Termination w.r.t. Q proof of TRS/nontermin/AG01#4.20.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(f(x)) → f(x)
g(0) → g(f(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f(f(x)) → f(x)
g(0) → g(f(0))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(0) → G(f(0))
G(0) → F(0)

The TRS R consists of the following rules:

f(f(x)) → f(x)
g(0) → g(f(0))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

G(0) → G(f(0))
G(0) → F(0)

The TRS R consists of the following rules:

f(f(x)) → f(x)
g(0) → g(f(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

G(0) → G(f(0))
G(0) → F(0)

The TRS R consists of the following rules:

f(f(x)) → f(x)
g(0) → g(f(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.