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:

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), l))

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

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

+₂ > s₁ > nil
app₂ > cons₂ > h > a₄ > s₁ > nil
app₂ > cons₂ > h > 1 > nil
sum₁ > cons₂ > h > a₄ > s₁ > nil
sum₁ > cons₂ > h > 1 > nil

Status:

cons₂: [2,1]
a₄: [1,2,3,4]
sum₁: [1]
s₁: [1]
app₂: [1,2]
h: []
1: []
nil: []
+₂: [1,2]

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

a(h, h, h, x) → s(x)
a(l, x, s(y), h) → a(l, x, y, s(h))
a(l, x, s(y), s(z)) → a(l, x, y, a(l, x, s(y), z))
a(l, s(x), h, z) → a(l, x, z, z)
a(s(l), h, h, z) → a(l, z, h, z)
+(x, h) → x
+(h, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
s(h) → 1
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, h), 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