NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_04/z105.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 QTRSRRRProof (⇔, 42 ms)
↳4 QTRS
↳5 DependencyPairsProof (⇔, 0 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 UsableRulesProof (⇔, 0 ms)
        ↳11 QDP
    ↳12 QDP
    ↳13 QDP
        ↳14 UsableRulesProof (⇔, 11 ms)
        ↳15 QDP
        ↳16 NonTerminationLoopProof (⇒, 40 ms)
        ↳17 NO

(0) Obligation:

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

Begin(a(x)) → Wait(Right1(x))
Begin(b(x)) → Wait(Right2(x))
Begin(c(x)) → Wait(Right3(x))
Begin(d(d(x))) → Wait(Right4(x))
Begin(d(x)) → Wait(Right5(x))
Right1(a(End(x))) → Left(b(b(b(End(x)))))
Right2(b(End(x))) → Left(c(c(c(End(x)))))
Right3(c(End(x))) → Left(d(d(d(End(x)))))
Right4(c(End(x))) → Left(a(End(x)))
Right5(c(d(End(x)))) → Left(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))
Right5(a(x)) → Aa(Right5(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))
Right5(b(x)) → Ab(Right5(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Right4(d(x)) → Ad(Right4(x))
Right5(d(x)) → Ad(Right5(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ad(Left(x)) → Left(d(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
a(a(x)) → b(b(b(x)))
a(x) → d(c(d(x)))
b(b(x)) → c(c(c(x)))
c(c(x)) → d(d(d(x)))
c(d(d(x))) → 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(Begin(x)) → Right1(Wait(x))
b(Begin(x)) → Right2(Wait(x))
c(Begin(x)) → Right3(Wait(x))
d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
End(a(Right1(x))) → End(b(b(b(Left(x)))))
End(b(Right2(x))) → End(c(c(c(Left(x)))))
End(c(Right3(x))) → End(d(d(d(Left(x)))))
End(c(Right4(x))) → End(a(Left(x)))
End(d(c(Right5(x)))) → End(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))
a(Right5(x)) → Right5(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))
b(Right5(x)) → Right5(Ab(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(a(x)) → b(b(b(x)))
a(x) → d(c(d(x)))
b(b(x)) → c(c(c(x)))
c(c(x)) → d(d(d(x)))
d(d(c(x))) → a(x)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(Aa(x1)) = 106 + x1   
POL(Ab(x1)) = 70 + x1   
POL(Ac(x1)) = 46 + x1   
POL(Ad(x1)) = 30 + x1   
POL(Begin(x1)) = x1   
POL(End(x1)) = x1   
POL(Left(x1)) = x1   
POL(Right1(x1)) = 105 + x1   
POL(Right2(x1)) = 69 + x1   
POL(Right3(x1)) = 45 + x1   
POL(Right4(x1)) = 60 + x1   
POL(Right5(x1)) = 30 + x1   
POL(Wait(x1)) = x1   
POL(a(x1)) = 106 + x1   
POL(b(x1)) = 70 + x1   
POL(c(x1)) = 46 + x1   
POL(d(x1)) = 30 + x1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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


(4) Obligation:

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

d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
End(c(Right4(x))) → End(a(Left(x)))
End(d(c(Right5(x)))) → End(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))
a(Right5(x)) → Right5(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))
b(Right5(x)) → Right5(Ab(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(x) → d(c(d(x)))
d(d(c(x))) → a(x)

Q is empty.

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

END(c(Right4(x))) → END(a(Left(x)))
END(c(Right4(x))) → A(Left(x))
END(c(Right4(x))) → LEFT(x)
END(d(c(Right5(x)))) → END(a(Left(x)))
END(d(c(Right5(x)))) → A(Left(x))
END(d(c(Right5(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)
LEFT(Ad(x)) → D(Left(x))
LEFT(Ad(x)) → LEFT(x)
LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
A(x) → D(c(d(x)))
A(x) → C(d(x))
A(x) → D(x)
D(d(c(x))) → A(x)

The TRS R consists of the following rules:

d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
End(c(Right4(x))) → End(a(Left(x)))
End(d(c(Right5(x)))) → End(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))
a(Right5(x)) → Right5(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))
b(Right5(x)) → Right5(Ab(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(x) → d(c(d(x)))
d(d(c(x))) → a(x)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 9 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

D(d(c(x))) → A(x)
A(x) → D(c(d(x)))
A(x) → D(x)

The TRS R consists of the following rules:

d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
End(c(Right4(x))) → End(a(Left(x)))
End(d(c(Right5(x)))) → End(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))
a(Right5(x)) → Right5(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))
b(Right5(x)) → Right5(Ab(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(x) → d(c(d(x)))
d(d(c(x))) → a(x)

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

(10) 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.

(11) Obligation:

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

D(d(c(x))) → A(x)
A(x) → D(c(d(x)))
A(x) → D(x)

The TRS R consists of the following rules:

d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
d(d(c(x))) → a(x)
a(x) → d(c(d(x)))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(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(Right5(x)) → Right5(Aa(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:

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)

The TRS R consists of the following rules:

d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
End(c(Right4(x))) → End(a(Left(x)))
End(d(c(Right5(x)))) → End(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))
a(Right5(x)) → Right5(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))
b(Right5(x)) → Right5(Ab(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(x) → d(c(d(x)))
d(d(c(x))) → a(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:

END(d(c(Right5(x)))) → END(a(Left(x)))
END(c(Right4(x))) → END(a(Left(x)))

The TRS R consists of the following rules:

d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
End(c(Right4(x))) → End(a(Left(x)))
End(d(c(Right5(x)))) → End(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))
a(Right5(x)) → Right5(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))
b(Right5(x)) → Right5(Ab(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(x) → d(c(d(x)))
d(d(c(x))) → a(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:

END(d(c(Right5(x)))) → END(a(Left(x)))
END(c(Right4(x))) → END(a(Left(x)))

The TRS R consists of the following rules:

Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(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(Right5(x)) → Right5(Aa(x))
d(d(c(x))) → a(x)
a(x) → d(c(d(x)))
d(d(Begin(x))) → Right4(Wait(x))
d(Begin(x)) → Right5(Wait(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
d(Right4(x)) → Right4(Ad(x))
d(Right5(x)) → Right5(Ad(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(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))
b(Right5(x)) → Right5(Ab(x))

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

(16) 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 = END(a(Left(Wait(x)))) evaluates to t =END(a(Left(Wait(x))))

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



Rewriting sequence

END(a(Left(Wait(x)))) → END(a(Begin(x)))
with rule Left(Wait(x')) → Begin(x') at position [0,0] and matcher [x' / x]

END(a(Begin(x))) → END(d(c(d(Begin(x)))))
with rule a(x') → d(c(d(x'))) at position [0] and matcher [x' / Begin(x)]

END(d(c(d(Begin(x))))) → END(d(c(Right5(Wait(x)))))
with rule d(Begin(x')) → Right5(Wait(x')) at position [0,0,0] and matcher [x' / x]

END(d(c(Right5(Wait(x))))) → END(a(Left(Wait(x))))
with rule END(d(c(Right5(x)))) → END(a(Left(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.



(17) NO