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) QTRSRRRProof (EQUIVALENT transformation)

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

[f₁, c₃] > [a, b₂]

Status:

a: []
f₁: [1]
b₂: [2,1]
c₃: [1,3,2]

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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))

(2) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(3) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(4) TRUE