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(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

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

The TRS R consists of the following rules:

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

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:

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

The TRS R consists of the following rules:

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

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.

ID_INC(c(x, y)) → ID_INC(y)
ID_INC(c(x, y)) → ID_INC(x)

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

Lexicographic Path Order [LPO].
Precedence:

[f, g]

The following usable rules [FROCOS05] were oriented:

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

(7) Obligation:

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

ID_INC(s(x)) → ID_INC(x)

The TRS R consists of the following rules:

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

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.

ID_INC(s(x)) → ID_INC(x)

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

Lexicographic Path Order [LPO].
Precedence:

[f, g]
id_inc1 > s₁

The following usable rules [FROCOS05] were oriented:

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

(9) Obligation:

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

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

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

The TRS R consists of the following rules:

f(s(x)) → f(id_inc(c(x, x)))
f(c(s(x), y)) → g(c(x, y))
g(c(s(x), y)) → g(c(y, x))
g(c(x, s(y))) → g(c(y, x))
g(c(x, x)) → f(x)
id_inc(c(x, y)) → c(id_inc(x), id_inc(y))
id_inc(s(x)) → s(id_inc(x))
id_inc(0) → 0
id_inc(0) → s(0)

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