NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Waldmann_07_size12/size-12-alpha-3-num-494.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 47 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔, 0 ms)
        ↳9 YES
    ↳10 QDP
        ↳11 UsableRulesProof (⇔, 0 ms)
        ↳12 QDP
    ↳13 QDP
        ↳14 UsableRulesProof (⇔, 4 ms)
        ↳15 QDP
        ↳16 QDPSizeChangeProof (⇔, 0 ms)
        ↳17 YES
    ↳18 QDP
        ↳19 UsableRulesProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 QDPSizeChangeProof (⇔, 0 ms)
        ↳22 YES
    ↳23 QDP
        ↳24 UsableRulesProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 15 ms)
        ↳30 QDP
        ↳31 NonTerminationLoopProof (⇒, 97 ms)
        ↳32 NO

(0) Obligation:

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

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

The TRS R consists of the following rules:

Begin(a(x)) → Wait(Right1(x))
Begin(c(x)) → Wait(Right2(x))
Begin(b(x)) → Wait(Right3(x))
Right1(a(End(x))) → Left(b(End(x)))
Right2(b(End(x))) → Left(a(End(x)))
Right3(c(End(x))) → Left(a(b(c(c(End(x))))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(x)) → b(x)
b(c(x)) → a(x)
c(b(x)) → a(b(c(c(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 6 SCCs with 22 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

B(c(x)) → A(x)
A(a(x)) → B(x)

The TRS R consists of the following rules:

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

B(c(x)) → A(x)
A(a(x)) → B(x)

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

(8) 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:

  • A(a(x)) → B(x)
    The graph contains the following edges 1 > 1

  • B(c(x)) → A(x)
    The graph contains the following edges 1 > 1

(9) YES

(10) Obligation:

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

C(b(x)) → C(x)
C(b(x)) → C(c(x))

The TRS R consists of the following rules:

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

C(b(x)) → C(x)
C(b(x)) → C(c(x))

The TRS R consists of the following rules:

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

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

(13) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(14) 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.

(15) Obligation:

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

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

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

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

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

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

(17) YES

(18) Obligation:

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

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

The TRS R consists of the following rules:

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

RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
RIGHT2(c(x)) → RIGHT2(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:

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

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

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

(22) YES

(23) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

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

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

(27) YES

(28) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(29) 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.

(30) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(31) 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(a(b(c(c(End(x')))))) evaluates to t =BEGIN(a(b(c(c(End(x'))))))

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



Rewriting sequence

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

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

BEGIN(b(c(End(x')))) → WAIT(Right3(c(End(x'))))
with rule BEGIN(b(x)) → WAIT(Right3(x)) at position [] and matcher [x / c(End(x'))]

WAIT(Right3(c(End(x')))) → WAIT(Left(a(b(c(c(End(x')))))))
with rule Right3(c(End(x''))) → Left(a(b(c(c(End(x'')))))) at position [0] and matcher [x'' / x']

WAIT(Left(a(b(c(c(End(x'))))))) → BEGIN(a(b(c(c(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.



(32) NO