NO Termination w.r.t. Q proof of Payet_24_payet-nonloop-4.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 UsableRulesProof (⇔, 0 ms)
↳6 QDP
↳7 QReductionProof (⇔, 0 ms)
↳8 QDP
↳9 TransformationProof (⇔, 0 ms)
↳10 QDP
↳11 TransformationProof (⇔, 0 ms)
↳12 QDP
↳13 TransformationProof (⇔, 2 ms)
↳14 QDP
↳15 TransformationProof (⇔, 0 ms)
↳16 QDP
↳17 TransformationProof (⇔, 0 ms)
↳18 QDP
↳19 QDPOrderProof (⇔, 1558 ms)
↳20 QDP
↳21 NonTerminationLoopProof (⇒, 0 ms)
↳22 NO

(0) Obligation:

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

f(x, z, s(s(s(s(s(s(s(y))))))), z) → f(x, 0, s(s(y)), z)
f(x, z, s(s(s(s(s(s(0)))))), z) → f(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

f(x, z, s(s(s(s(s(s(s(y))))))), z) → f(x, 0, s(s(y)), z)
f(x, z, s(s(s(s(s(s(0)))))), z) → f(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

The signature Sigma is {f}

(2) Obligation:

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

f(x, z, s(s(s(s(s(s(s(y))))))), z) → f(x, 0, s(s(y)), z)
f(x, z, s(s(s(s(s(s(0)))))), z) → f(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

The set Q consists of the following terms:

f(x0, x1, s(s(s(s(s(s(s(x2))))))), x1)
f(x0, x1, s(s(s(s(s(s(0)))))), x1)

(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:

F(x, z, s(s(s(s(s(s(s(y))))))), z) → F(x, 0, s(s(y)), z)
F(x, z, s(s(s(s(s(s(0)))))), z) → F(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

The TRS R consists of the following rules:

f(x, z, s(s(s(s(s(s(s(y))))))), z) → f(x, 0, s(s(y)), z)
f(x, z, s(s(s(s(s(s(0)))))), z) → f(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

The set Q consists of the following terms:

f(x0, x1, s(s(s(s(s(s(s(x2))))))), x1)
f(x0, x1, s(s(s(s(s(s(0)))))), x1)

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

(5) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(6) Obligation:

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

F(x, z, s(s(s(s(s(s(s(y))))))), z) → F(x, 0, s(s(y)), z)
F(x, z, s(s(s(s(s(s(0)))))), z) → F(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

R is empty.
The set Q consists of the following terms:

f(x0, x1, s(s(s(s(s(s(s(x2))))))), x1)
f(x0, x1, s(s(s(s(s(s(0)))))), x1)

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

(7) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

f(x0, x1, s(s(s(s(s(s(s(x2))))))), x1)
f(x0, x1, s(s(s(s(s(s(0)))))), x1)

(8) Obligation:

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

F(x, z, s(s(s(s(s(s(s(y))))))), z) → F(x, 0, s(s(y)), z)
F(x, z, s(s(s(s(s(s(0)))))), z) → F(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)

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

(9) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, z, s(s(s(s(s(s(s(y))))))), z) → F(x, 0, s(s(y)), z) we obtained the following new rules [LPAR04]:

F(z0, 0, s(s(s(s(s(s(s(x2))))))), 0) → F(z0, 0, s(s(x2)), 0) → F(z0, 0, s(s(s(s(s(s(s(x2))))))), 0) → F(z0, 0, s(s(x2)), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(z0))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(z0))), 0)

(10) Obligation:

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

F(x, z, s(s(s(s(s(s(0)))))), z) → F(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0)
F(z0, 0, s(s(s(s(s(s(s(x2))))))), 0) → F(z0, 0, s(s(x2)), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(z0))), 0)

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

(11) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(x, z, s(s(s(s(s(s(0)))))), z) → F(s(s(s(s(s(s(s(s(0)))))))), z, s(s(s(s(s(s(s(s(x)))))))), 0) we obtained the following new rules [LPAR04]:

F(z0, 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(z0, 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)

(12) Obligation:

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

F(z0, 0, s(s(s(s(s(s(s(x2))))))), 0) → F(z0, 0, s(s(x2)), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(z0))), 0)
F(z0, 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)

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

(13) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(z0, 0, s(s(s(s(s(s(s(x2))))))), 0) → F(z0, 0, s(s(x2)), 0) we obtained the following new rules [LPAR04]:

F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0) → F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)

(14) Obligation:

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

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(z0))), 0)
F(z0, 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)

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

(15) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(z0))), 0) we obtained the following new rules [LPAR04]:

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_1))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_1))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)

(16) Obligation:

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

F(z0, 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_1))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)

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

(17) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(z0, 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(z0)))))))), 0) we obtained the following new rules [LPAR04]:

F(s(s(s(s(y_1)))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(s(s(s(s(y_1)))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(y_0))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(y_0))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(0))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(0))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)
F(s(s(s(s(s(s(s(s(s(y_0))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(s(y_0))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))))))), 0)
F(s(s(s(s(s(s(s(s(s(s(y_0)))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(s(s(y_0)))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)

(18) Obligation:

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

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_1))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)
F(s(s(s(s(y_1)))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(y_0))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(0))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)
F(s(s(s(s(s(s(s(s(s(y_0))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))))))), 0)
F(s(s(s(s(s(s(s(s(s(s(y_0)))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(s(s(s(0))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(F(x1, x2, x3, x4)) = 0 +
[0,1]
·x1 +
[0,0]
·x2 +
[0,0]
·x3 +
[0,0]
·x4

POL(s(x1)) =
/0\
\0/
 +
/01\
\10/
·x1

POL(0) =
/1\
\0/

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(20) Obligation:

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

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(y_0)))))))), 0)
F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_1))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)
F(s(s(s(s(y_1)))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0)
F(s(s(s(s(s(y_0))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))), 0)
F(s(s(s(s(s(s(s(s(s(y_0))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0))))))))))))))))), 0)
F(s(s(s(s(s(s(s(s(s(s(y_0)))))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(y_0)))))))))))))))))), 0)

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

(21) 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 = F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) evaluates to t =F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)

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



Rewriting sequence

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0)
with rule F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(s(y_1)))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(s(y_1))))))), 0) at position [] and matcher [x0 / s(s(s(s(s(s(s(s(0)))))))), y_1 / s(s(s(s(0))))]

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0)
with rule F(x0, 0, s(s(s(s(s(s(s(s(s(s(s(0))))))))))), 0) → F(x0, 0, s(s(s(s(s(s(0)))))), 0) at position [] and matcher [x0 / s(s(s(s(s(s(s(s(0))))))))]

F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)
with rule F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(0)))))), 0) → F(s(s(s(s(s(s(s(s(0)))))))), 0, s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(0)))))))))))))))), 0)

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.



(22) NO