Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema

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

0 QTRS

↳1 DependencyPairsProof (⇔, 40 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 AND

    ↳5 QDP

        ↳6 UsableRulesProof (⇔, 0 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 42 ms)

        ↳9 QDP

        ↳10 DependencyGraphProof (⇔, 0 ms)

        ↳11 AND

            ↳12 QDP

                ↳13 UsableRulesProof (⇔, 0 ms)

                ↳14 QDP

                ↳15 QDPSizeChangeProof (⇔, 0 ms)

                ↳16 YES

            ↳17 QDP

                ↳18 MRRProof (⇔, 1 ms)

                ↳19 QDP

                ↳20 PisEmptyProof (⇔, 0 ms)

                ↳21 YES

            ↳22 QDP

                ↳23 QDPOrderProof (⇔, 0 ms)

                ↳24 QDP

                ↳25 DependencyGraphProof (⇔, 0 ms)

                ↳26 TRUE

    ↳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 QDPSizeChangeProof (⇔, 0 ms)

        ↳86 YES

    ↳87 QDP

        ↳88 UsableRulesProof (⇔, 0 ms)

        ↳89 QDP

        ↳90 NonTerminationLoopProof (⇒, 240 ms)

        ↳91 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) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

A²(c(b(x))) → B¹(c(A(x)))
A²(c(b(x))) → C¹(A(x))
A²(c(b(x))) → A²(x)
A¹(b(c(x))) → A¹(x)

Strictly oriented rules of the TRS R:

A(c(b(x))) → b(c(A(x)))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = 2·x₁
POL(A¹(x₁)) = x₁
POL(A²(x₁)) = 2·x₁
POL(B(x₁)) = 2·x₁
POL(B¹(x₁)) = 2 + 2·x₁
POL(B²(x₁)) = 2·x₁
POL(C(x₁)) = x₁
POL(C¹(x₁)) = x₁
POL(C²(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = 2 + 2·x₁
POL(c(x₁)) = x₁

(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))
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)))
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) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 6 less nodes.

(11) Complex Obligation (AND)

(12) Obligation:

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

C¹(A(B(x))) → C¹(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)))
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.

(13) UsableRulesProof (EQUIVALENT transformation)

(14) Obligation:

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

C¹(A(B(x))) → C¹(x)

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

C¹(A(B(x))) → C¹(x)
The graph contains the following edges 1 > 1

(16) YES

(17) Obligation:

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

B²(C(a(x))) → C²(B(x))
C²(B(A(x))) → B²(C(x))
B²(C(a(x))) → B²(x)
C²(B(A(x))) → C²(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)))
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) MRRProof (EQUIVALENT transformation)

B²(C(a(x))) → C²(B(x))
C²(B(A(x))) → B²(C(x))
B²(C(a(x))) → B²(x)
C²(B(A(x))) → C²(x)

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = 2 + x₁
POL(B(x₁)) = 2·x₁
POL(B²(x₁)) = 1 + 3·x₁
POL(C(x₁)) = 2 + x₁
POL(C²(x₁)) = 2·x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 3 + 3·x₁
POL(c(x₁)) = 2 + 3·x₁

(19) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

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

(20) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(21) YES

(22) Obligation:

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

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)))
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.

(23) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = 0
POL(A¹(x₁)) = x₁
POL(B(x₁)) = 0
POL(B¹(x₁)) = x₁
POL(C(x₁)) = 1 + x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = 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)))
a(b(c(x))) → c(b(a(x)))
C(B(A(x))) → A(B(C(x)))
c(A(B(x))) → B(A(c(x)))

(24) Obligation:

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

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)))
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.

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(26) TRUE

(27) 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.

(28) UsableRulesProof (EQUIVALENT transformation)

(29) 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.

(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:

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

(31) YES

(32) 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.

(33) UsableRulesProof (EQUIVALENT transformation)

(34) 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.

(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:

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

(36) YES

(37) 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.

(38) UsableRulesProof (EQUIVALENT transformation)

(39) 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.

(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:

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

(41) YES

(42) 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.

(43) UsableRulesProof (EQUIVALENT transformation)

(44) 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.

(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:

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

(46) YES

(47) 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.

(48) UsableRulesProof (EQUIVALENT transformation)

(49) 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.

(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:

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

(51) YES

(52) 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.

(53) UsableRulesProof (EQUIVALENT transformation)

(54) 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.

(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:

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

(56) YES

(57) 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.

(58) UsableRulesProof (EQUIVALENT transformation)

(59) 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.

(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:

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

(61) YES

(62) 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.

(63) UsableRulesProof (EQUIVALENT transformation)

(64) 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.

(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:

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

(66) YES

(67) 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.

(68) UsableRulesProof (EQUIVALENT transformation)

(69) 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.

(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:

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

(71) YES

(72) 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.

(73) UsableRulesProof (EQUIVALENT transformation)

(74) 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.

(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:

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

(76) YES

(77) 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.

(78) UsableRulesProof (EQUIVALENT transformation)

(79) 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.

(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:

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

(81) YES

(82) 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.

(83) UsableRulesProof (EQUIVALENT transformation)

(84) 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.

(85) 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

(86) YES

(87) 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.

(88) UsableRulesProof (EQUIVALENT transformation)

(89) 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.

(90) 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.