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:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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:

h(e(x), y) → h(d(x, y), s(y))
d(g(g(0, x), y), s(z)) → g(e(x), d(g(g(0, x), y), z))
d(g(g(0, x), y), 0) → e(y)
d(g(0, x), y) → e(x)
d(g(x, y), z) → g(d(x, z), e(y))
g(e(x), e(y)) → e(g(x, y))

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

(4) BOUNDS(n^1, INF)