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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 UsableRulesProof (⇔)
↳8 QDP
↳9 QReductionProof (⇔)
↳10 QDP
↳11 Rewriting (⇔)
↳12 QDP
↳13 UsableRulesProof (⇔)
↳14 QDP
↳15 QReductionProof (⇔)
↳16 QDP
↳17 NonTerminationProof (⇔)
↳18 NO

(0) Obligation:

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

f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

The set Q consists of the following terms:

f(0)
f(s(0))
p(s(0))

(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(0) → F(s(0))
F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))

The TRS R consists of the following rules:

f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

The set Q consists of the following terms:

f(0)
f(s(0))
p(s(0))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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

F(s(0)) → F(p(s(0)))
F(0) → F(s(0))

The TRS R consists of the following rules:

f(0) → cons(0, f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0

The set Q consists of the following terms:

f(0)
f(s(0))
p(s(0))

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

(7) 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.

(8) Obligation:

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

F(s(0)) → F(p(s(0)))
F(0) → F(s(0))

The TRS R consists of the following rules:

p(s(0)) → 0

The set Q consists of the following terms:

f(0)
f(s(0))
p(s(0))

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

(9) 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(0)
f(s(0))

(10) Obligation:

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

F(s(0)) → F(p(s(0)))
F(0) → F(s(0))

The TRS R consists of the following rules:

p(s(0)) → 0

The set Q consists of the following terms:

p(s(0))

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

(11) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule F(s(0)) → F(p(s(0))) at position [0] we obtained the following new rules [LPAR04]:

F(s(0)) → F(0)

(12) Obligation:

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

F(0) → F(s(0))
F(s(0)) → F(0)

The TRS R consists of the following rules:

p(s(0)) → 0

The set Q consists of the following terms:

p(s(0))

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

(13) 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.

(14) Obligation:

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

F(0) → F(s(0))
F(s(0)) → F(0)

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

p(s(0))

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

(15) 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].

p(s(0))

(16) Obligation:

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

F(0) → F(s(0))
F(s(0)) → F(0)

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

(17) NonTerminationProof (EQUIVALENT 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(0)) evaluates to t =F(s(0))

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



Rewriting sequence

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

F(0) → F(s(0))
with rule F(0) → F(s(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.



(18) NO