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:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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:

G(x, x, x) → G(c, d, e)
G(x, y, x) → G(c, d, e)
S(f(x, y)) → F(y, f(s(s(x)), a))
S(f(x, y)) → F(s(s(x)), a)
S(f(x, y)) → S(s(x))
S(f(x, y)) → S(x)
H(h(x, a), y) → H(h(a, y), h(a, x))
H(h(x, a), y) → H(a, y)
H(h(x, a), y) → H(a, x)
F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))
F(x, f(y, f(x, y))) → F(x, f(y, b))
F(x, f(y, f(x, y))) → F(y, b)
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(h(a, y), g(x, b, a)) → F(x, s(y))
F(h(a, y), g(x, b, a)) → S(y)
F(h(a, y), g(x, b, a)) → S(b)
H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
H(f(x, s(y)), b) → G(y, a, f(s(x), a))
H(f(x, s(y)), b) → F(s(x), a)
H(f(x, s(y)), b) → S(x)
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(x, g(x, a, f(s(x), y))) → H(x, b)
F(x, g(x, a, f(s(x), y))) → G(a, b, y)

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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 4 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

F(x, f(y, f(x, y))) → F(a, f(x, f(y, b)))

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(7) Obligation:

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

H(h(x, a), y) → H(h(a, y), h(a, x))

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(8) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))
F(x, g(x, a, f(s(x), y))) → H(x, b)

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(x, g(x, a, f(s(x), y))) → H(x, b)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
H(x1, x2)  =  H(x1, x2)
f(x1, x2)  =  f
s(x1)  =  x1
b  =  b
F(x1, x2)  =  F(x1, x2)
a  =  a
g(x1, x2, x3)  =  g
h(x1, x2)  =  x1
c  =  c
d  =  d
e  =  e

Recursive Path Order [RPO].
Precedence:

[H₂, F₂] > [f, g] > [b, a, c]
d > [b, a, c]
e > [b, a, c]

The following usable rules [FROCOS05] were oriented:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

(10) Obligation:

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

H(f(x, s(y)), b) → F(a, g(y, a, f(s(x), a)))
F(x, g(x, a, f(s(x), y))) → F(h(x, b), g(a, b, y))
F(h(a, y), g(x, b, a)) → H(f(x, s(y)), s(b))

The TRS R consists of the following rules:

g(x, x, x) → g(c, d, e)
g(x, y, x) → g(c, d, e)
s(f(x, y)) → f(y, f(s(s(x)), a))
h(h(x, a), y) → h(h(a, y), h(a, x))
f(x, f(y, f(x, y))) → f(a, f(x, f(y, b)))
f(h(a, y), g(x, b, a)) → h(f(x, s(y)), s(b))
h(f(x, s(y)), b) → f(a, g(y, a, f(s(x), a)))
f(x, g(x, a, f(s(x), y))) → f(h(x, b), g(a, b, y))
s(y) → b

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