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(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

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(x) → if(ge(x, s(s(0))), x)
if(false, x) → 0
if(true, x) → s(half(p(p(x))))
p(0) → 0
p(s(x)) → x
ge(x, 0) → true
ge(0, s(x)) → false
ge(s(x), s(y)) → ge(x, y)
log(0) → 0
log(s(x)) → s(log(half(s(x))))

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
half(s(s(x97_1))) →⁺ s(half(x97_1))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x97_1 / s(s(x97_1))].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)