Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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:

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



↳ QTRS
  ↳ DependencyPairsProof

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:

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


Using Dependency Pairs [1,15] we result in the following initial DP problem:
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:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof

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

F(0) → F(s(0))
F(s(0)) → F(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:

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

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof

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

F(0) → F(s(0))
F(s(0)) → F(p(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))
f(s(0))
f(0)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

f(s(0))
f(0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting

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

F(0) → F(s(0))
F(s(0)) → F(p(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.
By rewriting [15] the rule F(s(0)) → F(p(s(0))) at position [0] we obtained the following new rules:

F(s(0)) → F(0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ UsableRulesProof

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof

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

p(s(0))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesProof
            ↳ QDP
              ↳ QReductionProof
                ↳ QDP
                  ↳ Rewriting
                    ↳ QDP
                      ↳ UsableRulesProof
                        ↳ QDP
                          ↳ QReductionProof
                            ↳ QDP
                              ↳ NonTerminationProof

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.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:none


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:



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.