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)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
maplog(xs) → mapIter(xs, nil)
mapIter(xs, ys) → ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys) → ys
ifmap(false, xs, ys) → mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil) → true
isempty(cons(x, xs)) → false
last(nil) → error
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
droplast(nil) → nil
droplast(cons(x, nil)) → nil
droplast(cons(x, cons(y, xs))) → cons(x, droplast(cons(y, xs)))
a → b
a → c

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)
double(0) → 0
double(s(x)) → s(s(double(x)))
log(0) → logError
log(s(x)) → loop(s(x), s(0), 0)
loop(x, s(y), z) → if(le(x, s(y)), x, s(y), z)
if(true, x, y, z) → z
if(false, x, y, z) → loop(x, double(y), s(z))
maplog(xs) → mapIter(xs, nil)
mapIter(xs, ys) → ifmap(isempty(xs), xs, ys)
ifmap(true, xs, ys) → ys
ifmap(false, xs, ys) → mapIter(droplast(xs), cons(log(last(xs)), ys))
isempty(nil) → true
isempty(cons(x, xs)) → false
last(nil) → error
last(cons(x, nil)) → x
last(cons(x, cons(y, xs))) → last(cons(y, xs))
droplast(nil) → nil
droplast(cons(x, nil)) → nil
droplast(cons(x, cons(y, xs))) → cons(x, droplast(cons(y, xs)))
a → b
a → c

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)