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:

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

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:

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

The TRS R consists of the following rules:

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

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 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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.

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
B(x1, x2)  =  B(x1, x2)
b(x1, x2)  =  b(x1, x2)
c(x1, x2, x3)  =  c(x1, x2)
a  =  a
f(x1)  =  f(x1)
C(x1, x2, x3)  =  C(x1, x3)

Recursive Path Order [RPO].
Precedence:

[B₂, b₂, c₂, C₂] > [a, f₁]

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

F(c(x, y, z)) → F(b(y, z))

The TRS R consists of the following rules:

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

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