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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

f₂ > empty > cons₂
f₂ > g₂ > cons₂

Status:

cons₂: multiset
g₂: [1,2]
f₂: [2,1]
empty: multiset

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

f(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

(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