NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Trafo_06/un05.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 59 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 18 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 61 ms)
        ↳13 QDP
        ↳14 MRRProof (⇔, 69 ms)
        ↳15 QDP
        ↳16 MRRProof (⇔, 26 ms)
        ↳17 QDP
        ↳18 MRRProof (⇔, 25 ms)
        ↳19 QDP
        ↳20 QDPOrderProof (⇔, 1806 ms)
        ↳21 QDP
        ↳22 DependencyGraphProof (⇔, 0 ms)
        ↳23 TRUE
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 0 ms)
        ↳26 QDP
        ↳27 QDPSizeChangeProof (⇔, 0 ms)
        ↳28 YES
    ↳29 QDP
        ↳30 UsableRulesProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 QDPOrderProof (⇔, 34 ms)
        ↳33 QDP
        ↳34 MRRProof (⇔, 134 ms)
        ↳35 QDP
        ↳36 NonTerminationLoopProof (⇒, 1651 ms)
        ↳37 NO

(0) Obligation:

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

Begin(b(b(b(x)))) → Wait(Right1(x))
Begin(b(b(x))) → Wait(Right2(x))
Begin(b(x)) → Wait(Right3(x))
Begin(a(b(b(a(b(x)))))) → Wait(Right4(x))
Begin(b(b(a(b(x))))) → Wait(Right5(x))
Begin(b(a(b(x)))) → Wait(Right6(x))
Begin(a(b(x))) → Wait(Right7(x))
Begin(b(x)) → Wait(Right8(x))
Right1(b(End(x))) → Left(b(b(b(a(b(End(x)))))))
Right2(b(b(End(x)))) → Left(b(b(b(a(b(End(x)))))))
Right3(b(b(b(End(x))))) → Left(b(b(b(a(b(End(x)))))))
Right4(b(End(x))) → Left(b(b(a(b(b(End(x)))))))
Right5(b(a(End(x)))) → Left(b(b(a(b(b(End(x)))))))
Right6(b(a(b(End(x))))) → Left(b(b(a(b(b(End(x)))))))
Right7(b(a(b(b(End(x)))))) → Left(b(b(a(b(b(End(x)))))))
Right8(b(a(b(b(a(End(x))))))) → Left(b(b(a(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))
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))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(b(b(b(x)))) → b(b(b(a(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(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(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(Wait(x))
End(b(Right1(x))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(Right2(x)))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(b(Right3(x))))) → End(b(a(b(b(b(Left(x)))))))
End(b(Right4(x))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(Right5(x)))) → End(b(b(a(b(b(Left(x)))))))
End(b(a(b(Right6(x))))) → End(b(b(a(b(b(Left(x)))))))
End(b(b(a(b(Right7(x)))))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(b(a(b(Right8(x))))))) → End(b(b(a(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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(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(b(a(b(b(b(Left(x)))))))
END(b(Right1(x))) → B(a(b(b(b(Left(x))))))
END(b(Right1(x))) → A(b(b(b(Left(x)))))
END(b(Right1(x))) → B(b(b(Left(x))))
END(b(Right1(x))) → B(b(Left(x)))
END(b(Right1(x))) → B(Left(x))
END(b(Right1(x))) → LEFT(x)
END(b(b(Right2(x)))) → END(b(a(b(b(b(Left(x)))))))
END(b(b(Right2(x)))) → B(a(b(b(b(Left(x))))))
END(b(b(Right2(x)))) → A(b(b(b(Left(x)))))
END(b(b(Right2(x)))) → B(b(b(Left(x))))
END(b(b(Right2(x)))) → B(b(Left(x)))
END(b(b(Right2(x)))) → B(Left(x))
END(b(b(Right2(x)))) → LEFT(x)
END(b(b(b(Right3(x))))) → END(b(a(b(b(b(Left(x)))))))
END(b(b(b(Right3(x))))) → B(a(b(b(b(Left(x))))))
END(b(b(b(Right3(x))))) → A(b(b(b(Left(x)))))
END(b(b(b(Right3(x))))) → B(b(b(Left(x))))
END(b(b(b(Right3(x))))) → B(b(Left(x)))
END(b(b(b(Right3(x))))) → B(Left(x))
END(b(b(b(Right3(x))))) → LEFT(x)
END(b(Right4(x))) → END(b(b(a(b(b(Left(x)))))))
END(b(Right4(x))) → B(b(a(b(b(Left(x))))))
END(b(Right4(x))) → B(a(b(b(Left(x)))))
END(b(Right4(x))) → A(b(b(Left(x))))
END(b(Right4(x))) → B(b(Left(x)))
END(b(Right4(x))) → B(Left(x))
END(b(Right4(x))) → LEFT(x)
END(a(b(Right5(x)))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(Right5(x)))) → B(b(a(b(b(Left(x))))))
END(a(b(Right5(x)))) → B(a(b(b(Left(x)))))
END(a(b(Right5(x)))) → A(b(b(Left(x))))
END(a(b(Right5(x)))) → B(b(Left(x)))
END(a(b(Right5(x)))) → B(Left(x))
END(a(b(Right5(x)))) → LEFT(x)
END(b(a(b(Right6(x))))) → END(b(b(a(b(b(Left(x)))))))
END(b(a(b(Right6(x))))) → B(b(a(b(b(Left(x))))))
END(b(a(b(Right6(x))))) → B(a(b(b(Left(x)))))
END(b(a(b(Right6(x))))) → A(b(b(Left(x))))
END(b(a(b(Right6(x))))) → B(b(Left(x)))
END(b(a(b(Right6(x))))) → B(Left(x))
END(b(a(b(Right6(x))))) → LEFT(x)
END(b(b(a(b(Right7(x)))))) → END(b(b(a(b(b(Left(x)))))))
END(b(b(a(b(Right7(x)))))) → B(b(a(b(b(Left(x))))))
END(b(b(a(b(Right7(x)))))) → B(a(b(b(Left(x)))))
END(b(b(a(b(Right7(x)))))) → A(b(b(Left(x))))
END(b(b(a(b(Right7(x)))))) → B(b(Left(x)))
END(b(b(a(b(Right7(x)))))) → B(Left(x))
END(b(b(a(b(Right7(x)))))) → LEFT(x)
END(a(b(b(a(b(Right8(x))))))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(b(a(b(Right8(x))))))) → B(b(a(b(b(Left(x))))))
END(a(b(b(a(b(Right8(x))))))) → B(a(b(b(Left(x)))))
END(a(b(b(a(b(Right8(x))))))) → A(b(b(Left(x))))
END(a(b(b(a(b(Right8(x))))))) → B(b(Left(x)))
END(a(b(b(a(b(Right8(x))))))) → B(Left(x))
END(a(b(b(a(b(Right8(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)
B(b(b(b(x)))) → B(a(b(b(b(x)))))
B(b(b(b(x)))) → A(b(b(b(x))))
B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(a(b(b(a(b(x)))))) → B(a(b(b(x))))
B(a(b(b(a(b(x)))))) → A(b(b(x)))
B(a(b(b(a(b(x)))))) → B(b(x))

The TRS R consists of the following rules:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(Wait(x))
End(b(Right1(x))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(Right2(x)))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(b(Right3(x))))) → End(b(a(b(b(b(Left(x)))))))
End(b(Right4(x))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(Right5(x)))) → End(b(b(a(b(b(Left(x)))))))
End(b(a(b(Right6(x))))) → End(b(b(a(b(b(Left(x)))))))
End(b(b(a(b(Right7(x)))))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(b(a(b(Right8(x))))))) → End(b(b(a(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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(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 52 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))
B(a(b(b(a(b(x)))))) → B(a(b(b(x))))
B(a(b(b(a(b(x)))))) → B(b(x))

The TRS R consists of the following rules:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(Wait(x))
End(b(Right1(x))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(Right2(x)))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(b(Right3(x))))) → End(b(a(b(b(b(Left(x)))))))
End(b(Right4(x))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(Right5(x)))) → End(b(b(a(b(b(Left(x)))))))
End(b(a(b(Right6(x))))) → End(b(b(a(b(b(Left(x)))))))
End(b(b(a(b(Right7(x)))))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(b(a(b(Right8(x))))))) → End(b(b(a(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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(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:

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))
B(a(b(b(a(b(x)))))) → B(a(b(b(x))))
B(a(b(b(a(b(x)))))) → B(b(x))

The TRS R consists of the following rules:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(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(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))
b(b(b(b(x)))) → b(a(b(b(b(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))

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(a(b(b(a(b(x)))))) → B(a(b(b(x))))
B(a(b(b(a(b(x)))))) → B(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(B(x1)) = x1   
POL(Begin(x1)) = x1   
POL(Right1(x1)) = 0   
POL(Right2(x1)) = 0   
POL(Right3(x1)) = 0   
POL(Right4(x1)) = 0   
POL(Right5(x1)) = 0   
POL(Right6(x1)) = 1 + x1   
POL(Right7(x1)) = 0   
POL(Right8(x1)) = 1 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = 1 + x1   

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

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(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(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(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))

(11) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))

The TRS R consists of the following rules:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(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(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))
b(b(b(b(x)))) → b(a(b(b(b(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))

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

(12) 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:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(B(x1)) = 2·x1   
POL(Begin(x1)) = 3 + 2·x1   
POL(Right1(x1)) = 2·x1   
POL(Right2(x1)) = 2·x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = 2·x1   
POL(Right5(x1)) = 2·x1   
POL(Right6(x1)) = 2·x1   
POL(Right7(x1)) = 2·x1   
POL(Right8(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   

(13) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))

The TRS R consists of the following rules:

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

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

(14) 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:

b(Right6(x)) → Right6(Ab(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(B(x1)) = x1   
POL(Right1(x1)) = x1   
POL(Right2(x1)) = 2·x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = 2·x1   
POL(Right5(x1)) = 2·x1   
POL(Right6(x1)) = 3 + 2·x1   
POL(Right7(x1)) = x1   
POL(Right8(x1)) = 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 2·x1   

(15) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))

The TRS R consists of the following rules:

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(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))
b(b(b(b(x)))) → b(a(b(b(b(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))

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

(16) 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:

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(Right7(x)) → Right7(Ab(x))
b(Right8(x)) → Right8(Ab(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = 2·x1   
POL(B(x1)) = 2·x1   
POL(Right1(x1)) = 2 + 2·x1   
POL(Right2(x1)) = 2 + x1   
POL(Right3(x1)) = 2 + 2·x1   
POL(Right4(x1)) = 2 + 2·x1   
POL(Right5(x1)) = 2·x1   
POL(Right6(x1)) = 2·x1   
POL(Right7(x1)) = 1 + 2·x1   
POL(Right8(x1)) = 1 + 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 2·x1   

(17) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))

The TRS R consists of the following rules:

b(Right5(x)) → Right5(Ab(x))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))
b(b(b(b(x)))) → b(a(b(b(b(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))

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

(18) 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:

b(Right5(x)) → Right5(Ab(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(x1)) = 1 + 2·x1   
POL(B(x1)) = x1   
POL(Right1(x1)) = 2·x1   
POL(Right2(x1)) = 2·x1   
POL(Right3(x1)) = 2·x1   
POL(Right4(x1)) = 2·x1   
POL(Right5(x1)) = 2 + x1   
POL(Right6(x1)) = x1   
POL(Right7(x1)) = x1   
POL(Right8(x1)) = 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = 2·x1   

(19) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))
B(b(b(b(x)))) → B(a(b(b(b(x)))))

The TRS R consists of the following rules:

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

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

(20) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


B(b(b(b(x)))) → B(a(b(b(b(x)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(B(x1)) = 0A +
[0A,-I,0A]
·x1

POL(a(x1)) =
/0A\
|0A|
\0A/
 +
/-I0A-I\
|0A0A1A|
\-I1A0A/
·x1

POL(b(x1)) =
/0A\
|0A|
\0A/
 +
/1A-I0A\
|-I-I0A|
\0A0A-I/
·x1

POL(Right1(x1)) =
/0A\
|-I|
\0A/
 +
/-I-I-I\
|-I0A-I|
\0A0A-I/
·x1

POL(Aa(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I-I\
|0A0A-I|
\0A-I-I/
·x1

POL(Right2(x1)) =
/0A\
|-I|
\0A/
 +
/0A-I-I\
|0A-I-I|
\-I-I-I/
·x1

POL(Right3(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|0A-I-I|
\0A0A0A/
·x1

POL(Right4(x1)) =
/0A\
|-I|
\1A/
 +
/-I-I0A\
|0A-I0A|
\1A0A-I/
·x1

POL(Right5(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I-I\
|0A-I-I|
\0A-I-I/
·x1

POL(Right6(x1)) =
/1A\
|1A|
\-I/
 +
/-I0A0A\
|0A1A-I|
\0A0A0A/
·x1

POL(Right7(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1

POL(Right8(x1)) =
/0A\
|-I|
\-I/
 +
/0A-I-I\
|0A-I0A|
\0A-I0A/
·x1

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

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

(21) Obligation:

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

B(a(b(b(a(b(x)))))) → B(b(a(b(b(x)))))

The TRS R consists of the following rules:

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

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

(22) DependencyGraphProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

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

The TRS R consists of the following rules:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(Wait(x))
End(b(Right1(x))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(Right2(x)))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(b(Right3(x))))) → End(b(a(b(b(b(Left(x)))))))
End(b(Right4(x))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(Right5(x)))) → End(b(b(a(b(b(Left(x)))))))
End(b(a(b(Right6(x))))) → End(b(b(a(b(b(Left(x)))))))
End(b(b(a(b(Right7(x)))))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(b(a(b(Right8(x))))))) → End(b(b(a(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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))

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

(25) 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.

(26) Obligation:

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

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

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

(27) 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

(28) YES

(29) Obligation:

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

END(b(b(Right2(x)))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right1(x))) → END(b(a(b(b(b(Left(x)))))))
END(b(b(b(Right3(x))))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right4(x))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(Right5(x)))) → END(b(b(a(b(b(Left(x)))))))
END(b(a(b(Right6(x))))) → END(b(b(a(b(b(Left(x)))))))
END(b(b(a(b(Right7(x)))))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(b(a(b(Right8(x))))))) → END(b(b(a(b(b(Left(x)))))))

The TRS R consists of the following rules:

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(Wait(x))
End(b(Right1(x))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(Right2(x)))) → End(b(a(b(b(b(Left(x)))))))
End(b(b(b(Right3(x))))) → End(b(a(b(b(b(Left(x)))))))
End(b(Right4(x))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(Right5(x)))) → End(b(b(a(b(b(Left(x)))))))
End(b(a(b(Right6(x))))) → End(b(b(a(b(b(Left(x)))))))
End(b(b(a(b(Right7(x)))))) → End(b(b(a(b(b(Left(x)))))))
End(a(b(b(a(b(Right8(x))))))) → End(b(b(a(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))
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))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))

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

(30) 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.

(31) Obligation:

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

END(b(b(Right2(x)))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right1(x))) → END(b(a(b(b(b(Left(x)))))))
END(b(b(b(Right3(x))))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right4(x))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(Right5(x)))) → END(b(b(a(b(b(Left(x)))))))
END(b(a(b(Right6(x))))) → END(b(b(a(b(b(Left(x)))))))
END(b(b(a(b(Right7(x)))))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(b(a(b(Right8(x))))))) → END(b(b(a(b(b(Left(x)))))))

The TRS R consists of the following rules:

Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(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(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))
b(b(b(b(x)))) → b(a(b(b(b(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))

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(a(b(Right5(x)))) → END(b(b(a(b(b(Left(x)))))))
END(a(b(b(a(b(Right8(x))))))) → END(b(b(a(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(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(Right7(x1)) = 0   
POL(Right8(x1)) = 0   
POL(Wait(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1   

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

b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(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(b(b(b(x)))) → b(a(b(b(b(x)))))
b(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))

(33) Obligation:

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

END(b(b(Right2(x)))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right1(x))) → END(b(a(b(b(b(Left(x)))))))
END(b(b(b(Right3(x))))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right4(x))) → END(b(b(a(b(b(Left(x)))))))
END(b(a(b(Right6(x))))) → END(b(b(a(b(b(Left(x)))))))
END(b(b(a(b(Right7(x)))))) → END(b(b(a(b(b(Left(x)))))))

The TRS R consists of the following rules:

Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Wait(x)) → Begin(x)
b(b(b(Begin(x)))) → Right1(Wait(x))
b(b(Begin(x))) → Right2(Wait(x))
b(Begin(x)) → Right3(Wait(x))
b(a(b(b(a(Begin(x)))))) → Right4(Wait(x))
b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(a(b(Begin(x)))) → Right6(Wait(x))
b(a(Begin(x))) → Right7(Wait(x))
b(Begin(x)) → Right8(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(a(b(b(a(b(x)))))) → b(b(a(b(b(x)))))
b(b(b(b(x)))) → b(a(b(b(b(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))

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

(34) 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:

b(a(b(b(Begin(x))))) → Right5(Wait(x))
b(Begin(x)) → Right8(Wait(x))

Used ordering: Polynomial interpretation [POLO]:

POL(Aa(x1)) = x1   
POL(Ab(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(Right2(x1)) = 1 + 2·x1   
POL(Right3(x1)) = 1 + 2·x1   
POL(Right4(x1)) = 1 + 2·x1   
POL(Right5(x1)) = 2·x1   
POL(Right6(x1)) = 1 + 2·x1   
POL(Right7(x1)) = 1 + 2·x1   
POL(Right8(x1)) = x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   

(35) Obligation:

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

END(b(b(Right2(x)))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right1(x))) → END(b(a(b(b(b(Left(x)))))))
END(b(b(b(Right3(x))))) → END(b(a(b(b(b(Left(x)))))))
END(b(Right4(x))) → END(b(b(a(b(b(Left(x)))))))
END(b(a(b(Right6(x))))) → END(b(b(a(b(b(Left(x)))))))
END(b(b(a(b(Right7(x)))))) → END(b(b(a(b(b(Left(x)))))))

The TRS R consists of the following rules:

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

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

(36) 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(a(b(b(b(Left(Aa(Ab(Wait(x)))))))))) evaluates to t =END(b(a(b(b(b(Left(Aa(Ab(Wait(x))))))))))

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



Rewriting sequence

END(b(a(b(b(b(Left(Aa(Ab(Wait(x)))))))))) → END(b(a(b(b(b(a(Left(Ab(Wait(x))))))))))
with rule Left(Aa(x')) → a(Left(x')) at position [0,0,0,0,0,0] and matcher [x' / Ab(Wait(x))]

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

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

END(b(a(b(b(b(a(b(Begin(x))))))))) → END(b(a(b(b(Right6(Wait(x)))))))
with rule b(a(b(Begin(x')))) → Right6(Wait(x')) at position [0,0,0,0,0] and matcher [x' / x]

END(b(a(b(b(Right6(Wait(x))))))) → END(b(a(b(Right6(Ab(Wait(x)))))))
with rule b(Right6(x')) → Right6(Ab(x')) at position [0,0,0,0] and matcher [x' / Wait(x)]

END(b(a(b(Right6(Ab(Wait(x))))))) → END(b(b(a(b(b(Left(Ab(Wait(x)))))))))
with rule END(b(a(b(Right6(x'))))) → END(b(b(a(b(b(Left(x'))))))) at position [] and matcher [x' / Ab(Wait(x))]

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

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

END(b(b(a(b(b(b(Begin(x)))))))) → END(b(b(a(b(Right2(Wait(x)))))))
with rule b(b(Begin(x'))) → Right2(Wait(x')) at position [0,0,0,0,0] and matcher [x' / x]

END(b(b(a(b(Right2(Wait(x))))))) → END(b(b(a(Right2(Ab(Wait(x)))))))
with rule b(Right2(x')) → Right2(Ab(x')) at position [0,0,0,0] and matcher [x' / Wait(x)]

END(b(b(a(Right2(Ab(Wait(x))))))) → END(b(b(Right2(Aa(Ab(Wait(x)))))))
with rule a(Right2(x')) → Right2(Aa(x')) at position [0,0,0] and matcher [x' / Ab(Wait(x))]

END(b(b(Right2(Aa(Ab(Wait(x))))))) → END(b(a(b(b(b(Left(Aa(Ab(Wait(x))))))))))
with rule END(b(b(Right2(x)))) → END(b(a(b(b(b(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.



(37) NO