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 proven:

0 QTRS

↳1 QTRS Reverse (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 116 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 QDP

        ↳8 UsableRulesProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 213 ms)

        ↳11 QDP

        ↳12 QDPOrderProof (⇔, 1327 ms)

        ↳13 QDP

        ↳14 QDPOrderProof (⇔, 161 ms)

        ↳15 QDP

        ↳16 QDPOrderProof (⇔, 1371 ms)

        ↳17 QDP

        ↳18 QDPOrderProof (⇔, 152 ms)

        ↳19 QDP

        ↳20 PisEmptyProof (⇔, 0 ms)

        ↳21 YES

    ↳22 QDP

        ↳23 UsableRulesProof (⇔, 0 ms)

        ↳24 QDP

        ↳25 QDPSizeChangeProof (⇔, 0 ms)

        ↳26 YES

    ↳27 QDP

        ↳28 UsableRulesProof (⇔, 10 ms)

        ↳29 QDP

        ↳30 QDPOrderProof (⇔, 52 ms)

        ↳31 QDP

        ↳32 QDPOrderProof (⇔, 0 ms)

        ↳33 QDP

        ↳34 QDPOrderProof (⇔, 24 ms)

        ↳35 QDP

        ↳36 MRRProof (⇔, 76 ms)

        ↳37 QDP

        ↳38 MRRProof (⇔, 53 ms)

        ↳39 QDP

        ↳40 QDPOrderProof (⇔, 4646 ms)

        ↳41 QDP

        ↳42 PisEmptyProof (⇔, 0 ms)

        ↳43 YES

(0) Obligation:

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

Begin(a(b(a(c(b(a(x))))))) → Wait(Right1(x))
Begin(b(a(c(b(a(x)))))) → Wait(Right2(x))
Begin(a(c(b(a(x))))) → Wait(Right3(x))
Begin(c(b(a(x)))) → Wait(Right4(x))
Begin(b(a(x))) → Wait(Right5(x))
Begin(a(x)) → Wait(Right6(x))
Right1(b(End(x))) → Left(a(b(a(c(b(b(a(b(a(c(End(x))))))))))))
Right2(b(a(End(x)))) → Left(a(b(a(c(b(b(a(b(a(c(End(x))))))))))))
Right3(b(a(b(End(x))))) → Left(a(b(a(c(b(b(a(b(a(c(End(x))))))))))))
Right4(b(a(b(a(End(x)))))) → Left(a(b(a(c(b(b(a(b(a(c(End(x))))))))))))
Right5(b(a(b(a(c(End(x))))))) → Left(a(b(a(c(b(b(a(b(a(c(End(x))))))))))))
Right6(b(a(b(a(c(b(End(x)))))))) → Left(a(b(a(c(b(b(a(b(a(c(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))
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))
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))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
b(a(b(a(c(b(a(x))))))) → a(b(a(c(b(b(a(b(a(c(x))))))))))

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(x))
End(b(Right1(x))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(Right2(x)))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(a(b(Right3(x))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(a(b(Right4(x)))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(c(a(b(a(b(Right5(x))))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(c(a(b(a(b(Right6(x)))))))) → End(c(a(b(a(b(b(c(a(b(a(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))
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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(x))))))))))

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

END(b(Right1(x))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(Right1(x))) → C(a(b(a(b(b(c(a(b(a(Left(x)))))))))))
END(b(Right1(x))) → A(b(a(b(b(c(a(b(a(Left(x))))))))))
END(b(Right1(x))) → B(a(b(b(c(a(b(a(Left(x)))))))))
END(b(Right1(x))) → A(b(b(c(a(b(a(Left(x))))))))
END(b(Right1(x))) → B(b(c(a(b(a(Left(x)))))))
END(b(Right1(x))) → B(c(a(b(a(Left(x))))))
END(b(Right1(x))) → C(a(b(a(Left(x)))))
END(b(Right1(x))) → A(b(a(Left(x))))
END(b(Right1(x))) → B(a(Left(x)))
END(b(Right1(x))) → A(Left(x))
END(b(Right1(x))) → LEFT(x)
END(a(b(Right2(x)))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(Right2(x)))) → C(a(b(a(b(b(c(a(b(a(Left(x)))))))))))
END(a(b(Right2(x)))) → A(b(a(b(b(c(a(b(a(Left(x))))))))))
END(a(b(Right2(x)))) → B(a(b(b(c(a(b(a(Left(x)))))))))
END(a(b(Right2(x)))) → A(b(b(c(a(b(a(Left(x))))))))
END(a(b(Right2(x)))) → B(b(c(a(b(a(Left(x)))))))
END(a(b(Right2(x)))) → B(c(a(b(a(Left(x))))))
END(a(b(Right2(x)))) → C(a(b(a(Left(x)))))
END(a(b(Right2(x)))) → A(b(a(Left(x))))
END(a(b(Right2(x)))) → B(a(Left(x)))
END(a(b(Right2(x)))) → A(Left(x))
END(a(b(Right2(x)))) → LEFT(x)
END(b(a(b(Right3(x))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(a(b(Right3(x))))) → C(a(b(a(b(b(c(a(b(a(Left(x)))))))))))
END(b(a(b(Right3(x))))) → A(b(a(b(b(c(a(b(a(Left(x))))))))))
END(b(a(b(Right3(x))))) → B(a(b(b(c(a(b(a(Left(x)))))))))
END(b(a(b(Right3(x))))) → A(b(b(c(a(b(a(Left(x))))))))
END(b(a(b(Right3(x))))) → B(b(c(a(b(a(Left(x)))))))
END(b(a(b(Right3(x))))) → B(c(a(b(a(Left(x))))))
END(b(a(b(Right3(x))))) → C(a(b(a(Left(x)))))
END(b(a(b(Right3(x))))) → A(b(a(Left(x))))
END(b(a(b(Right3(x))))) → B(a(Left(x)))
END(b(a(b(Right3(x))))) → A(Left(x))
END(b(a(b(Right3(x))))) → LEFT(x)
END(a(b(a(b(Right4(x)))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(a(b(Right4(x)))))) → C(a(b(a(b(b(c(a(b(a(Left(x)))))))))))
END(a(b(a(b(Right4(x)))))) → A(b(a(b(b(c(a(b(a(Left(x))))))))))
END(a(b(a(b(Right4(x)))))) → B(a(b(b(c(a(b(a(Left(x)))))))))
END(a(b(a(b(Right4(x)))))) → A(b(b(c(a(b(a(Left(x))))))))
END(a(b(a(b(Right4(x)))))) → B(b(c(a(b(a(Left(x)))))))
END(a(b(a(b(Right4(x)))))) → B(c(a(b(a(Left(x))))))
END(a(b(a(b(Right4(x)))))) → C(a(b(a(Left(x)))))
END(a(b(a(b(Right4(x)))))) → A(b(a(Left(x))))
END(a(b(a(b(Right4(x)))))) → B(a(Left(x)))
END(a(b(a(b(Right4(x)))))) → A(Left(x))
END(a(b(a(b(Right4(x)))))) → LEFT(x)
END(c(a(b(a(b(Right5(x))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(c(a(b(a(b(Right5(x))))))) → C(a(b(a(b(b(c(a(b(a(Left(x)))))))))))
END(c(a(b(a(b(Right5(x))))))) → A(b(a(b(b(c(a(b(a(Left(x))))))))))
END(c(a(b(a(b(Right5(x))))))) → B(a(b(b(c(a(b(a(Left(x)))))))))
END(c(a(b(a(b(Right5(x))))))) → A(b(b(c(a(b(a(Left(x))))))))
END(c(a(b(a(b(Right5(x))))))) → B(b(c(a(b(a(Left(x)))))))
END(c(a(b(a(b(Right5(x))))))) → B(c(a(b(a(Left(x))))))
END(c(a(b(a(b(Right5(x))))))) → C(a(b(a(Left(x)))))
END(c(a(b(a(b(Right5(x))))))) → A(b(a(Left(x))))
END(c(a(b(a(b(Right5(x))))))) → B(a(Left(x)))
END(c(a(b(a(b(Right5(x))))))) → A(Left(x))
END(c(a(b(a(b(Right5(x))))))) → LEFT(x)
END(b(c(a(b(a(b(Right6(x)))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → C(a(b(a(b(b(c(a(b(a(Left(x)))))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → A(b(a(b(b(c(a(b(a(Left(x))))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → B(a(b(b(c(a(b(a(Left(x)))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → A(b(b(c(a(b(a(Left(x))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → B(b(c(a(b(a(Left(x)))))))
END(b(c(a(b(a(b(Right6(x)))))))) → B(c(a(b(a(Left(x))))))
END(b(c(a(b(a(b(Right6(x)))))))) → C(a(b(a(Left(x)))))
END(b(c(a(b(a(b(Right6(x)))))))) → A(b(a(Left(x))))
END(b(c(a(b(a(b(Right6(x)))))))) → B(a(Left(x)))
END(b(c(a(b(a(b(Right6(x)))))))) → A(Left(x))
END(b(c(a(b(a(b(Right6(x)))))))) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
A(b(c(a(b(a(b(x))))))) → C(a(b(a(b(b(c(a(b(a(x))))))))))
A(b(c(a(b(a(b(x))))))) → A(b(a(b(b(c(a(b(a(x)))))))))
A(b(c(a(b(a(b(x))))))) → B(a(b(b(c(a(b(a(x))))))))
A(b(c(a(b(a(b(x))))))) → A(b(b(c(a(b(a(x)))))))
A(b(c(a(b(a(b(x))))))) → B(b(c(a(b(a(x))))))
A(b(c(a(b(a(b(x))))))) → B(c(a(b(a(x)))))
A(b(c(a(b(a(b(x))))))) → C(a(b(a(x))))
A(b(c(a(b(a(b(x))))))) → A(b(a(x)))
A(b(c(a(b(a(b(x))))))) → B(a(x))
A(b(c(a(b(a(b(x))))))) → A(x)

The TRS R consists of the following rules:

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(x))
End(b(Right1(x))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(Right2(x)))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(a(b(Right3(x))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(a(b(Right4(x)))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(c(a(b(a(b(Right5(x))))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(c(a(b(a(b(Right6(x)))))))) → End(c(a(b(a(b(b(c(a(b(a(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))
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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(x))))))))))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 75 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(x))
End(b(Right1(x))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(Right2(x)))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(a(b(Right3(x))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(a(b(Right4(x)))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(c(a(b(a(b(Right5(x))))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(c(a(b(a(b(Right6(x)))))))) → End(c(a(b(a(b(b(c(a(b(a(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))
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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(x))))))))))

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

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

The TRS R consists of the following rules:

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(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(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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))
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))

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

(10) 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(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(A(x₁)) = 2·x₁
POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Begin(x₁)) = 1 + 2·x₁
POL(Right1(x₁)) = 2·x₁
POL(Right2(x₁)) = x₁
POL(Right3(x₁)) = x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = 2·x₁
POL(Right6(x₁)) = 2·x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁

(11) Obligation:

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

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

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

(12) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

A(b(c(a(b(a(b(x))))))) → A(x)

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

POL(A(x₁)) = 0A +
[ 0A, 0A, -I ]

· x₁

POL(b(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ -I 0A -I \

| 0A -I -I |

\ 0A -I 0A /

· x₁

POL(c(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ -I 1A 0A \

| -I -I -I |

\ 0A 0A 1A /

· x₁

POL(a(x₁)) =
/ 0A \

| 1A |

\ 1A /

+
/ -I 1A -I \

| 0A -I -I |

\ -I 0A -I /

· x₁

POL(Right1(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A 0A 0A \

| -I -I 0A |

\ 1A 0A 0A /

· x₁

POL(Aa(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ -I -I -I \

| -I -I -I |

\ -I -I -I /

· x₁

POL(Right2(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| -I -I 0A |

\ 0A 0A 1A /

· x₁

POL(Right3(x₁)) =
/ 0A \

| -I |

\ 1A /

+
/ 0A 0A 0A \

| -I -I 0A |

\ 1A 0A 1A /

· x₁

POL(Right4(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I -I 0A \

| -I -I 0A |

\ 0A 0A 1A /

· x₁

POL(Right5(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ -I -I 0A \

| -I -I 0A |

\ -I -I 1A /

· x₁

POL(Right6(x₁)) =
/ 0A \

| -I |

\ 1A /

+
/ 0A 0A -I \

| -I -I -I |

\ 0A 0A 0A /

· x₁

POL(Ab(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I -I -I \

| -I -I -I |

\ -I -I 0A /

· x₁

POL(Ac(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 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:

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

(13) Obligation:

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

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

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

(14) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

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

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

POL( A(x₁) ) = 2x₁ + 2

POL( b(x₁) ) = max{0, x₁ - 1}

POL( a(x₁) ) = 2

POL( Right1(x₁) ) = max{0, -2}

POL( Aa(x₁) ) = 0

POL( Right2(x₁) ) = max{0, -2}

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( c(x₁) ) = 2

POL( Ab(x₁) ) = 0

POL( Ac(x₁) ) = 0

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

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

(15) Obligation:

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

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

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

(16) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic integers [ARCTIC,STERNAGEL_THIEMANN_RTA14]:

POL(A(x₁)) = 0A +
[ -1A, -1A, -1A ]

· x₁

POL(b(x₁)) =
/ 1A \

| 2A |

\ 2A /

+
/ -I -1A -I \

| -I -I -I |

\ -1A -I 0A /

· x₁

POL(c(x₁)) =
/ 1A \

| -1A |

\ 2A /

+
/ -1A -1A -I \

| -1A -1A 2A |

\ -1A -I -I /

· x₁

POL(a(x₁)) =
/ 1A \

| -I |

\ -1A /

+
/ -1A -1A -I \

| 1A -1A -I |

\ -1A -I 0A /

· x₁

POL(Right1(x₁)) =
/ 0A \

| -1A |

\ -I /

+
/ -I -1A -I \

| -1A -I -1A |

\ -I -I -I /

· x₁

POL(Aa(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ -1A -I -I \

| -1A -I -I |

\ -1A -1A -1A /

· x₁

POL(Right2(x₁)) =
/ 0A \

| -1A |

\ -I /

+
/ -I -I -I \

| -I -I -I |

\ -I -I -I /

· x₁

POL(Right3(x₁)) =
/ 0A \

| -1A |

\ -I /

+
/ -1A -1A -I \

| -1A -I 0A |

\ -I -1A -I /

· x₁

POL(Right4(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ -I -I -I \

| -1A -I -I |

\ -I -I -I /

· x₁

POL(Right5(x₁)) =
/ 1A \

| -1A |

\ -I /

+
/ -1A -1A -1A \

| 0A -I -I |

\ -I -I -1A /

· x₁

POL(Right6(x₁)) =
/ -1A \

| -I |

\ 2A /

+
/ 0A -1A -I \

| -1A -I -1A |

\ 0A 1A -I /

· x₁

POL(Ab(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ -I -I -I \

| -1A -I -I |

\ -I -I -I /

· x₁

POL(Ac(x₁)) =
/ 0A \

| -1A |

\ -1A /

+
/ -1A -I -I \

| -I -I -I |

\ -I -I -1A /

· x₁

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

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

(17) Obligation:

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

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

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

(18) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

A(b(c(a(b(a(b(x))))))) → A(b(a(x)))

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

POL( A(x₁) ) = max{0, x₁ - 2}

POL( b(x₁) ) = x₁ + 2

POL( a(x₁) ) = x₁

POL( Right1(x₁) ) = max{0, -2}

POL( Aa(x₁) ) = 0

POL( Right2(x₁) ) = max{0, -2}

POL( Right3(x₁) ) = 0

POL( Right4(x₁) ) = max{0, -2}

POL( Right5(x₁) ) = max{0, -2}

POL( Right6(x₁) ) = max{0, -2}

POL( c(x₁) ) = max{0, x₁ - 2}

POL( Ab(x₁) ) = 0

POL( Ac(x₁) ) = 0

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

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

(19) Obligation:

Q DP problem:
P is empty.
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(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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))
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))

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:

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

The TRS R consists of the following rules:

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(x))
End(b(Right1(x))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(Right2(x)))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(a(b(Right3(x))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(a(b(Right4(x)))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(c(a(b(a(b(Right5(x))))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(c(a(b(a(b(Right6(x)))))))) → End(c(a(b(a(b(b(c(a(b(a(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))
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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(x))))))))))

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

(23) UsableRulesProof (EQUIVALENT transformation)

(24) Obligation:

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

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

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

(25) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

LEFT(Aa(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

(26) YES

(27) Obligation:

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

END(a(b(Right2(x)))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(Right1(x))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(a(b(Right3(x))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(a(b(Right4(x)))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(c(a(b(a(b(Right5(x))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))

The TRS R consists of the following rules:

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(b(Begin(x))) → Right5(Wait(x))
a(Begin(x)) → Right6(Wait(x))
End(b(Right1(x))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(Right2(x)))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(a(b(Right3(x))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(a(b(a(b(Right4(x)))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(c(a(b(a(b(Right5(x))))))) → End(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
End(b(c(a(b(a(b(Right6(x)))))))) → End(c(a(b(a(b(b(c(a(b(a(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))
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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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:

END(a(b(Right2(x)))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(Right1(x))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(a(b(Right3(x))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(a(b(Right4(x)))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(c(a(b(a(b(Right5(x))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))

The TRS R consists of the following rules:

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

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

(30) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(b(a(b(Right3(x))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(c(a(b(a(b(Right6(x)))))))) → END(c(a(b(a(b(b(c(a(b(a(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(Begin(x₁)) = x₁
POL(END(x₁)) = x₁
POL(Left(x₁)) = 0
POL(Right1(x₁)) = 0
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 0
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 1
POL(b(x₁)) = 1 + x₁
POL(c(x₁)) = 1

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

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))

(31) Obligation:

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

END(a(b(Right2(x)))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(b(Right1(x))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(a(b(Right4(x)))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(c(a(b(a(b(Right5(x))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))

The TRS R consists of the following rules:

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

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

(32) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(b(Right1(x))) → END(c(a(b(a(b(b(c(a(b(a(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(Begin(x₁)) = x₁
POL(END(x₁)) = x₁
POL(Left(x₁)) = 0
POL(Right1(x₁)) = 0
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = 0
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 0
POL(b(x₁)) = 1
POL(c(x₁)) = 0

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

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))

(33) Obligation:

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

END(a(b(Right2(x)))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(a(b(Right4(x)))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(c(a(b(a(b(Right5(x))))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))

The TRS R consists of the following rules:

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

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

(34) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

END(a(b(Right2(x)))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))
END(a(b(a(b(Right4(x)))))) → END(c(a(b(a(b(b(c(a(b(a(Left(x))))))))))))

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

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = 0
POL(Ac(x₁)) = 0
POL(Begin(x₁)) = x₁
POL(END(x₁)) = x₁
POL(Left(x₁)) = 0
POL(Right1(x₁)) = 0
POL(Right2(x₁)) = 0
POL(Right3(x₁)) = 0
POL(Right4(x₁)) = 0
POL(Right5(x₁)) = 0
POL(Right6(x₁)) = x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = 1 + x₁
POL(b(x₁)) = 1
POL(c(x₁)) = 0

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

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))

(35) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(36) MRRProof (EQUIVALENT transformation)

a(b(c(a(b(a(Begin(x))))))) → Right1(Wait(x))
a(b(c(a(b(Begin(x)))))) → Right2(Wait(x))
a(b(c(Begin(x)))) → Right4(Wait(x))
a(Begin(x)) → Right6(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Begin(x₁)) = 1 + x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = 1 + x₁
POL(Right1(x₁)) = x₁
POL(Right2(x₁)) = x₁
POL(Right3(x₁)) = 1 + x₁
POL(Right4(x₁)) = x₁
POL(Right5(x₁)) = 1 + x₁
POL(Right6(x₁)) = x₁
POL(Wait(x₁)) = x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁

(37) Obligation:

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

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

The TRS R consists of the following rules:

Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(c(a(Begin(x))))) → Right3(Wait(x))
a(b(Begin(x))) → Right5(Wait(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(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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))
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))

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

(38) MRRProof (EQUIVALENT transformation)

a(b(c(a(Begin(x))))) → Right3(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x₁)) = x₁
POL(Ab(x₁)) = x₁
POL(Ac(x₁)) = x₁
POL(Begin(x₁)) = 1 + 2·x₁
POL(END(x₁)) = 2·x₁
POL(Left(x₁)) = 1 + x₁
POL(Right1(x₁)) = x₁
POL(Right2(x₁)) = x₁
POL(Right3(x₁)) = x₁
POL(Right4(x₁)) = 2·x₁
POL(Right5(x₁)) = 1 + x₁
POL(Right6(x₁)) = 2·x₁
POL(Wait(x₁)) = 2·x₁
POL(a(x₁)) = x₁
POL(b(x₁)) = x₁
POL(c(x₁)) = x₁

(39) Obligation:

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

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

The TRS R consists of the following rules:

Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(Begin(x))) → Right5(Wait(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(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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))
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))

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

(40) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

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

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

POL(END(x₁)) = -I +
[ 0A, 0A, -I ]

· x₁

POL(c(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A -I -I \

| -I -I -I |

\ -I -I 0A /

· x₁

POL(a(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A 0A -I \

| 0A 1A -I |

\ 0A 0A 0A /

· x₁

POL(b(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A 0A -I \

| -I 0A -I |

\ 0A 0A 0A /

· x₁

POL(Right5(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 1A 0A 0A \

| -I 1A 1A |

\ -I -I -I /

· x₁

POL(Left(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A -I -I \

| -I 0A 0A |

\ 0A -I -I /

· x₁

POL(Ab(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| -I 0A 0A |

\ -I 0A 0A /

· x₁

POL(Aa(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 1A |

\ 0A 1A 0A /

· x₁

POL(Ac(x₁)) =
/ -I \

| -I |

\ -I /

+
/ 0A -I -I \

| -I -I -I |

\ -I -I -I /

· x₁

POL(Wait(x₁)) =
/ 0A \

| 1A |

\ 0A /

+
/ -I 0A -I \

| 0A 1A -I |

\ -I 0A 0A /

· x₁

POL(Begin(x₁)) =
/ 0A \

| 1A |

\ 0A /

+
/ -I 0A -I \

| 0A 1A 0A |

\ -I -I -I /

· x₁

POL(Right1(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ 1A -I 0A \

| -I 1A 1A |

\ -I 0A 0A /

· x₁

POL(Right2(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 1A 0A 0A \

| -I 1A 1A |

\ -I -I -I /

· x₁

POL(Right3(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ 0A -I -I \

| -I 0A 0A |

\ 0A -I -I /

· x₁

POL(Right4(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A -I -I \

| -I 0A 0A |

\ -I -I -I /

· x₁

POL(Right6(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A -I -I \

| -I 0A 0A |

\ 0A -I -I /

· x₁

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

Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(Begin(x))) → Right5(Wait(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(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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))
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))

(41) Obligation:

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

Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(b(Begin(x))) → Right5(Wait(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(b(c(a(b(a(b(x))))))) → c(a(b(a(b(b(c(a(b(a(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))
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))

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

(42) PisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) QTRS Reverse (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) UsableRulesProof (EQUIVALENT transformation)

(9) Obligation:

(10) MRRProof (EQUIVALENT transformation)

(11) Obligation:

(12) QDPOrderProof (EQUIVALENT transformation)

(13) Obligation:

(14) QDPOrderProof (EQUIVALENT transformation)

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) QDPOrderProof (EQUIVALENT transformation)

(19) Obligation:

(20) PisEmptyProof (EQUIVALENT transformation)

(21) YES

(22) Obligation:

(23) UsableRulesProof (EQUIVALENT transformation)

(24) Obligation:

(25) QDPSizeChangeProof (EQUIVALENT transformation)

(26) YES

(27) Obligation:

(28) UsableRulesProof (EQUIVALENT transformation)

(29) Obligation:

(30) QDPOrderProof (EQUIVALENT transformation)

(31) Obligation:

(32) QDPOrderProof (EQUIVALENT transformation)

(33) Obligation:

(34) QDPOrderProof (EQUIVALENT transformation)

(35) Obligation:

(36) MRRProof (EQUIVALENT transformation)

(37) Obligation:

(38) MRRProof (EQUIVALENT transformation)

(39) Obligation:

(40) QDPOrderProof (EQUIVALENT transformation)

(41) Obligation:

(42) PisEmptyProof (EQUIVALENT transformation)

(43) YES