Termination w.r.t. Q proof of /tmp/tmpJUSQpv/#4.7.trs

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

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

The set Q consists of the following terms:

0
f(s(x0))

↳ 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

The set Q consists of the following terms:

0
f(s(x0))

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

The set Q consists of the following terms:

0
f(s(x0))

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

The set Q consists of the following terms:

0
f(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.