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:
Combined order from the following AFS and order.
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sum(x1)  =  x1
a(x1, x2, x3)  =  a(x1, x2, x3)
h  =  h
s(x1)  =  s(x1)

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

app₂ > [cons₂, h] > nil > s₁
app₂ > [cons₂, h] > a₃ > s₁

Status:

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

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, 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:
The TRS R consists of the following rules:

sum(cons(x, nil)) → cons(x, nil)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(cons(x₁, x₂)) = x₁ + x₂   
POL(nil) = 0   
POL(sum(x₁)) = 1 + x₁   

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

sum(cons(x, nil)) → cons(x, nil)

(4) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(5) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(6) TRUE

(7) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(8) TRUE

(9) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(10) TRUE