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) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(s(s(x))) → P(h(g(x)))
F(s(s(x))) → H(g(x))
F(s(s(x))) → G(x)
G(s(x)) → H(g(x))
G(s(x)) → G(x)
H(x) → +¹(p(x), q(x))
H(x) → P(x)
H(x) → Q(x)
+¹(x, s(y)) → +¹(x, y)
F(s(s(x))) → +¹(p(g(x)), q(g(x)))
F(s(s(x))) → P(g(x))
F(s(s(x))) → Q(g(x))
G(s(x)) → +¹(p(g(x)), q(g(x)))
G(s(x)) → P(g(x))
G(s(x)) → Q(g(x))

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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 13 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

+¹(x, s(y)) → +¹(x, y)

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.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(x, s(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(s(x)) → G(x)

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.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

G(s(x)) → G(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:

s₁ > G₁

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE