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:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))

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:

g(x, 0) → 0
g(d, s(x)) → s(s(g(d, x)))
g(h, s(0)) → 0
g(h, s(s(x))) → s(g(h, x))
double(x) → g(d, x)
half(x) → g(h, x)
f(s(x), y) → f(half(s(x)), double(y))
f(s(0), y) → y
id(x) → f(x, s(0))

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
g(d, s(x)) →⁺ s(s(g(d, x)))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0].
The pumping substitution is [x / s(x)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)