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) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

[f₁, g₁] > h₁ > pair₂ > [s₁, p₁, q₁]
[f₁, g₁] > h₁ > +₂ > [s₁, p₁, q₁]
0 > pair₂ > [s₁, p₁, q₁]

Status:

f₁: multiset
0: multiset
s₁: [1]
p₁: multiset
h₁: multiset
g₁: multiset
pair₂: multiset
+₂: [1,2]
q₁: multiset

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

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(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