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(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(z)))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

f₃ > 0 > s₁

Status:

f₃: [1,2,3]
s₁: multiset
0: multiset

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

f(x, 0, 0) → s(x)
f(0, y, 0) → s(y)
f(0, 0, z) → s(z)
f(s(0), y, z) → f(0, s(y), s(z))
f(s(x), s(y), 0) → f(x, y, s(0))
f(s(x), 0, s(z)) → f(x, s(0), z)
f(0, s(0), s(0)) → s(s(0))
f(s(x), s(y), s(z)) → f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) → f(0, y, s(0))
f(0, s(0), s(s(z))) → f(0, s(0), z)
f(0, s(s(y)), s(s(z))) → f(0, y, f(0, s(s(y)), s(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