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

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

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

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 with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

F(s(x), s(y)) → F(x, s(c(s(y))))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

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.

F(s(x), s(y)) → F(x, s(c(s(y))))

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

Recursive path order with status [RPO].
Precedence:

F₂ > s₁ > c
f > s₁ > c

Status:

F₂: [1,2]
s₁: [1]
c: multiset
f: []

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

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

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

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(x, c(y)) → F(y, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

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(x, c(y)) → F(y, y)

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

Recursive path order with status [RPO].
Precedence:

c₁ > F₂
f > F₂
s > F₂

Status:

F₂: multiset
c₁: multiset
f: multiset
s: multiset

The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

The set Q consists of the following terms:

f(x0, c(x1))
f(s(x0), s(x1))

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