The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0, nil))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x), l))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
cons/1

(4) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
lastbit(0) → 0
lastbit(s(0)) → s(0)
lastbit(s(s(x))) → lastbit(x)
zero(0) → true
zero(s(x)) → false
conv(x) → conviter(x, cons(0))
conviter(x, l) → if(zero(x), x, l)
if(true, x, l) → l
if(false, x, l) → conviter(half(x), cons(lastbit(x)))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
half(s(s(x))) →⁺ s(half(x))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x / s(s(x))].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)