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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

[g₁, h₂] > +₂ > [s₁, f₁]
[g₁, h₂] > 0

Status:

f₁: multiset
g₁: [1]
h₂: [2,1]
s₁: multiset
0: multiset
+₂: [2,1]

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

(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