → Wait(Right8(x))
Begin(c(b(x))) → Wait(Right9(x))
Begin(b(x)) → Wait(Right10(x))
Begin(C(a(x))) → Wait(Right11(x))
Begin(a(x)) → Wait(Right12(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right12(B(C(End(x)))) → Left(a(C(B(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))
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))
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))
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))
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))
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))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
AC(Left(x)) → Left(C(x))
AB(Left(x)) → Left(B(x))
AA(Left(x)) → Left(A(x))
Wait(Left(x)) → Begin(x)
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
b(a(C(x))) → C(a(b(x)))
c(A(B(x))) → B(A(c(x)))
A(c(b(x))) → b(c(A(x)))
B(C(a(x))) → a(C(B(x)))

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

(83) 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.

(84) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(c(x))) → WAIT(Right1(x))
BEGIN(c(x)) → WAIT(Right2(x))
BEGIN(B(A(x))) → WAIT(Right3(x))
BEGIN(A(x)) → WAIT(Right4(x))
BEGIN(a(C(x))) → WAIT(Right5(x))
BEGIN(C(x)) → WAIT(Right6(x))
BEGIN(A(B(x))) → WAIT(Right7(x))
BEGIN(B(x)) → WAIT(Right8(x))
BEGIN(c(b(x))) → WAIT(Right9(x))
BEGIN(b(x)) → WAIT(Right10(x))
BEGIN(C(a(x))) → WAIT(Right11(x))
BEGIN(a(x)) → WAIT(Right12(x))

The TRS R consists of the following rules:

Right12(B(C(End(x)))) → Left(a(C(B(End(x)))))
Right12(a(x)) → Aa(Right12(x))
Right12(b(x)) → Ab(Right12(x))
Right12(c(x)) → Ac(Right12(x))
Right12(C(x)) → AC(Right12(x))
Right12(B(x)) → AB(Right12(x))
Right12(A(x)) → AA(Right12(x))
AA(Left(x)) → Left(A(x))
A(c(b(x))) → b(c(A(x)))
b(a(C(x))) → C(a(b(x)))
C(B(A(x))) → A(B(C(x)))
c(A(B(x))) → B(A(c(x)))
B(C(a(x))) → a(C(B(x)))
a(b(c(x))) → c(b(a(x)))
AB(Left(x)) → Left(B(x))
AC(Left(x)) → Left(C(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Right11(B(End(x))) → Left(a(C(B(End(x)))))
Right11(a(x)) → Aa(Right11(x))
Right11(b(x)) → Ab(Right11(x))
Right11(c(x)) → Ac(Right11(x))
Right11(C(x)) → AC(Right11(x))
Right11(B(x)) → AB(Right11(x))
Right11(A(x)) → AA(Right11(x))
Right10(A(c(End(x)))) → Left(b(c(A(End(x)))))
Right10(a(x)) → Aa(Right10(x))
Right10(b(x)) → Ab(Right10(x))
Right10(c(x)) → Ac(Right10(x))
Right10(C(x)) → AC(Right10(x))
Right10(B(x)) → AB(Right10(x))
Right10(A(x)) → AA(Right10(x))
Right9(A(End(x))) → Left(b(c(A(End(x)))))
Right9(a(x)) → Aa(Right9(x))
Right9(b(x)) → Ab(Right9(x))
Right9(c(x)) → Ac(Right9(x))
Right9(C(x)) → AC(Right9(x))
Right9(B(x)) → AB(Right9(x))
Right9(A(x)) → AA(Right9(x))
Right8(c(A(End(x)))) → Left(B(A(c(End(x)))))
Right8(a(x)) → Aa(Right8(x))
Right8(b(x)) → Ab(Right8(x))
Right8(c(x)) → Ac(Right8(x))
Right8(C(x)) → AC(Right8(x))
Right8(B(x)) → AB(Right8(x))
Right8(A(x)) → AA(Right8(x))
Right7(c(End(x))) → Left(B(A(c(End(x)))))
Right7(a(x)) → Aa(Right7(x))
Right7(b(x)) → Ab(Right7(x))
Right7(c(x)) → Ac(Right7(x))
Right7(C(x)) → AC(Right7(x))
Right7(B(x)) → AB(Right7(x))
Right7(A(x)) → AA(Right7(x))
Right6(b(a(End(x)))) → Left(C(a(b(End(x)))))
Right6(a(x)) → Aa(Right6(x))
Right6(b(x)) → Ab(Right6(x))
Right6(c(x)) → Ac(Right6(x))
Right6(C(x)) → AC(Right6(x))
Right6(B(x)) → AB(Right6(x))
Right6(A(x)) → AA(Right6(x))
Right5(b(End(x))) → Left(C(a(b(End(x)))))
Right5(a(x)) → Aa(Right5(x))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(C(x)) → AC(Right5(x))
Right5(B(x)) → AB(Right5(x))
Right5(A(x)) → AA(Right5(x))
Right4(C(B(End(x)))) → Left(A(B(C(End(x)))))
Right4(a(x)) → Aa(Right4(x))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(C(x)) → AC(Right4(x))
Right4(B(x)) → AB(Right4(x))
Right4(A(x)) → AA(Right4(x))
Right3(C(End(x))) → Left(A(B(C(End(x)))))
Right3(a(x)) → Aa(Right3(x))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(C(x)) → AC(Right3(x))
Right3(B(x)) → AB(Right3(x))
Right3(A(x)) → AA(Right3(x))
Right2(a(b(End(x)))) → Left(c(b(a(End(x)))))
Right2(a(x)) → Aa(Right2(x))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(C(x)) → AC(Right2(x))
Right2(B(x)) → AB(Right2(x))
Right2(A(x)) → AA(Right2(x))
Right1(a(End(x))) → Left(c(b(a(End(x)))))
Right1(a(x)) → Aa(Right1(x))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(C(x)) → AC(Right1(x))
Right1(B(x)) → AB(Right1(x))
Right1(A(x)) → AA(Right1(x))

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

(85) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = BEGIN(A(B(A(c(End(x)))))) evaluates to t =BEGIN(A(B(A(c(End(x))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

BEGIN(A(B(A(c(End(x)))))) → WAIT(Right7(A(c(End(x)))))
with rule BEGIN(A(B(x'))) → WAIT(Right7(x')) at position [] and matcher [x' / A(c(End(x)))]

WAIT(Right7(A(c(End(x))))) → WAIT(AA(Right7(c(End(x)))))
with rule Right7(A(x')) → AA(Right7(x')) at position [0] and matcher [x' / c(End(x))]

WAIT(AA(Right7(c(End(x))))) → WAIT(AA(Left(B(A(c(End(x)))))))
with rule Right7(c(End(x'))) → Left(B(A(c(End(x'))))) at position [0,0] and matcher [x' / x]

WAIT(AA(Left(B(A(c(End(x))))))) → WAIT(Left(A(B(A(c(End(x)))))))
with rule AA(Left(x')) → Left(A(x')) at position [0] and matcher [x' / B(A(c(End(x))))]

WAIT(Left(A(B(A(c(End(x))))))) → BEGIN(A(B(A(c(End(x))))))
with rule WAIT(Left(x)) → BEGIN(x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(86) NO