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:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(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:

le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(x), y) → plus(y, times(x, y))
log(x, 0) → baseError
log(x, s(0)) → baseError
log(0, s(s(b))) → logZeroError
log(s(x), s(s(b))) → loop(s(x), s(s(b)), s(0), 0)
loop(x, s(s(b)), s(y), z) → if(le(x, s(y)), x, s(s(b)), s(y), z)
if(true, x, b, y, z) → z
if(false, x, b, y, z) → loop(x, b, times(b, y), s(z))

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
le(s(x), s(y)) →⁺ le(x, y)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x / s(x), y / s(y)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)