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(s(x)) → f(g(x, x))
g(0, 1) → s(0)
0 → 1

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
0 → 1

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:

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

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
0 → 1

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:

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

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
0 → 1

Q is empty.
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
          ↳ UsableRulesReductionPairsProof

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

F(s(x)) → F(g(x, x))

The TRS R consists of the following rules:

f(s(x)) → f(g(x, x))
g(0, 1) → s(0)
0 → 1

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(1) = 0   
POL(F(x1)) = x1   
POL(g(x1, x2)) = x1 + x2   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ UsableRulesReductionPairsProof
            ↳ QDP
              ↳ NonTerminationProof

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

F(s(x)) → F(g(x, x))

The TRS R consists of the following rules:

g(0, 1) → s(0)
0 → 1

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(s(x)) → F(g(x, x))

The TRS R consists of the following rules:

g(0, 1) → s(0)
0 → 1


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

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



Rewriting sequence

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

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

F(s(0)) → F(g(0, 0))
with rule F(s(x)) → F(g(x, 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.