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:

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, z)

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

app₂ > cons₂ > sum₁ > nil
app₂ > cons₂ > sum₁ > h > s₁
app₂ > cons₂ > a₃ > h > s₁

Status:

cons₂: [2,1]
a₃: [1,2,3]
sum₁: [1]
s₁: [1]
app₂: [1,2]
h: []
nil: []

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

app(nil, k) → k
app(l, nil) → l
app(cons(x, l), k) → cons(x, app(l, k))
sum(cons(x, nil)) → cons(x, nil)
sum(cons(x, cons(y, l))) → sum(cons(a(x, y, h), l))
a(h, h, x) → s(x)
a(x, s(y), h) → a(x, y, s(h))
a(x, s(y), s(z)) → a(x, y, a(x, s(y), z))
a(s(x), h, z) → a(x, z, 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