Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

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

The TRS R consists of the following rules:

f(c(c(a, y, a), b(x, z), a)) → b(y, f(c(f(a), z, z)))
f(b(b(x, f(y)), z)) → c(z, x, f(b(b(f(a), y), y)))
c(b(a, a), b(y, z), x) → b(a, b(z, 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 5 less nodes.

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

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

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

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

[f₁, c₂] > [F₁, b₂, a]

Status:

F₁: [1]
b₂: [2,1]
f₁: [1]
a: multiset
c₂: [1,2]

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE