Termination w.r.t. Q proof of /tmp/tmpeQZWTU/tricky1.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(g(x), g(y)) → f(p(f(g(x), s(y))), g(s(p(x))))
p(0) → g(0)
g(s(p(x))) → p(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

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(g(x), g(y)) → P(f(g(x), s(y)))
P(0) → G(0)
F(g(x), g(y)) → P(x)
F(g(x), g(y)) → G(s(p(x)))
F(g(x), g(y)) → F(g(x), s(y))
F(g(x), g(y)) → F(p(f(g(x), s(y))), g(s(p(x))))

The TRS R consists of the following rules:

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

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(g(x), g(y)) → P(f(g(x), s(y)))
P(0) → G(0)
F(g(x), g(y)) → P(x)
F(g(x), g(y)) → G(s(p(x)))
F(g(x), g(y)) → F(g(x), s(y))
F(g(x), g(y)) → F(p(f(g(x), s(y))), g(s(p(x))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 6 less nodes.