Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(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(g(a)) → F(s(g(b)))
F(g(a)) → G(b)
G(x) → F(g(x))
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(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 1 SCC with 1 less node.

(4) Obligation:

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

F(g(a)) → G(b)
G(x) → F(g(x))
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(g(a)) → G(b)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
g(x1)  =  g(x1)
a  =  a
G(x1)  =  G(x1)
b  =  b
f(x1)  =  f
s(x1)  =  s(x1)

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

[a, s₁] > [g₁, G₁] > [b, f]

Status:

a: multiset
f: []
g₁: multiset
G₁: multiset
b: multiset
s₁: [1]

The following usable rules [FROCOS05] were oriented:

g(x) → f(g(x))
f(f(x)) → b
f(g(a)) → f(s(g(b)))

(6) Obligation:

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

G(x) → F(g(x))
G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

G(x) → G(x)

The TRS R consists of the following rules:

f(g(a)) → f(s(g(b)))
f(f(x)) → b
g(x) → f(g(x))

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