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-399.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 6 ms)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔, 0 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 108 ms)
        ↳9 QDP
    ↳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
        ↳28 NonTerminationLoopProof (⇒, 92 ms)
        ↳29 NO

(0) Obligation:

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

Begin(b(x)) → Wait(Right1(x))
Begin(a(c(x))) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(c(End(x)))
Right2(c(End(x))) → Left(a(b(c(a(End(x))))))
Right3(c(a(End(x)))) → Left(a(b(c(a(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(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(b(c(a(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(x)) → WAIT(Right1(x))
BEGIN(b(x)) → RIGHT1(x)
BEGIN(a(c(x))) → WAIT(Right2(x))
BEGIN(a(c(x))) → RIGHT2(x)
BEGIN(c(x)) → WAIT(Right3(x))
BEGIN(c(x)) → RIGHT3(x)
RIGHT1(b(End(x))) → C(End(x))
RIGHT2(c(End(x))) → A(b(c(a(End(x)))))
RIGHT2(c(End(x))) → B(c(a(End(x))))
RIGHT2(c(End(x))) → C(a(End(x)))
RIGHT2(c(End(x))) → A(End(x))
RIGHT3(c(a(End(x)))) → A(b(c(a(End(x)))))
RIGHT3(c(a(End(x)))) → B(c(a(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(x) → B(x)
B(b(x)) → C(x)
C(a(c(x))) → A(b(c(a(x))))
C(a(c(x))) → B(c(a(x)))
C(a(c(x))) → C(a(x))
C(a(c(x))) → A(x)

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right1(x))
Begin(a(c(x))) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(c(End(x)))
Right2(c(End(x))) → Left(a(b(c(a(End(x))))))
Right3(c(a(End(x)))) → Left(a(b(c(a(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(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(b(c(a(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 22 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

Begin(b(x)) → Wait(Right1(x))
Begin(a(c(x))) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(c(End(x)))
Right2(c(End(x))) → Left(a(b(c(a(End(x))))))
Right3(c(a(End(x)))) → Left(a(b(c(a(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(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(b(c(a(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(b(x)) → C(x)
C(a(c(x))) → A(b(c(a(x))))
A(x) → B(x)
C(a(c(x))) → B(c(a(x)))
C(a(c(x))) → C(a(x))
C(a(c(x))) → A(x)

The TRS R consists of the following rules:

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

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

(8) 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 dependency pairs:

C(a(c(x))) → B(c(a(x)))
C(a(c(x))) → C(a(x))
C(a(c(x))) → A(x)


Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = 1 + x1   
POL(B(x1)) = 1 + x1   
POL(C(x1)) = 2 + x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 2 + x1   

(9) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(10) 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(b(x)) → Wait(Right1(x))
Begin(a(c(x))) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(c(End(x)))
Right2(c(End(x))) → Left(a(b(c(a(End(x))))))
Right3(c(a(End(x)))) → Left(a(b(c(a(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(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(b(c(a(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(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.

(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(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

(14) YES

(15) 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(b(x)) → Wait(Right1(x))
Begin(a(c(x))) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(c(End(x)))
Right2(c(End(x))) → Left(a(b(c(a(End(x))))))
Right3(c(a(End(x)))) → Left(a(b(c(a(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(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(b(c(a(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(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.

(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(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

(19) YES

(20) 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(b(x)) → Wait(Right1(x))
Begin(a(c(x))) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Right1(b(End(x))) → Left(c(End(x)))
Right2(c(End(x))) → Left(a(b(c(a(End(x))))))
Right3(c(a(End(x)))) → Left(a(b(c(a(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(x) → b(x)
b(b(x)) → c(x)
c(a(c(x))) → a(b(c(a(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(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.

(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(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

(24) YES

(25) Obligation:

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

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

The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

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

(28) 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(a(End(x')))))) evaluates to t =BEGIN(a(b(c(a(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(a(End(x')))))) → BEGIN(b(b(c(a(End(x'))))))
with rule a(x) → b(x) at position [0] and matcher [x / b(c(a(End(x'))))]

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

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

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

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



(29) NO