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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(s(x)) → G(x)
F(g(x), s(0), y) → F(y, y, g(x))

The TRS R consists of the following rules:

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

Q is empty.
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:

G(s(x)) → G(x)
F(g(x), s(0), y) → F(y, y, g(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

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

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

G(s(x)) → G(x)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

G(s(x)) → G(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ ForwardInstantiation

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

F(g(x), s(0), y) → F(y, y, g(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(g(x), s(0), y) → F(y, y, g(x)) we obtained the following new rules:

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



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ ForwardInstantiation
              ↳ QDP
                ↳ NonTerminationProof

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


s = F(g(x0), g(s(0)), g(y_0)) evaluates to t =F(g(y_0), g(y_0), g(x0))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

F(g(s(0)), g(s(0)), g(s(0))) → F(g(s(0)), s(g(0)), g(s(0)))
with rule g(s(x)) → s(g(x)) at position [1] and matcher [x / 0]

F(g(s(0)), s(g(0)), g(s(0))) → F(g(s(0)), s(0), g(s(0)))
with rule g(0) → 0 at position [1,0] and matcher [ ]

F(g(s(0)), s(0), g(s(0))) → F(g(s(0)), g(s(0)), g(s(0)))
with rule F(g(x0), s(0), g(y_0)) → F(g(y_0), g(y_0), g(x0))

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.