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(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, x)))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

f₁ > 1 > s₁
f₁ > g₂ > +₂ > s₁
0 > 1 > s₁

Status:

f₁: multiset
g₂: [1,2]
s₁: multiset
0: multiset
1: multiset
+₂: multiset

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(x, s(x))
g(0, y) → y
g(s(x), y) → g(x, +(y, s(x)))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
g(s(x), y) → g(x, s(+(y, 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