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 QTRS Reverse (⇔, 0 ms)

↳2 QTRS

↳3 QTRSRRRProof (⇔, 123 ms)

↳4 QTRS

↳5 DependencyPairsProof (⇔, 59 ms)

↳6 QDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 AND

    ↳9 QDP

        ↳10 UsableRulesProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 DependencyGraphProof (⇔, 0 ms)

        ↳13 AND

            ↳14 QDP

                ↳15 QDPOrderProof (⇔, 39 ms)

                ↳16 QDP

                ↳17 DependencyGraphProof (⇔, 0 ms)

                ↳18 QDP

                ↳19 QDPOrderProof (⇔, 1 ms)

                ↳20 QDP

                ↳21 DependencyGraphProof (⇔, 0 ms)

                ↳22 QDP

                ↳23 MRRProof (⇔, 57 ms)

                ↳24 QDP

                ↳25 MRRProof (⇔, 10 ms)

                ↳26 QDP

                ↳27 MRRProof (⇔, 0 ms)

                ↳28 QDP

                ↳29 QDPOrderProof (⇔, 100 ms)

                ↳30 QDP

                ↳31 PisEmptyProof (⇔, 0 ms)

                ↳32 YES

            ↳33 QDP

                ↳34 UsableRulesProof (⇔, 0 ms)

                ↳35 QDP

                ↳36 QDPSizeChangeProof (⇔, 0 ms)

                ↳37 YES

    ↳38 QDP

        ↳39 UsableRulesProof (⇔, 0 ms)

        ↳40 QDP

        ↳41 QDPSizeChangeProof (⇔, 0 ms)

        ↳42 YES

    ↳43 QDP

        ↳44 UsableRulesProof (⇔, 0 ms)

        ↳45 QDP

        ↳46 QDPOrderProof (⇔, 54 ms)

        ↳47 QDP

        ↳48 QDPOrderProof (⇔, 19 ms)

        ↳49 QDP

        ↳50 QDPOrderProof (⇔, 37 ms)

        ↳51 QDP

        ↳52 MRRProof (⇔, 60 ms)

        ↳53 QDP

        ↳54 QDPOrderProof (⇔, 171 ms)

        ↳55 QDP

        ↳56 MRRProof (⇔, 97 ms)

        ↳57 QDP

        ↳58 MRRProof (⇔, 102 ms)

        ↳59 QDP

        ↳60 MRRProof (⇔, 82 ms)

        ↳61 QDP

        ↳62 MRRProof (⇔, 67 ms)

        ↳63 QDP

        ↳64 MRRProof (⇔, 62 ms)

        ↳65 QDP

        ↳66 NonTerminationLoopProof (⇒, 3309 ms)

        ↳67 NO

(0) Obligation:

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

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

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

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

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 2 + x₁
POL(Ab(x₁)) = 2 + x₁
POL(Ac(x₁)) = 2 + x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = x₁
POL(End(x₁)) = x₁
POL(Left(x₁)) = x₁
POL(Right1(x₁)) = 2 + x₁
POL(Right10(x₁)) = 4 + x₁
POL(Right11(x₁)) = 2 + x₁
POL(Right2(x₁)) = 2 + x₁
POL(Right3(x₁)) = 3 + x₁
POL(Right4(x₁)) = 1 + x₁
POL(Right5(x₁)) = x₁
POL(Right6(x₁)) = 4 + x₁
POL(Right7(x₁)) = 2 + x₁
POL(Right8(x₁)) = 2 + x₁
POL(Right9(x₁)) = 2 + x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 2 + x₁
POL(b(x₁)) = 2 + x₁
POL(c(x₁)) = 2 + x₁
POL(d(x₁)) = x₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

c(a(Begin(x))) → Right3(Wait(x))
c(Begin(x)) → Right4(Wait(x))
End(b(Right3(x))) → End(c(b(Left(x))))
End(a(b(Right4(x)))) → End(c(b(Left(x))))
c(a(b(x))) → c(b(x))

(4) Obligation:

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

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

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(b(Right1(x))) → B(d(c(Left(x))))
END(b(Right1(x))) → D(c(Left(x)))
END(b(Right1(x))) → C(Left(x))
END(b(Right1(x))) → LEFT(x)
END(d(b(Right2(x)))) → END(b(d(c(Left(x)))))
END(d(b(Right2(x)))) → B(d(c(Left(x))))
END(d(b(Right2(x)))) → D(c(Left(x)))
END(d(b(Right2(x)))) → C(Left(x))
END(d(b(Right2(x)))) → LEFT(x)
END(a(Right5(x))) → END(c(d(Left(x))))
END(a(Right5(x))) → C(d(Left(x)))
END(a(Right5(x))) → D(Left(x))
END(a(Right5(x))) → LEFT(x)
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(Right6(x))) → C(b(a(Left(x))))
END(b(Right6(x))) → B(a(Left(x)))
END(b(Right6(x))) → A(Left(x))
END(b(Right6(x))) → LEFT(x)
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → C(b(a(Left(x))))
END(b(b(Right7(x)))) → B(a(Left(x)))
END(b(b(Right7(x)))) → A(Left(x))
END(b(b(Right7(x)))) → LEFT(x)
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right8(x))) → D(b(Left(x)))
END(d(Right8(x))) → B(Left(x))
END(d(Right8(x))) → LEFT(x)
END(d(Right9(x))) → END(d(b(d(Left(x)))))
END(d(Right9(x))) → D(b(d(Left(x))))
END(d(Right9(x))) → B(d(Left(x)))
END(d(Right9(x))) → D(Left(x))
END(d(Right9(x))) → LEFT(x)
END(d(Right10(x))) → END(b(b(Left(x))))
END(d(Right10(x))) → B(b(Left(x)))
END(d(Right10(x))) → B(Left(x))
END(d(Right10(x))) → LEFT(x)
END(a(d(Right11(x)))) → END(b(b(Left(x))))
END(a(d(Right11(x)))) → B(b(Left(x)))
END(a(d(Right11(x)))) → B(Left(x))
END(a(d(Right11(x)))) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
LEFT(Ad(x)) → D(Left(x))
LEFT(Ad(x)) → LEFT(x)
LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
B(d(b(x))) → B(d(c(x)))
B(d(b(x))) → D(c(x))
B(d(b(x))) → C(x)
D(a(x)) → C(d(x))
D(a(x)) → D(x)
B(b(b(x))) → C(b(a(x)))
B(b(b(x))) → B(a(x))
B(b(b(x))) → A(x)
C(d(x)) → D(b(x))
C(d(x)) → B(x)
C(d(x)) → D(b(d(x)))
C(d(x)) → B(d(x))
C(a(d(x))) → B(b(x))
C(a(d(x))) → B(x)

The TRS R consists of the following rules:

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

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 37 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

B(d(b(x))) → D(c(x))
D(a(x)) → C(d(x))
C(d(x)) → D(b(x))
D(a(x)) → D(x)
C(d(x)) → B(x)
B(d(b(x))) → B(d(c(x)))
B(d(b(x))) → C(x)
C(d(x)) → D(b(d(x)))
C(d(x)) → B(d(x))
B(b(b(x))) → C(b(a(x)))
C(a(d(x))) → B(b(x))
B(b(b(x))) → B(a(x))
C(a(d(x))) → B(x)

The TRS R consists of the following rules:

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

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

(10) 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.

(11) Obligation:

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

B(d(b(x))) → D(c(x))
D(a(x)) → C(d(x))
C(d(x)) → D(b(x))
D(a(x)) → D(x)
C(d(x)) → B(x)
B(d(b(x))) → B(d(c(x)))
B(d(b(x))) → C(x)
C(d(x)) → D(b(d(x)))
C(d(x)) → B(d(x))
B(b(b(x))) → C(b(a(x)))
C(a(d(x))) → B(b(x))
B(b(b(x))) → B(a(x))
C(a(d(x))) → B(x)

The TRS R consists of the following rules:

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

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

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(13) Complex Obligation (AND)

(14) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(15) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

B(d(b(x))) → C(x)
B(b(b(x))) → B(a(x))
C(a(d(x))) → B(x)

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

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = 1
POL(Ac(x₁)) = 1 + x₁
POL(Ad(x₁)) = x₁
POL(B(x₁)) = x₁
POL(Begin(x₁)) = x₁
POL(C(x₁)) = x₁
POL(Right1(x₁)) = 1 + x₁
POL(Right10(x₁)) = 0
POL(Right11(x₁)) = 0
POL(Right2(x₁)) = 1 + x₁
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 1 + x₁
POL(Right7(x₁)) = 1 + x₁
POL(Right8(x₁)) = 1 + x₁
POL(Right9(x₁)) = 0
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = 1 + x₁
POL(c(x₁)) = 1 + x₁
POL(d(x₁)) = x₁

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

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(17) DependencyGraphProof (EQUIVALENT transformation)

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

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(19) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

C(d(x)) → B(x)
C(d(x)) → B(d(x))

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

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(B(x₁)) = 0
POL(Begin(x₁)) = x₁
POL(C(x₁)) = x₁
POL(Right1(x₁)) = 0
POL(Right10(x₁)) = 0
POL(Right11(x₁)) = 0
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 0
POL(Right7(x₁)) = 0
POL(Right8(x₁)) = 0
POL(Right9(x₁)) = 0
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 0
POL(b(x₁)) = 0
POL(c(x₁)) = x₁
POL(d(x₁)) = 1

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

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

(20) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

B(d(b(x))) → B(d(c(x)))

The TRS R consists of the following rules:

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

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

(23) 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 rules of the TRS R:

a(Right7(x)) → Right7(Aa(x))
a(Right8(x)) → Right8(Aa(x))
a(Right10(x)) → Right10(Aa(x))
b(d(Begin(x))) → Right1(Wait(x))
b(Begin(x)) → Right2(Wait(x))
b(b(Begin(x))) → Right6(Wait(x))
b(Begin(x)) → Right7(Wait(x))
b(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
b(Right10(x)) → Right10(Ab(x))
c(Begin(x)) → Right8(Wait(x))
c(Begin(x)) → Right9(Wait(x))
c(a(Begin(x))) → Right10(Wait(x))
c(Begin(x)) → Right11(Wait(x))
c(Right7(x)) → Right7(Ac(x))
c(Right8(x)) → Right8(Ac(x))
c(Right10(x)) → Right10(Ac(x))
d(Begin(x)) → Right5(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(B(x₁)) = 2·x₁
POL(Begin(x₁)) = 3 + 2·x₁
POL(Right1(x₁)) = x₁
POL(Right10(x₁)) = 2 + 2·x₁
POL(Right11(x₁)) = x₁
POL(Right2(x₁)) = 2·x₁
POL(Right3(x₁)) = 2·x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = 2·x₁
POL(Right6(x₁)) = 2·x₁
POL(Right7(x₁)) = 3 + x₁
POL(Right8(x₁)) = 3 + 2·x₁
POL(Right9(x₁)) = x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(24) Obligation:

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

B(d(b(x))) → B(d(c(x)))

The TRS R consists of the following rules:

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

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

(25) MRRProof (EQUIVALENT transformation)

a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
a(Right6(x)) → Right6(Aa(x))
a(Right11(x)) → Right11(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(Right6(x)) → Right6(Ab(x))
b(Right11(x)) → Right11(Ab(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
c(Right6(x)) → Right6(Ac(x))
c(Right11(x)) → Right11(Ac(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(B(x₁)) = 2·x₁
POL(Right1(x₁)) = 1 + 2·x₁
POL(Right10(x₁)) = 2·x₁
POL(Right11(x₁)) = 2 + 2·x₁
POL(Right2(x₁)) = 2 + x₁
POL(Right3(x₁)) = 1 + 2·x₁
POL(Right4(x₁)) = 2 + 2·x₁
POL(Right5(x₁)) = 2 + x₁
POL(Right6(x₁)) = 1 + 2·x₁
POL(Right7(x₁)) = x₁
POL(Right8(x₁)) = x₁
POL(Right9(x₁)) = x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(26) Obligation:

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

B(d(b(x))) → B(d(c(x)))

The TRS R consists of the following rules:

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

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

(27) MRRProof (EQUIVALENT transformation)

a(Right9(x)) → Right9(Aa(x))
b(Right9(x)) → Right9(Ab(x))
c(Right9(x)) → Right9(Ac(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 2·x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = 1 + 2·x₁
POL(Ad(x₁)) = x₁
POL(B(x₁)) = 2·x₁
POL(Right1(x₁)) = 2·x₁
POL(Right10(x₁)) = 2·x₁
POL(Right11(x₁)) = 2·x₁
POL(Right2(x₁)) = x₁
POL(Right3(x₁)) = x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = 2·x₁
POL(Right6(x₁)) = x₁
POL(Right7(x₁)) = 2·x₁
POL(Right8(x₁)) = 2·x₁
POL(Right9(x₁)) = 2 + x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(28) Obligation:

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

B(d(b(x))) → B(d(c(x)))

The TRS R consists of the following rules:

b(b(b(x))) → c(b(a(x)))
c(d(x)) → d(b(x))
d(a(x)) → c(d(x))
c(d(x)) → d(b(d(x)))
c(a(d(x))) → b(b(x))
b(d(b(x))) → b(d(c(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
d(Right6(x)) → Right6(Ad(x))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(x))
d(Right9(x)) → Right9(Ad(x))
d(Right10(x)) → Right10(Ad(x))
d(Right11(x)) → Right11(Ad(x))

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

(29) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

B(d(b(x))) → B(d(c(x)))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x₁)) = -I +
[ 0A, -I, -I ]

· x₁

POL(d(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ -I 0A -I \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

POL(b(x₁)) =
/ 0A \

| 1A |

\ -I /

+
/ 0A 0A 0A \

| 1A 0A 1A |

\ -I -I -I /

· x₁

POL(c(x₁)) =
/ -I \

| 0A |

\ -I /

+
/ 0A -I 1A \

| 0A -I 0A |

\ -I -I 1A /

· x₁

POL(a(x₁)) =
/ 0A \

| 1A |

\ 1A /

+
/ 1A 1A 0A \

| 1A 1A 1A |

\ 0A 0A 0A /

· x₁

POL(Right1(x₁)) =
/ 0A \

| 1A |

\ 1A /

+
/ -I 0A 0A \

| 0A 0A 0A |

\ 1A 1A 1A /

· x₁

POL(Ad(x₁)) =
/ -I \

| 0A |

\ -I /

+
/ 0A 0A -I \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

POL(Right2(x₁)) =
/ 1A \

| 1A |

\ -I /

+
/ -I -I -I \

| -I -I 0A |

\ 1A 1A 1A /

· x₁

POL(Right3(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I -I -I \

| -I -I -I |

\ 0A 0A 0A /

· x₁

POL(Right4(x₁)) =
/ -I \

| -I |

\ 1A /

+
/ 0A 0A 0A \

| 0A 1A 0A |

\ 1A 1A 1A /

· x₁

POL(Right5(x₁)) =
/ -I \

| 0A |

\ 0A /

+
/ 0A -I -I \

| 0A 0A -I |

\ -I 0A 0A /

· x₁

POL(Right6(x₁)) =
/ 0A \

| 1A |

\ 1A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 1A 1A 1A /

· x₁

POL(Right7(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A -I \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

POL(Right8(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ -I -I -I /

· x₁

POL(Right9(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· x₁

POL(Right10(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ -I 0A -I \

| 0A -I 0A |

\ 0A -I 0A /

· x₁

POL(Right11(x₁)) =
/ 0A \

| -I |

\ 1A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ -I -I -I /

· x₁

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

c(d(x)) → d(b(x))
d(a(x)) → c(d(x))
c(d(x)) → d(b(d(x)))
c(a(d(x))) → b(b(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
d(Right6(x)) → Right6(Ad(x))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(x))
d(Right9(x)) → Right9(Ad(x))
d(Right10(x)) → Right10(Ad(x))
d(Right11(x)) → Right11(Ad(x))
b(d(b(x))) → b(d(c(x)))
b(b(b(x))) → c(b(a(x)))

(30) Obligation:

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

b(b(b(x))) → c(b(a(x)))
c(d(x)) → d(b(x))
d(a(x)) → c(d(x))
c(d(x)) → d(b(d(x)))
c(a(d(x))) → b(b(x))
b(d(b(x))) → b(d(c(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
d(Right6(x)) → Right6(Ad(x))
d(Right7(x)) → Right7(Ad(x))
d(Right8(x)) → Right8(Ad(x))
d(Right9(x)) → Right9(Ad(x))
d(Right10(x)) → Right10(Ad(x))
d(Right11(x)) → Right11(Ad(x))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) YES

(33) Obligation:

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

D(a(x)) → D(x)

The TRS R consists of the following rules:

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

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

(34) UsableRulesProof (EQUIVALENT transformation)

(35) Obligation:

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

D(a(x)) → D(x)

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

(36) 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:

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

(37) YES

(38) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(39) UsableRulesProof (EQUIVALENT transformation)

(40) Obligation:

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

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

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

(41) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

LEFT(Ad(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Ab(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Ac(x)) → LEFT(x)
The graph contains the following edges 1 > 1
LEFT(Aa(x)) → LEFT(x)
The graph contains the following edges 1 > 1

(42) YES

(43) Obligation:

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

END(d(b(Right2(x)))) → END(b(d(c(Left(x)))))
END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(a(Right5(x))) → END(c(d(Left(x))))
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))
END(d(Right10(x))) → END(b(b(Left(x))))
END(a(d(Right11(x)))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(44) UsableRulesProof (EQUIVALENT transformation)

(45) Obligation:

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

END(d(b(Right2(x)))) → END(b(d(c(Left(x)))))
END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(a(Right5(x))) → END(c(d(Left(x))))
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))
END(d(Right10(x))) → END(b(b(Left(x))))
END(a(d(Right11(x)))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(46) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(a(d(Right11(x)))) → END(b(b(Left(x))))

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

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = x₁
POL(END(x₁)) = x₁
POL(Left(x₁)) = 0
POL(Right1(x₁)) = 0
POL(Right10(x₁)) = 1
POL(Right11(x₁)) = 1
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 0
POL(Right7(x₁)) = 0
POL(Right8(x₁)) = 1
POL(Right9(x₁)) = 1
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = 1
POL(c(x₁)) = 1
POL(d(x₁)) = x₁

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

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

(47) Obligation:

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

END(d(b(Right2(x)))) → END(b(d(c(Left(x)))))
END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(a(Right5(x))) → END(c(d(Left(x))))
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))
END(d(Right10(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(48) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(a(Right5(x))) → END(c(d(Left(x))))

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

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = x₁
POL(END(x₁)) = x₁
POL(Left(x₁)) = 0
POL(Right1(x₁)) = 0
POL(Right10(x₁)) = 0
POL(Right11(x₁)) = 0
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 0
POL(Right7(x₁)) = 0
POL(Right8(x₁)) = 0
POL(Right9(x₁)) = 0
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 1
POL(b(x₁)) = 0
POL(c(x₁)) = 0
POL(d(x₁)) = 0

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

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

(49) Obligation:

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

END(d(b(Right2(x)))) → END(b(d(c(Left(x)))))
END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))
END(d(Right10(x))) → END(b(b(Left(x))))

The TRS R consists of the following rules:

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

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

(50) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(d(b(Right2(x)))) → END(b(d(c(Left(x)))))
END(d(Right10(x))) → END(b(b(Left(x))))

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

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = x₁
POL(END(x₁)) = x₁
POL(Left(x₁)) = 0
POL(Right1(x₁)) = 0
POL(Right10(x₁)) = 0
POL(Right11(x₁)) = 0
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 0
POL(Right7(x₁)) = 0
POL(Right8(x₁)) = 0
POL(Right9(x₁)) = 0
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 0
POL(b(x₁)) = 0
POL(c(x₁)) = x₁
POL(d(x₁)) = 1

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

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

(51) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(52) MRRProof (EQUIVALENT transformation)

b(Begin(x)) → Right2(Wait(x))
c(a(Begin(x))) → Right10(Wait(x))
c(Begin(x)) → Right11(Wait(x))
d(Begin(x)) → Right5(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = 1 + 2·x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = 1 + 2·x₁
POL(Right1(x₁)) = 1 + 2·x₁
POL(Right10(x₁)) = x₁
POL(Right11(x₁)) = x₁
POL(Right2(x₁)) = x₁
POL(Right3(x₁)) = x₁
POL(Right4(x₁)) = 2·x₁
POL(Right5(x₁)) = 2·x₁
POL(Right6(x₁)) = 1 + 2·x₁
POL(Right7(x₁)) = 1 + 2·x₁
POL(Right8(x₁)) = 1 + 2·x₁
POL(Right9(x₁)) = 1 + 2·x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁
POL(d(x₁)) = x₁

(53) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(54) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(b(Right6(x))) → END(c(b(a(Left(x)))))
END(b(b(Right7(x)))) → END(c(b(a(Left(x)))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( END(x₁) ) = x₁

POL( b(x₁) ) = 1

POL( d(x₁) ) = max{0, -2}

POL( a(x₁) ) = 2

POL( c(x₁) ) = max{0, x₁ - 1}

POL( Left(x₁) ) = 0

POL( Ab(x₁) ) = 0

POL( Ad(x₁) ) = 0

POL( Ac(x₁) ) = 0

POL( Aa(x₁) ) = 0

POL( Wait(x₁) ) = 0

POL( Begin(x₁) ) = 0

POL( Right8(x₁) ) = max{0, -2}

POL( Right9(x₁) ) = max{0, -2}

POL( Right1(x₁) ) = max{0, -2}

POL( Right2(x₁) ) = 0

POL( Right3(x₁) ) = max{0, -2}

POL( Right4(x₁) ) = max{0, -2}

POL( Right5(x₁) ) = max{0, -2}

POL( Right6(x₁) ) = max{0, -2}

POL( Right7(x₁) ) = max{0, -2}

POL( Right10(x₁) ) = max{0, -2}

POL( Right11(x₁) ) = 0

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

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

(55) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(56) MRRProof (EQUIVALENT transformation)

b(Right4(x)) → Right4(Ab(x))
b(Right10(x)) → Right10(Ab(x))
b(Right11(x)) → Right11(Ab(x))
a(Right4(x)) → Right4(Aa(x))
a(Right10(x)) → Right10(Aa(x))
a(Right11(x)) → Right11(Aa(x))
c(Right4(x)) → Right4(Ac(x))
c(Right10(x)) → Right10(Ac(x))
c(Right11(x)) → Right11(Ac(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 2·x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = 2·x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = 2·x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = x₁
POL(Right1(x₁)) = 2·x₁
POL(Right10(x₁)) = 1 + x₁
POL(Right11(x₁)) = 1 + 2·x₁
POL(Right2(x₁)) = 2·x₁
POL(Right3(x₁)) = x₁
POL(Right4(x₁)) = 1 + x₁
POL(Right5(x₁)) = x₁
POL(Right6(x₁)) = 2·x₁
POL(Right7(x₁)) = 2·x₁
POL(Right8(x₁)) = 2·x₁
POL(Right9(x₁)) = 2·x₁
POL(Wait(x₁)) = 2·x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(57) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(58) MRRProof (EQUIVALENT transformation)

b(Right5(x)) → Right5(Ab(x))
a(Right5(x)) → Right5(Aa(x))
c(Right5(x)) → Right5(Ac(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 2·x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = 2·x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = x₁
POL(Right1(x₁)) = 2·x₁
POL(Right10(x₁)) = x₁
POL(Right11(x₁)) = 2·x₁
POL(Right2(x₁)) = 2·x₁
POL(Right3(x₁)) = 2·x₁
POL(Right4(x₁)) = 2·x₁
POL(Right5(x₁)) = 3 + x₁
POL(Right6(x₁)) = 2·x₁
POL(Right7(x₁)) = 2·x₁
POL(Right8(x₁)) = 2·x₁
POL(Right9(x₁)) = 2·x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(59) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(60) MRRProof (EQUIVALENT transformation)

b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 2·x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = 2·x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = 2·x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = x₁
POL(Right1(x₁)) = 2·x₁
POL(Right10(x₁)) = 2·x₁
POL(Right11(x₁)) = 2·x₁
POL(Right2(x₁)) = 1 + 2·x₁
POL(Right3(x₁)) = 1 + x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = 2·x₁
POL(Right6(x₁)) = 2·x₁
POL(Right7(x₁)) = 2·x₁
POL(Right8(x₁)) = 2·x₁
POL(Right9(x₁)) = 2·x₁
POL(Wait(x₁)) = 2·x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(61) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(62) MRRProof (EQUIVALENT transformation)

b(b(Begin(x))) → Right6(Wait(x))
b(Right6(x)) → Right6(Ab(x))
b(Right7(x)) → Right7(Ab(x))
a(Right6(x)) → Right6(Aa(x))
a(Right7(x)) → Right7(Aa(x))
c(Right6(x)) → Right6(Ac(x))
c(Right7(x)) → Right7(Ac(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 2·x₁
POL(Ab(x₁)) = 2·x₁
POL(Ac(x₁)) = 2·x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = 2 + x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = x₁
POL(Right1(x₁)) = 2·x₁
POL(Right10(x₁)) = x₁
POL(Right11(x₁)) = 2·x₁
POL(Right2(x₁)) = 2·x₁
POL(Right3(x₁)) = x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = x₁
POL(Right6(x₁)) = 1 + 3·x₁
POL(Right7(x₁)) = 2 + x₁
POL(Right8(x₁)) = 2·x₁
POL(Right9(x₁)) = 2·x₁
POL(Wait(x₁)) = 2 + x₁
POL(a(x₁)) = 2·x₁
POL(b(x₁)) = 2·x₁
POL(c(x₁)) = 2·x₁
POL(d(x₁)) = x₁

(63) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(64) MRRProof (EQUIVALENT transformation)

b(Begin(x)) → Right7(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = 1 + 2·x₁
POL(Ab(x₁)) = 1 + 2·x₁
POL(Ac(x₁)) = 1 + 2·x₁
POL(Ad(x₁)) = x₁
POL(Begin(x₁)) = 2·x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = x₁
POL(Right1(x₁)) = 1 + 2·x₁
POL(Right10(x₁)) = 2·x₁
POL(Right11(x₁)) = 2·x₁
POL(Right2(x₁)) = x₁
POL(Right3(x₁)) = 2·x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = 2·x₁
POL(Right6(x₁)) = 2·x₁
POL(Right7(x₁)) = 2·x₁
POL(Right8(x₁)) = 1 + 2·x₁
POL(Right9(x₁)) = 1 + 2·x₁
POL(Wait(x₁)) = 2·x₁
POL(a(x₁)) = 1 + 2·x₁
POL(b(x₁)) = 1 + 2·x₁
POL(c(x₁)) = 1 + 2·x₁
POL(d(x₁)) = x₁

(65) Obligation:

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

END(b(Right1(x))) → END(b(d(c(Left(x)))))
END(d(Right8(x))) → END(d(b(Left(x))))
END(d(Right9(x))) → END(d(b(d(Left(x)))))

The TRS R consists of the following rules:

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

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

(66) 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 = END(b(d(c(Left(Ad(Ad(Wait(x)))))))) evaluates to t =END(b(d(c(Left(Ad(Ad(Wait(x))))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Matcher: [ ]
Semiunifier: [ ]

Rewriting sequence

END(b(d(c(Left(Ad(Ad(Wait(x)))))))) → END(b(d(c(d(Left(Ad(Wait(x))))))))
with rule Left(Ad(x')) → d(Left(x')) at position [0,0,0,0] and matcher [x' / Ad(Wait(x))]

END(b(d(c(d(Left(Ad(Wait(x)))))))) → END(b(d(c(d(d(Left(Wait(x))))))))
with rule Left(Ad(x')) → d(Left(x')) at position [0,0,0,0,0] and matcher [x' / Wait(x)]

END(b(d(c(d(d(Left(Wait(x)))))))) → END(b(d(c(d(d(Begin(x)))))))
with rule Left(Wait(x')) → Begin(x') at position [0,0,0,0,0,0] and matcher [x' / x]

END(b(d(c(d(d(Begin(x))))))) → END(b(d(d(b(d(Begin(x)))))))
with rule c(d(x')) → d(b(x')) at position [0,0,0] and matcher [x' / d(Begin(x))]

END(b(d(d(b(d(Begin(x))))))) → END(b(d(d(Right1(Wait(x))))))
with rule b(d(Begin(x'))) → Right1(Wait(x')) at position [0,0,0,0] and matcher [x' / x]

END(b(d(d(Right1(Wait(x)))))) → END(b(d(Right1(Ad(Wait(x))))))
with rule d(Right1(x')) → Right1(Ad(x')) at position [0,0,0] and matcher [x' / Wait(x)]

END(b(d(Right1(Ad(Wait(x)))))) → END(b(Right1(Ad(Ad(Wait(x))))))
with rule d(Right1(x')) → Right1(Ad(x')) at position [0,0] and matcher [x' / Ad(Wait(x))]

END(b(Right1(Ad(Ad(Wait(x)))))) → END(b(d(c(Left(Ad(Ad(Wait(x))))))))
with rule END(b(Right1(x))) → END(b(d(c(Left(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.

(0) Obligation:

(1) QTRS Reverse (EQUIVALENT transformation)

(2) Obligation:

(3) QTRSRRRProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) UsableRulesProof (EQUIVALENT transformation)

(11) Obligation:

(12) DependencyGraphProof (EQUIVALENT transformation)

(13) Complex Obligation (AND)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) DependencyGraphProof (EQUIVALENT transformation)

(18) Obligation:

(19) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) DependencyGraphProof (EQUIVALENT transformation)

(22) Obligation:

(23) MRRProof (EQUIVALENT transformation)

(24) Obligation:

(25) MRRProof (EQUIVALENT transformation)

(26) Obligation:

(27) MRRProof (EQUIVALENT transformation)

(28) Obligation:

(29) QDPOrderProof (EQUIVALENT transformation)

(30) Obligation:

(31) PisEmptyProof (EQUIVALENT transformation)

(32) YES

(33) Obligation:

(34) UsableRulesProof (EQUIVALENT transformation)

(35) Obligation:

(36) QDPSizeChangeProof (EQUIVALENT transformation)

(37) YES

(38) Obligation:

(39) UsableRulesProof (EQUIVALENT transformation)

(40) Obligation:

(41) QDPSizeChangeProof (EQUIVALENT transformation)

(42) YES

(43) Obligation:

(44) UsableRulesProof (EQUIVALENT transformation)

(45) Obligation:

(46) QDPOrderProof (EQUIVALENT transformation)

(47) Obligation:

(48) QDPOrderProof (EQUIVALENT transformation)

(49) Obligation:

(50) QDPOrderProof (EQUIVALENT transformation)

(51) Obligation:

(52) MRRProof (EQUIVALENT transformation)

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) MRRProof (EQUIVALENT transformation)

(57) Obligation:

(58) MRRProof (EQUIVALENT transformation)

(59) Obligation:

(60) MRRProof (EQUIVALENT transformation)

(61) Obligation:

(62) MRRProof (EQUIVALENT transformation)

(63) Obligation:

(64) MRRProof (EQUIVALENT transformation)

(65) Obligation:

(66) NonTerminationLoopProof (COMPLETE transformation)

(67) NO