The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

Runtime Complexity TRS:
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)

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

Runtime Complexity Relative TRS:
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)

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
a(s(s(x22014_0)), s(h), h) →⁺ a(x22014_0, a(s(x22014_0), s(h), h), a(s(x22014_0), s(h), h))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [x22014_0 / s(x22014_0)].
The result substitution is [ ].

The rewrite sequence
a(s(s(x22014_0)), s(h), h) →⁺ a(x22014_0, a(s(x22014_0), s(h), h), a(s(x22014_0), s(h), h))
gives rise to a decreasing loop by considering the right hand sides subterm at position [2].
The pumping substitution is [x22014_0 / s(x22014_0)].
The result substitution is [ ].

(4) BOUNDS(2^n, INF)