NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Gebhardt_06/14.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 30 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPOrderProof (⇔, 74 ms)
        ↳9 QDP
        ↳10 DependencyGraphProof (⇔, 0 ms)
        ↳11 QDP
    ↳12 QDP
        ↳13 UsableRulesProof (⇔, 0 ms)
        ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔, 0 ms)
        ↳16 YES
    ↳17 QDP
        ↳18 UsableRulesProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 QDPSizeChangeProof (⇔, 0 ms)
        ↳21 YES
    ↳22 QDP
        ↳23 UsableRulesProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 QDPSizeChangeProof (⇔, 0 ms)
        ↳26 YES
    ↳27 QDP
        ↳28 UsableRulesProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 QDPSizeChangeProof (⇔, 0 ms)
        ↳31 YES
    ↳32 QDP
        ↳33 UsableRulesProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 QDPSizeChangeProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QDPSizeChangeProof (⇔, 0 ms)
        ↳41 YES
    ↳42 QDP
        ↳43 UsableRulesProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 NonTerminationLoopProof (⇒, 150 ms)
        ↳46 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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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

(2) Obligation:

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

BEGIN(0(0(0(x)))) → WAIT(Right1(x))
BEGIN(0(0(0(x)))) → RIGHT1(x)
BEGIN(0(0(x))) → WAIT(Right2(x))
BEGIN(0(0(x))) → RIGHT2(x)
BEGIN(0(x)) → WAIT(Right3(x))
BEGIN(0(x)) → RIGHT3(x)
BEGIN(1(0(1(x)))) → WAIT(Right4(x))
BEGIN(1(0(1(x)))) → RIGHT4(x)
BEGIN(0(1(x))) → WAIT(Right5(x))
BEGIN(0(1(x))) → RIGHT5(x)
BEGIN(1(x)) → WAIT(Right6(x))
BEGIN(1(x)) → RIGHT6(x)
RIGHT1(0(End(x))) → 01(1(1(1(End(x)))))
RIGHT1(0(End(x))) → 11(1(1(End(x))))
RIGHT1(0(End(x))) → 11(1(End(x)))
RIGHT1(0(End(x))) → 11(End(x))
RIGHT2(0(0(End(x)))) → 01(1(1(1(End(x)))))
RIGHT2(0(0(End(x)))) → 11(1(1(End(x))))
RIGHT2(0(0(End(x)))) → 11(1(End(x)))
RIGHT2(0(0(End(x)))) → 11(End(x))
RIGHT3(0(0(0(End(x))))) → 01(1(1(1(End(x)))))
RIGHT3(0(0(0(End(x))))) → 11(1(1(End(x))))
RIGHT3(0(0(0(End(x))))) → 11(1(End(x)))
RIGHT3(0(0(0(End(x))))) → 11(End(x))
RIGHT4(1(End(x))) → 01(0(0(0(End(x)))))
RIGHT4(1(End(x))) → 01(0(0(End(x))))
RIGHT4(1(End(x))) → 01(0(End(x)))
RIGHT4(1(End(x))) → 01(End(x))
RIGHT5(1(1(End(x)))) → 01(0(0(0(End(x)))))
RIGHT5(1(1(End(x)))) → 01(0(0(End(x))))
RIGHT5(1(1(End(x)))) → 01(0(End(x)))
RIGHT5(1(1(End(x)))) → 01(End(x))
RIGHT6(1(1(0(End(x))))) → 01(0(0(0(End(x)))))
RIGHT6(1(1(0(End(x))))) → 01(0(0(End(x))))
RIGHT6(1(1(0(End(x))))) → 01(0(End(x)))
RIGHT1(0(x)) → A01(Right1(x))
RIGHT1(0(x)) → RIGHT1(x)
RIGHT2(0(x)) → A01(Right2(x))
RIGHT2(0(x)) → RIGHT2(x)
RIGHT3(0(x)) → A01(Right3(x))
RIGHT3(0(x)) → RIGHT3(x)
RIGHT4(0(x)) → A01(Right4(x))
RIGHT4(0(x)) → RIGHT4(x)
RIGHT5(0(x)) → A01(Right5(x))
RIGHT5(0(x)) → RIGHT5(x)
RIGHT6(0(x)) → A01(Right6(x))
RIGHT6(0(x)) → RIGHT6(x)
RIGHT1(1(x)) → A11(Right1(x))
RIGHT1(1(x)) → RIGHT1(x)
RIGHT2(1(x)) → A11(Right2(x))
RIGHT2(1(x)) → RIGHT2(x)
RIGHT3(1(x)) → A11(Right3(x))
RIGHT3(1(x)) → RIGHT3(x)
RIGHT4(1(x)) → A11(Right4(x))
RIGHT4(1(x)) → RIGHT4(x)
RIGHT5(1(x)) → A11(Right5(x))
RIGHT5(1(x)) → RIGHT5(x)
RIGHT6(1(x)) → A11(Right6(x))
RIGHT6(1(x)) → RIGHT6(x)
A01(Left(x)) → 01(x)
A11(Left(x)) → 11(x)
WAIT(Left(x)) → BEGIN(x)
01(0(0(0(x)))) → 01(1(1(1(x))))
01(0(0(0(x)))) → 11(1(1(x)))
01(0(0(0(x)))) → 11(1(x))
01(0(0(0(x)))) → 11(x)
11(1(0(1(x)))) → 01(0(0(0(x))))
11(1(0(1(x)))) → 01(0(0(x)))
11(1(0(1(x)))) → 01(0(x))
11(1(0(1(x)))) → 01(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 8 SCCs with 43 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(6) UsableRulesProof (EQUIVALENT transformation)

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

(7) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(8) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL(0(x1)) = 1 + x1   
POL(01(x1)) = x1   
POL(1(x1)) = 1 + x1   
POL(11(x1)) = x1   

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

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

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) DependencyGraphProof (EQUIVALENT transformation)

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

(11) Obligation:

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

01(0(0(0(x)))) → 01(1(1(1(x))))

The TRS R consists of the following rules:

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

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

(12) Obligation:

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

RIGHT6(1(x)) → RIGHT6(x)
RIGHT6(0(x)) → RIGHT6(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(13) 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.

(14) Obligation:

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

RIGHT6(1(x)) → RIGHT6(x)
RIGHT6(0(x)) → RIGHT6(x)

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • RIGHT6(1(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

  • RIGHT6(0(x)) → RIGHT6(x)
    The graph contains the following edges 1 > 1

(16) YES

(17) Obligation:

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

RIGHT5(1(x)) → RIGHT5(x)
RIGHT5(0(x)) → RIGHT5(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(18) 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.

(19) Obligation:

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

RIGHT5(1(x)) → RIGHT5(x)
RIGHT5(0(x)) → RIGHT5(x)

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

(20) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • RIGHT5(1(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

  • RIGHT5(0(x)) → RIGHT5(x)
    The graph contains the following edges 1 > 1

(21) YES

(22) Obligation:

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

RIGHT4(1(x)) → RIGHT4(x)
RIGHT4(0(x)) → RIGHT4(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(23) 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.

(24) Obligation:

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

RIGHT4(1(x)) → RIGHT4(x)
RIGHT4(0(x)) → RIGHT4(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:

  • RIGHT4(1(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

  • RIGHT4(0(x)) → RIGHT4(x)
    The graph contains the following edges 1 > 1

(26) YES

(27) Obligation:

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

RIGHT3(1(x)) → RIGHT3(x)
RIGHT3(0(x)) → RIGHT3(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(28) 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.

(29) Obligation:

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

RIGHT3(1(x)) → RIGHT3(x)
RIGHT3(0(x)) → RIGHT3(x)

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

(30) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • RIGHT3(1(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

  • RIGHT3(0(x)) → RIGHT3(x)
    The graph contains the following edges 1 > 1

(31) YES

(32) Obligation:

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

RIGHT2(1(x)) → RIGHT2(x)
RIGHT2(0(x)) → RIGHT2(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(33) 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.

(34) Obligation:

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

RIGHT2(1(x)) → RIGHT2(x)
RIGHT2(0(x)) → RIGHT2(x)

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

(35) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • RIGHT2(1(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

  • RIGHT2(0(x)) → RIGHT2(x)
    The graph contains the following edges 1 > 1

(36) YES

(37) Obligation:

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

RIGHT1(1(x)) → RIGHT1(x)
RIGHT1(0(x)) → RIGHT1(x)

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(38) 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.

(39) Obligation:

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

RIGHT1(1(x)) → RIGHT1(x)
RIGHT1(0(x)) → RIGHT1(x)

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

(40) QDPSizeChangeProof (EQUIVALENT transformation)

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

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

  • RIGHT1(1(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

  • RIGHT1(0(x)) → RIGHT1(x)
    The graph contains the following edges 1 > 1

(41) YES

(42) Obligation:

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

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

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(1(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(0(0(End(x))))))
Right5(1(1(End(x)))) → Left(0(0(0(0(End(x))))))
Right6(1(1(0(End(x))))) → Left(0(0(0(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(1(0(1(x)))) → 0(0(0(0(x))))

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

(43) 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.

(44) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(45) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = BEGIN(0(0(0(0(End(x')))))) evaluates to t =BEGIN(0(0(0(0(End(x'))))))

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



Rewriting sequence

BEGIN(0(0(0(0(End(x')))))) → BEGIN(0(1(1(1(End(x'))))))
with rule 0(0(0(0(x)))) → 0(1(1(1(x)))) at position [0] and matcher [x / End(x')]

BEGIN(0(1(1(1(End(x')))))) → WAIT(Right5(1(1(End(x')))))
with rule BEGIN(0(1(x))) → WAIT(Right5(x)) at position [] and matcher [x / 1(1(End(x')))]

WAIT(Right5(1(1(End(x'))))) → WAIT(Left(0(0(0(0(End(x')))))))
with rule Right5(1(1(End(x'')))) → Left(0(0(0(0(End(x'')))))) at position [0] and matcher [x'' / x']

WAIT(Left(0(0(0(0(End(x'))))))) → BEGIN(0(0(0(0(End(x'))))))
with rule WAIT(Left(x)) → BEGIN(x)

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(46) NO