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:
Lexicographic path order with status [LPO].
Precedence:

nil > g₂ > norm₁ > 0
nil > g₂ > norm₁ > s₁
f₂ > g₂ > norm₁ > 0
f₂ > g₂ > norm₁ > s₁

Status:

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

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