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(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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(c(s(x), y)) → F(c(x, s(y)))
G(c(x, s(y))) → G(c(s(x), y))
G(s(f(x))) → G(f(x))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

G(c(x, s(y))) → G(c(s(x), y))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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.

G(c(x, s(y))) → G(c(s(x), y))

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

Recursive path order with status [RPO].
Precedence:

f > c₂ > s₁

Status:

c₂: [2,1]
s₁: multiset
f: []

The following usable rules [FROCOS05] were oriented:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(x))

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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:

F(c(s(x), y)) → F(c(x, s(y)))

The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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.

F(c(s(x), y)) → F(c(x, s(y)))

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

Recursive path order with status [RPO].
Precedence:

F₁ > f₁
g > c₁ > f₁
g > s₁ > f₁

Status:

F₁: [1]
c₁: [1]
s₁: [1]
f₁: multiset
g: []

The following usable rules [FROCOS05] were oriented:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(x))

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(c(s(x), y)) → f(c(x, s(y)))
g(c(x, s(y))) → g(c(s(x), y))
g(s(f(x))) → g(f(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