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:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

sum₁ > +₂ > s₁
sum₁ > 0
nil > 0
nil > s₁
cons₂ > *₂ > +₂ > s₁
cons₂ > *₂ > 0
cons₂ > prod₁ > 0
cons₂ > prod₁ > s₁

Status:

cons₂: [1,2]
*₂: [1,2]
prod₁: [1]
sum₁: [1]
s₁: [1]
0: []
nil: []
+₂: [1,2]

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

(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