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

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

The TRS R consists of the following rules:

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

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 1 SCC with 2 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

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

Lexicographic path order with status [LPO].
Quasi-Precedence:

[B₁, c₃] > F₁ > f₁ > a

Status:

B₁: [1]
a: []
F₁: [1]
f₁: [1]
c₃: [1,3,2]

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

The TRS R consists of the following rules:

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

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