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:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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:

log(x, s(s(y))) → cond(le(x, s(s(y))), x, y)
cond(true, x, y) → s(0)
cond(false, x, y) → double(log(x, square(s(s(y)))))
le(0, v) → true
le(s(u), 0) → false
le(s(u), s(v)) → le(u, v)
double(0) → 0
double(s(x)) → s(s(double(x)))
square(0) → 0
square(s(x)) → s(plus(square(x), double(x)))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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

(4) BOUNDS(n^1, INF)