NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Gebhardt_06/10.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 (⇔, 72 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 7 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 31 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 38 ms)
        ↳13 QDP
        ↳14 MRRProof (⇔, 22 ms)
        ↳15 QDP
        ↳16 MRRProof (⇔, 28 ms)
        ↳17 QDP
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 9 ms)
        ↳25 QDP
        ↳26 NonTerminationLoopProof (⇒, 2966 ms)
        ↳27 NO

(0) Obligation:

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

Begin(0(0(0(x)))) → Wait(Right1(x))
Begin(0(0(x))) → Wait(Right2(x))
Begin(0(x)) → Wait(Right3(x))
Begin(0(0(1(x)))) → Wait(Right4(x))
Begin(0(1(x))) → Wait(Right5(x))
Begin(1(x)) → Wait(Right6(x))
Right1(0(End(x))) → Left(0(1(1(1(End(x))))))
Right2(0(0(End(x)))) → Left(0(1(1(1(End(x))))))
Right3(0(0(0(End(x))))) → Left(0(1(1(1(End(x))))))
Right4(1(End(x))) → Left(0(0(1(0(End(x))))))
Right5(1(0(End(x)))) → Left(0(0(1(0(End(x))))))
Right6(1(0(0(End(x))))) → Left(0(0(1(0(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
Wait(Left(x)) → Begin(x)
0(0(0(0(x)))) → 0(1(1(1(x))))
1(0(0(1(x)))) → 0(0(1(0(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:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
End(0(Right1(x))) → End(1(1(1(0(Left(x))))))
End(0(0(Right2(x)))) → End(1(1(1(0(Left(x))))))
End(0(0(0(Right3(x))))) → End(1(1(1(0(Left(x))))))
End(1(Right4(x))) → End(0(1(0(0(Left(x))))))
End(0(1(Right5(x)))) → End(0(1(0(0(Left(x))))))
End(0(0(1(Right6(x))))) → End(0(1(0(0(Left(x))))))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))
Left(A0(x)) → 0(Left(x))
Left(A1(x)) → 1(Left(x))
Left(Wait(x)) → Begin(x)
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(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(0(Right1(x))) → END(1(1(1(0(Left(x))))))
END(0(Right1(x))) → 11(1(1(0(Left(x)))))
END(0(Right1(x))) → 11(1(0(Left(x))))
END(0(Right1(x))) → 11(0(Left(x)))
END(0(Right1(x))) → 01(Left(x))
END(0(Right1(x))) → LEFT(x)
END(0(0(Right2(x)))) → END(1(1(1(0(Left(x))))))
END(0(0(Right2(x)))) → 11(1(1(0(Left(x)))))
END(0(0(Right2(x)))) → 11(1(0(Left(x))))
END(0(0(Right2(x)))) → 11(0(Left(x)))
END(0(0(Right2(x)))) → 01(Left(x))
END(0(0(Right2(x)))) → LEFT(x)
END(0(0(0(Right3(x))))) → END(1(1(1(0(Left(x))))))
END(0(0(0(Right3(x))))) → 11(1(1(0(Left(x)))))
END(0(0(0(Right3(x))))) → 11(1(0(Left(x))))
END(0(0(0(Right3(x))))) → 11(0(Left(x)))
END(0(0(0(Right3(x))))) → 01(Left(x))
END(0(0(0(Right3(x))))) → LEFT(x)
END(1(Right4(x))) → END(0(1(0(0(Left(x))))))
END(1(Right4(x))) → 01(1(0(0(Left(x)))))
END(1(Right4(x))) → 11(0(0(Left(x))))
END(1(Right4(x))) → 01(0(Left(x)))
END(1(Right4(x))) → 01(Left(x))
END(1(Right4(x))) → LEFT(x)
END(0(1(Right5(x)))) → END(0(1(0(0(Left(x))))))
END(0(1(Right5(x)))) → 01(1(0(0(Left(x)))))
END(0(1(Right5(x)))) → 11(0(0(Left(x))))
END(0(1(Right5(x)))) → 01(0(Left(x)))
END(0(1(Right5(x)))) → 01(Left(x))
END(0(1(Right5(x)))) → LEFT(x)
END(0(0(1(Right6(x))))) → END(0(1(0(0(Left(x))))))
END(0(0(1(Right6(x))))) → 01(1(0(0(Left(x)))))
END(0(0(1(Right6(x))))) → 11(0(0(Left(x))))
END(0(0(1(Right6(x))))) → 01(0(Left(x)))
END(0(0(1(Right6(x))))) → 01(Left(x))
END(0(0(1(Right6(x))))) → LEFT(x)
LEFT(A0(x)) → 01(Left(x))
LEFT(A0(x)) → LEFT(x)
LEFT(A1(x)) → 11(Left(x))
LEFT(A1(x)) → LEFT(x)
01(0(0(0(x)))) → 11(1(1(0(x))))
01(0(0(0(x)))) → 11(1(0(x)))
01(0(0(0(x)))) → 11(0(x))
11(0(0(1(x)))) → 01(1(0(0(x))))
11(0(0(1(x)))) → 11(0(0(x)))
11(0(0(1(x)))) → 01(0(x))
11(0(0(1(x)))) → 01(x)

The TRS R consists of the following rules:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
End(0(Right1(x))) → End(1(1(1(0(Left(x))))))
End(0(0(Right2(x)))) → End(1(1(1(0(Left(x))))))
End(0(0(0(Right3(x))))) → End(1(1(1(0(Left(x))))))
End(1(Right4(x))) → End(0(1(0(0(Left(x))))))
End(0(1(Right5(x)))) → End(0(1(0(0(Left(x))))))
End(0(0(1(Right6(x))))) → End(0(1(0(0(Left(x))))))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))
Left(A0(x)) → 0(Left(x))
Left(A1(x)) → 1(Left(x))
Left(Wait(x)) → Begin(x)
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(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 32 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

11(0(0(1(x)))) → 01(1(0(0(x))))
01(0(0(0(x)))) → 11(1(1(0(x))))
11(0(0(1(x)))) → 11(0(0(x)))
11(0(0(1(x)))) → 01(0(x))
01(0(0(0(x)))) → 11(1(0(x)))
11(0(0(1(x)))) → 01(x)
01(0(0(0(x)))) → 11(0(x))

The TRS R consists of the following rules:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
End(0(Right1(x))) → End(1(1(1(0(Left(x))))))
End(0(0(Right2(x)))) → End(1(1(1(0(Left(x))))))
End(0(0(0(Right3(x))))) → End(1(1(1(0(Left(x))))))
End(1(Right4(x))) → End(0(1(0(0(Left(x))))))
End(0(1(Right5(x)))) → End(0(1(0(0(Left(x))))))
End(0(0(1(Right6(x))))) → End(0(1(0(0(Left(x))))))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))
Left(A0(x)) → 0(Left(x))
Left(A1(x)) → 1(Left(x))
Left(Wait(x)) → Begin(x)
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(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:

11(0(0(1(x)))) → 01(1(0(0(x))))
01(0(0(0(x)))) → 11(1(1(0(x))))
11(0(0(1(x)))) → 11(0(0(x)))
11(0(0(1(x)))) → 01(0(x))
01(0(0(0(x)))) → 11(1(0(x)))
11(0(0(1(x)))) → 01(x)
01(0(0(0(x)))) → 11(0(x))

The TRS R consists of the following rules:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(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.


11(0(0(1(x)))) → 11(0(0(x)))
11(0(0(1(x)))) → 01(0(x))
01(0(0(0(x)))) → 11(1(0(x)))
11(0(0(1(x)))) → 01(x)
01(0(0(0(x)))) → 11(0(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 1 + x1   
POL(01(x1)) = 1 + x1   
POL(1(x1)) = 1 + x1   
POL(11(x1)) = 1 + x1   
POL(A0(x1)) = x1   
POL(A1(x1)) = x1   
POL(Begin(x1)) = x1   
POL(Right1(x1)) = 0   
POL(Right2(x1)) = 0   
POL(Right3(x1)) = 1   
POL(Right4(x1)) = 0   
POL(Right5(x1)) = 0   
POL(Right6(x1)) = 0   
POL(Wait(x1)) = x1   

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

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))

(11) Obligation:

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

11(0(0(1(x)))) → 01(1(0(0(x))))
01(0(0(0(x)))) → 11(1(1(0(x))))

The TRS R consists of the following rules:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(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:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
0(Right1(x)) → Right1(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right6(x)) → Right6(A0(x))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right6(x)) → Right6(A1(x))

Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 2·x1   
POL(01(x1)) = 2·x1   
POL(1(x1)) = 2·x1   
POL(11(x1)) = 2·x1   
POL(A0(x1)) = x1   
POL(A1(x1)) = 2·x1   
POL(Begin(x1)) = 3 + x1   
POL(Right1(x1)) = 1 + 2·x1   
POL(Right2(x1)) = 2·x1   
POL(Right3(x1)) = 1 + x1   
POL(Right4(x1)) = x1   
POL(Right5(x1)) = x1   
POL(Right6(x1)) = 1 + x1   
POL(Wait(x1)) = 2·x1   

(13) Obligation:

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

11(0(0(1(x)))) → 01(1(0(0(x))))
01(0(0(0(x)))) → 11(1(1(0(x))))

The TRS R consists of the following rules:

0(Right2(x)) → Right2(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(Right2(x)) → Right2(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(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:

0(Right2(x)) → Right2(A0(x))
0(Right4(x)) → Right4(A0(x))
1(Right2(x)) → Right2(A1(x))
1(Right4(x)) → Right4(A1(x))

Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 2·x1   
POL(01(x1)) = 2·x1   
POL(1(x1)) = 2·x1   
POL(11(x1)) = 2·x1   
POL(A0(x1)) = x1   
POL(A1(x1)) = x1   
POL(Right2(x1)) = 1 + 2·x1   
POL(Right4(x1)) = 1 + 2·x1   
POL(Right5(x1)) = 2·x1   

(15) Obligation:

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

11(0(0(1(x)))) → 01(1(0(0(x))))
01(0(0(0(x)))) → 11(1(1(0(x))))

The TRS R consists of the following rules:

0(Right5(x)) → Right5(A0(x))
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(Right5(x)) → Right5(A1(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:

0(Right5(x)) → Right5(A0(x))
1(Right5(x)) → Right5(A1(x))

Used ordering: Polynomial interpretation [POLO]:

POL(0(x1)) = 2·x1   
POL(01(x1)) = 2·x1   
POL(1(x1)) = 2·x1   
POL(11(x1)) = 2·x1   
POL(A0(x1)) = 2·x1   
POL(A1(x1)) = 1 + x1   
POL(Right5(x1)) = 3 + x1   

(17) Obligation:

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

11(0(0(1(x)))) → 01(1(0(0(x))))
01(0(0(0(x)))) → 11(1(1(0(x))))

The TRS R consists of the following rules:

0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))

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

(18) Obligation:

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

LEFT(A1(x)) → LEFT(x)
LEFT(A0(x)) → LEFT(x)

The TRS R consists of the following rules:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
End(0(Right1(x))) → End(1(1(1(0(Left(x))))))
End(0(0(Right2(x)))) → End(1(1(1(0(Left(x))))))
End(0(0(0(Right3(x))))) → End(1(1(1(0(Left(x))))))
End(1(Right4(x))) → End(0(1(0(0(Left(x))))))
End(0(1(Right5(x)))) → End(0(1(0(0(Left(x))))))
End(0(0(1(Right6(x))))) → End(0(1(0(0(Left(x))))))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))
Left(A0(x)) → 0(Left(x))
Left(A1(x)) → 1(Left(x))
Left(Wait(x)) → Begin(x)
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))

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

(19) 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.

(20) Obligation:

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

LEFT(A1(x)) → LEFT(x)
LEFT(A0(x)) → LEFT(x)

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

(21) 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(A1(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(A0(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

(22) YES

(23) Obligation:

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

END(0(0(Right2(x)))) → END(1(1(1(0(Left(x))))))
END(0(Right1(x))) → END(1(1(1(0(Left(x))))))
END(0(0(0(Right3(x))))) → END(1(1(1(0(Left(x))))))
END(1(Right4(x))) → END(0(1(0(0(Left(x))))))
END(0(1(Right5(x)))) → END(0(1(0(0(Left(x))))))
END(0(0(1(Right6(x))))) → END(0(1(0(0(Left(x))))))

The TRS R consists of the following rules:

0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
End(0(Right1(x))) → End(1(1(1(0(Left(x))))))
End(0(0(Right2(x)))) → End(1(1(1(0(Left(x))))))
End(0(0(0(Right3(x))))) → End(1(1(1(0(Left(x))))))
End(1(Right4(x))) → End(0(1(0(0(Left(x))))))
End(0(1(Right5(x)))) → End(0(1(0(0(Left(x))))))
End(0(0(1(Right6(x))))) → End(0(1(0(0(Left(x))))))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))
Left(A0(x)) → 0(Left(x))
Left(A1(x)) → 1(Left(x))
Left(Wait(x)) → Begin(x)
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))

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

(24) 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.

(25) Obligation:

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

END(0(0(Right2(x)))) → END(1(1(1(0(Left(x))))))
END(0(Right1(x))) → END(1(1(1(0(Left(x))))))
END(0(0(0(Right3(x))))) → END(1(1(1(0(Left(x))))))
END(1(Right4(x))) → END(0(1(0(0(Left(x))))))
END(0(1(Right5(x)))) → END(0(1(0(0(Left(x))))))
END(0(0(1(Right6(x))))) → END(0(1(0(0(Left(x))))))

The TRS R consists of the following rules:

Left(A0(x)) → 0(Left(x))
Left(A1(x)) → 1(Left(x))
Left(Wait(x)) → Begin(x)
0(0(0(Begin(x)))) → Right1(Wait(x))
0(0(Begin(x))) → Right2(Wait(x))
0(Begin(x)) → Right3(Wait(x))
0(Right1(x)) → Right1(A0(x))
0(Right2(x)) → Right2(A0(x))
0(Right3(x)) → Right3(A0(x))
0(Right4(x)) → Right4(A0(x))
0(Right5(x)) → Right5(A0(x))
0(Right6(x)) → Right6(A0(x))
0(0(0(0(x)))) → 1(1(1(0(x))))
1(0(0(1(x)))) → 0(1(0(0(x))))
1(0(0(Begin(x)))) → Right4(Wait(x))
1(0(Begin(x))) → Right5(Wait(x))
1(Begin(x)) → Right6(Wait(x))
1(Right1(x)) → Right1(A1(x))
1(Right2(x)) → Right2(A1(x))
1(Right3(x)) → Right3(A1(x))
1(Right4(x)) → Right4(A1(x))
1(Right5(x)) → Right5(A1(x))
1(Right6(x)) → Right6(A1(x))

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

(26) 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(1(1(1(0(Left(A0(A1(Wait(x))))))))) evaluates to t =END(1(1(1(0(Left(A0(A1(Wait(x)))))))))

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



Rewriting sequence

END(1(1(1(0(Left(A0(A1(Wait(x))))))))) → END(1(1(1(0(0(Left(A1(Wait(x)))))))))
with rule Left(A0(x')) → 0(Left(x')) at position [0,0,0,0,0] and matcher [x' / A1(Wait(x))]

END(1(1(1(0(0(Left(A1(Wait(x))))))))) → END(1(1(1(0(0(1(Left(Wait(x)))))))))
with rule Left(A1(x')) → 1(Left(x')) at position [0,0,0,0,0,0] and matcher [x' / Wait(x)]

END(1(1(1(0(0(1(Left(Wait(x))))))))) → END(1(1(1(0(0(1(Begin(x))))))))
with rule Left(Wait(x')) → Begin(x') at position [0,0,0,0,0,0,0] and matcher [x' / x]

END(1(1(1(0(0(1(Begin(x)))))))) → END(1(1(0(1(0(0(Begin(x))))))))
with rule 1(0(0(1(x')))) → 0(1(0(0(x')))) at position [0,0,0] and matcher [x' / Begin(x)]

END(1(1(0(1(0(0(Begin(x)))))))) → END(1(1(0(Right4(Wait(x))))))
with rule 1(0(0(Begin(x')))) → Right4(Wait(x')) at position [0,0,0,0] and matcher [x' / x]

END(1(1(0(Right4(Wait(x)))))) → END(1(1(Right4(A0(Wait(x))))))
with rule 0(Right4(x')) → Right4(A0(x')) at position [0,0,0] and matcher [x' / Wait(x)]

END(1(1(Right4(A0(Wait(x)))))) → END(1(Right4(A1(A0(Wait(x))))))
with rule 1(Right4(x')) → Right4(A1(x')) at position [0,0] and matcher [x' / A0(Wait(x))]

END(1(Right4(A1(A0(Wait(x)))))) → END(0(1(0(0(Left(A1(A0(Wait(x)))))))))
with rule END(1(Right4(x'))) → END(0(1(0(0(Left(x')))))) at position [] and matcher [x' / A1(A0(Wait(x)))]

END(0(1(0(0(Left(A1(A0(Wait(x))))))))) → END(0(1(0(0(1(Left(A0(Wait(x)))))))))
with rule Left(A1(x')) → 1(Left(x')) at position [0,0,0,0,0] and matcher [x' / A0(Wait(x))]

END(0(1(0(0(1(Left(A0(Wait(x))))))))) → END(0(1(0(0(1(0(Left(Wait(x)))))))))
with rule Left(A0(x')) → 0(Left(x')) at position [0,0,0,0,0,0] and matcher [x' / Wait(x)]

END(0(1(0(0(1(0(Left(Wait(x))))))))) → END(0(1(0(0(1(0(Begin(x))))))))
with rule Left(Wait(x')) → Begin(x') at position [0,0,0,0,0,0,0] and matcher [x' / x]

END(0(1(0(0(1(0(Begin(x)))))))) → END(0(0(1(0(0(0(Begin(x))))))))
with rule 1(0(0(1(x')))) → 0(1(0(0(x')))) at position [0,0] and matcher [x' / 0(Begin(x))]

END(0(0(1(0(0(0(Begin(x)))))))) → END(0(0(1(Right1(Wait(x))))))
with rule 0(0(0(Begin(x')))) → Right1(Wait(x')) at position [0,0,0,0] and matcher [x' / x]

END(0(0(1(Right1(Wait(x)))))) → END(0(0(Right1(A1(Wait(x))))))
with rule 1(Right1(x')) → Right1(A1(x')) at position [0,0,0] and matcher [x' / Wait(x)]

END(0(0(Right1(A1(Wait(x)))))) → END(0(Right1(A0(A1(Wait(x))))))
with rule 0(Right1(x')) → Right1(A0(x')) at position [0,0] and matcher [x' / A1(Wait(x))]

END(0(Right1(A0(A1(Wait(x)))))) → END(1(1(1(0(Left(A0(A1(Wait(x)))))))))
with rule END(0(Right1(x))) → END(1(1(1(0(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.



(27) NO