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:

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, z)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

[nil, 0] > [g₂, rem₂] > [norm₁, s₁]
f₂ > [g₂, rem₂] > [norm₁, s₁]

Status:

norm₁: multiset
nil: multiset
0: multiset
g₂: [1,2]
s₁: multiset
f₂: multiset
rem₂: [2,1]

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

norm(nil) → 0
norm(g(x, y)) → s(norm(x))
f(x, nil) → g(nil, x)
f(x, g(y, z)) → g(f(x, y), z)
rem(nil, y) → nil
rem(g(x, y), 0) → g(x, y)
rem(g(x, y), s(z)) → rem(x, 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