NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z127.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 (⇔, 47 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 147 ms)
        ↳11 QDP
        ↳12 NonTerminationLoopProof (⇒, 56 ms)
        ↳13 NO
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 15 ms)
        ↳21 QDP

(0) Obligation:

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

Begin(a(a(x))) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(b(x))) → Wait(Right3(x))
Begin(b(x)) → Wait(Right4(x))
Right1(a(End(x))) → Left(a(b(b(End(x)))))
Right2(a(a(End(x)))) → Left(a(b(b(End(x)))))
Right3(b(End(x))) → Left(a(b(a(End(x)))))
Right4(b(a(End(x)))) → Left(a(b(a(End(x)))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Wait(Left(x)) → Begin(x)
a(a(a(x))) → a(b(b(x)))
b(a(b(x))) → a(b(a(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:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
End(a(Right1(x))) → End(b(b(a(Left(x)))))
End(a(a(Right2(x)))) → End(b(b(a(Left(x)))))
End(b(Right3(x))) → End(a(b(a(Left(x)))))
End(a(b(Right4(x)))) → End(a(b(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(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(a(Right1(x))) → END(b(b(a(Left(x)))))
END(a(Right1(x))) → B(b(a(Left(x))))
END(a(Right1(x))) → B(a(Left(x)))
END(a(Right1(x))) → A(Left(x))
END(a(Right1(x))) → LEFT(x)
END(a(a(Right2(x)))) → END(b(b(a(Left(x)))))
END(a(a(Right2(x)))) → B(b(a(Left(x))))
END(a(a(Right2(x)))) → B(a(Left(x)))
END(a(a(Right2(x)))) → A(Left(x))
END(a(a(Right2(x)))) → LEFT(x)
END(b(Right3(x))) → END(a(b(a(Left(x)))))
END(b(Right3(x))) → A(b(a(Left(x))))
END(b(Right3(x))) → B(a(Left(x)))
END(b(Right3(x))) → A(Left(x))
END(b(Right3(x))) → LEFT(x)
END(a(b(Right4(x)))) → END(a(b(a(Left(x)))))
END(a(b(Right4(x)))) → A(b(a(Left(x))))
END(a(b(Right4(x)))) → B(a(Left(x)))
END(a(b(Right4(x)))) → A(Left(x))
END(a(b(Right4(x)))) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
A(a(a(x))) → B(b(a(x)))
A(a(a(x))) → B(a(x))
B(a(b(x))) → A(b(a(x)))
B(a(b(x))) → B(a(x))
B(a(b(x))) → A(x)

The TRS R consists of the following rules:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
End(a(Right1(x))) → End(b(b(a(Left(x)))))
End(a(a(Right2(x)))) → End(b(b(a(Left(x)))))
End(b(Right3(x))) → End(a(b(a(Left(x)))))
End(a(b(Right4(x)))) → End(a(b(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(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 18 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

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

The TRS R consists of the following rules:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
End(a(Right1(x))) → End(b(b(a(Left(x)))))
End(a(a(Right2(x)))) → End(b(b(a(Left(x)))))
End(b(Right3(x))) → End(a(b(a(Left(x)))))
End(a(b(Right4(x)))) → End(a(b(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(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:

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

The TRS R consists of the following rules:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(x)))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(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.


B(a(b(x))) → B(a(x))
B(a(b(x))) → A(x)
A(a(a(x))) → B(a(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(Aa(x1)) = x1   
POL(Ab(x1)) = x1   
POL(B(x1)) = x1   
POL(Begin(x1)) = x1   
POL(Right1(x1)) = 0   
POL(Right2(x1)) = 0   
POL(Right3(x1)) = 0   
POL(Right4(x1)) = 0   
POL(Wait(x1)) = x1   
POL(a(x1)) = 1 + x1   
POL(b(x1)) = 1 + x1   

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

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(x)))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))

(11) Obligation:

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

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

The TRS R consists of the following rules:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(x)))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))

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

(12) 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(a(a(x')))))) evaluates to t =A(b(a(a(a(a(x'))))))

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



Rewriting sequence

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

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

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

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

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

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



(13) NO

(14) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)

The TRS R consists of the following rules:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
End(a(Right1(x))) → End(b(b(a(Left(x)))))
End(a(a(Right2(x)))) → End(b(b(a(Left(x)))))
End(b(Right3(x))) → End(a(b(a(Left(x)))))
End(a(b(Right4(x)))) → End(a(b(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(x)))

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

(15) 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.

(16) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)

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

(17) 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(Ab(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

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

(18) YES

(19) Obligation:

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

END(a(a(Right2(x)))) → END(b(b(a(Left(x)))))
END(a(Right1(x))) → END(b(b(a(Left(x)))))
END(b(Right3(x))) → END(a(b(a(Left(x)))))
END(a(b(Right4(x)))) → END(a(b(a(Left(x)))))

The TRS R consists of the following rules:

a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
End(a(Right1(x))) → End(b(b(a(Left(x)))))
End(a(a(Right2(x)))) → End(b(b(a(Left(x)))))
End(b(Right3(x))) → End(a(b(a(Left(x)))))
End(a(b(Right4(x)))) → End(a(b(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(x)))

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

(20) 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.

(21) Obligation:

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

END(a(a(Right2(x)))) → END(b(b(a(Left(x)))))
END(a(Right1(x))) → END(b(b(a(Left(x)))))
END(b(Right3(x))) → END(a(b(a(Left(x)))))
END(a(b(Right4(x)))) → END(a(b(a(Left(x)))))

The TRS R consists of the following rules:

Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Wait(x)) → Begin(x)
a(a(Begin(x))) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(a(a(x))) → b(b(a(x)))
b(a(b(x))) → a(b(a(x)))
b(a(Begin(x))) → Right3(Wait(x))
b(Begin(x)) → Right4(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))

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