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

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

The TRS R consists of the following rules:

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

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1)  =  F(x0)
G(x0, x1)  =  G(x1)

Tags:
F has argument tags [0,0] and root tag 1
G has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x₁)) = x₁   
POL(G(x₁)) = 0   
POL(c(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(4) 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))

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

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

(8) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
G(x0, x1)  =  G(x1)

Tags:
G has argument tags [1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(G(x₁)) = 0   
POL(c(x₁, x₂)) = 1 + x₂   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(9) 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)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

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

(13) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1)  =  F(x1)

Tags:
F has argument tags [1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(F(x₁)) = 0   
POL(c(x₁, x₂)) = 1 + x₁   
POL(s(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(14) 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)))
f(c(s(x), s(y))) → g(c(x, y))
g(c(x, s(y))) → g(c(s(x), y))
g(c(s(x), s(y))) → f(c(x, y))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE