Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

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

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive Path Order [RPO].
Precedence:

f₁ > g₁ > +₂ > s₁
[0, 1] > s₁

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE