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(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

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

The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

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

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

Recursive Path Order [RPO].
Precedence:

a > b > F₁
a > b > c

The following usable rules [FROCOS05] were oriented:

f(x, f(y, z)) → f(f(x, y), z)
f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))

(4) Obligation:

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

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

The TRS R consists of the following rules:

f(f(a, b), x) → f(b, f(a, f(c, f(b, f(a, x)))))
f(x, f(y, z)) → f(f(x, y), z)

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