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(f(x, y, z), u, f(x, y, v)) → f(x, y, f(z, u, v))
f(x, y, y) → y
f(x, y, g(y)) → x
f(x, x, y) → x
f(g(x), x, y) → y

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Recursive path order with status [RPO].
Precedence:

g₁ > f₃

Status:

f₃: [1,2,3]
g₁: multiset

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

f(f(x, y, z), u, f(x, y, v)) → f(x, y, f(z, u, v))
f(x, y, y) → y
f(x, y, g(y)) → x
f(x, x, y) → x
f(g(x), x, y) → y

(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