NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/loop2.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 5 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 NonTerminationLoopProof (⇒, 489 ms)
        ↳9 NO
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
        ↳13 QDPSizeChangeProof (⇔, 0 ms)
        ↳14 YES
    ↳15 QDP
        ↳16 UsableRulesProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔, 0 ms)
        ↳19 YES
    ↳20 QDP
        ↳21 UsableRulesProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 QDPSizeChangeProof (⇔, 0 ms)
        ↳24 YES
    ↳25 QDP
        ↳26 UsableRulesProof (⇔, 0 ms)
        ↳27 QDP

(0) Obligation:

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

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(a(a(End(x))))
Right2(b(b(End(x)))) → Left(a(a(End(x))))
Right3(a(End(x))) → Left(b(a(b(End(x)))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(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(b(b(x))) → WAIT(Right1(x))
BEGIN(b(b(x))) → RIGHT1(x)
BEGIN(b(x)) → WAIT(Right2(x))
BEGIN(b(x)) → RIGHT2(x)
BEGIN(a(x)) → WAIT(Right3(x))
BEGIN(a(x)) → RIGHT3(x)
RIGHT1(b(End(x))) → A(a(End(x)))
RIGHT1(b(End(x))) → A(End(x))
RIGHT2(b(b(End(x)))) → A(a(End(x)))
RIGHT2(b(b(End(x)))) → A(End(x))
RIGHT3(a(End(x))) → B(a(b(End(x))))
RIGHT3(a(End(x))) → A(b(End(x)))
RIGHT3(a(End(x))) → B(End(x))
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
AB(Left(x)) → B(x)
AA(Left(x)) → A(x)
WAIT(Left(x)) → BEGIN(x)
B(b(b(x))) → A(a(x))
B(b(b(x))) → A(x)
A(a(x)) → B(a(b(x)))
A(a(x)) → A(b(x))
A(a(x)) → B(x)

The TRS R consists of the following rules:

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(a(a(End(x))))
Right2(b(b(End(x)))) → Left(a(a(End(x))))
Right3(a(End(x))) → Left(b(a(b(End(x)))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(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 5 SCCs with 18 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

Begin(b(b(x))) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(a(a(End(x))))
Right2(b(b(End(x)))) → Left(a(a(End(x))))
Right3(a(End(x))) → Left(b(a(b(End(x)))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(b(b(x))) → a(a(x))
a(a(x)) → b(a(b(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:

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

The TRS R consists of the following rules:

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

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

(8) 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 = A(b(a(a(b(a(a(x))))))) evaluates to t =A(b(a(a(b(a(a(b(x))))))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [x / b(x)]
  • Semiunifier: [ ]



Rewriting sequence

A(b(a(a(b(a(a(x))))))) → A(b(a(a(b(b(a(b(x))))))))
with rule a(a(x')) → b(a(b(x'))) at position [0,0,0,0,0] and matcher [x' / x]

A(b(a(a(b(b(a(b(x)))))))) → A(b(b(a(b(b(b(a(b(x)))))))))
with rule a(a(x')) → b(a(b(x'))) at position [0,0] and matcher [x' / b(b(a(b(x))))]

A(b(b(a(b(b(b(a(b(x))))))))) → A(b(b(a(a(a(a(b(x))))))))
with rule b(b(b(x'))) → a(a(x')) at position [0,0,0,0] and matcher [x' / a(b(x))]

A(b(b(a(a(a(a(b(x)))))))) → A(b(b(b(a(b(a(a(b(x)))))))))
with rule a(a(x')) → b(a(b(x'))) at position [0,0,0] and matcher [x' / a(a(b(x)))]

A(b(b(b(a(b(a(a(b(x))))))))) → A(a(a(a(b(a(a(b(x))))))))
with rule b(b(b(x'))) → a(a(x')) at position [0] and matcher [x' / a(b(a(a(b(x)))))]

A(a(a(a(b(a(a(b(x)))))))) → A(b(a(a(b(a(a(b(x))))))))
with rule A(a(x)) → A(b(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.



(9) NO

(10) Obligation:

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

RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)

The TRS R consists of the following rules:

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

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

(11) 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.

(12) Obligation:

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

RIGHT3(a(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)

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

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

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

(14) YES

(15) Obligation:

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

RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)

The TRS R consists of the following rules:

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

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

(16) 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.

(17) Obligation:

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

RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)

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

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

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

(19) YES

(20) Obligation:

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

RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)

The TRS R consists of the following rules:

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

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

(21) 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.

(22) Obligation:

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

RIGHT1(a(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)

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

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

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

(24) YES

(25) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(x))) → WAIT(Right1(x))
BEGIN(b(x)) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right3(x))

The TRS R consists of the following rules:

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

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

(26) 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.

(27) Obligation:

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

WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(x))) → WAIT(Right1(x))
BEGIN(b(x)) → WAIT(Right2(x))
BEGIN(a(x)) → WAIT(Right3(x))

The TRS R consists of the following rules:

Right3(a(End(x))) → Left(b(a(b(End(x)))))
Right3(b(x)) → Ab(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Aa(Left(x)) → Left(a(x))
a(a(x)) → b(a(b(x)))
b(b(b(x))) → a(a(x))
Ab(Left(x)) → Left(b(x))
Right2(b(b(End(x)))) → Left(a(a(End(x))))
Right2(b(x)) → Ab(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(End(x))) → Left(a(a(End(x))))
Right1(b(x)) → Ab(Right1(x))
Right1(a(x)) → Aa(Right1(x))

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