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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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)))

The set Q consists of the following terms:

f(0)
f(s(x0))
g(x0)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

F(s(x)) → G(f(x))
F(s(x)) → F(x)

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)))

The set Q consists of the following terms:

f(0)
f(s(x0))
g(x0)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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

F(s(x)) → F(x)

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)))

The set Q consists of the following terms:

f(0)
f(s(x0))
g(x0)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(s(x)) → F(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
s(x1)  =  s(x1)
f(x1)  =  f(x1)
0  =  0
1  =  1
g(x1)  =  g(x1)
+(x1, x2)  =  x1

Recursive Path Order [RPO].
Precedence:

s₁ > f₁ > g₁
[0, 1] > g₁

The following usable rules [FROCOS05] were oriented:

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

(8) Obligation:

Q DP problem:
P is empty.
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)))

The set Q consists of the following terms:

f(0)
f(s(x0))
g(x0)

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

(9) PisEmptyProof (EQUIVALENT transformation)

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

(10) TRUE