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:

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

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

The TRS R consists of the following rules:

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

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 9 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

Recursive path order with status [RPO].
Quasi-Precedence:

c₃ > C₁ > b₂ > a
f > a

Status:

b₂: [2,1]
c₃: multiset

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

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

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

The TRS R consists of the following rules:

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

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