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:

f(x, y) → g(x, y)
g(h(x), y) → h(f(x, y))
g(h(x), y) → h(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:

f(x, y) → g(x, y)
g(h(x), y) → h(f(x, y))
g(h(x), y) → h(g(x, y))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
f/1
g/1

(4) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

f(x) → g(x)
g(h(x)) → h(f(x))
g(h(x)) → h(g(x))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
f(h(x2_0)) →⁺ h(f(x2_0))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0].
The pumping substitution is [x2_0 / h(x2_0)].
The result substitution is [ ].

(6) BOUNDS(n^1, INF)