The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

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

(2) BOUNDS(n^1, INF)

(3) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(4) Obligation:

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

f(g(X)) → g(f(f(X)))
f(h(X)) → h(g(X))

S is empty.
Rewrite Strategy: FULL

(5) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
h/0

(6) Obligation:

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

f(g(X)) → g(f(f(X)))
f(h) → h

S is empty.
Rewrite Strategy: FULL

(7) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)

Infered types.

(8) Obligation:

TRS:
Rules:
f(g(X)) → g(f(f(X)))
f(h) → h

Types:
f :: g:h → g:h
g :: g:h → g:h
h :: g:h
hole_g:h1_0 :: g:h
gen_g:h2_0 :: Nat → g:h

(9) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
f

(10) Obligation:

TRS:
Rules:
f(g(X)) → g(f(f(X)))
f(h) → h

Types:
f :: g:h → g:h
g :: g:h → g:h
h :: g:h
hole_g:h1_0 :: g:h
gen_g:h2_0 :: Nat → g:h

Generator Equations:
gen_g:h2_0(0) ⇔ h
gen_g:h2_0(+(x, 1)) ⇔ g(gen_g:h2_0(x))

The following defined symbols remain to be analysed:
f

(11) RewriteLemmaProof (EQUIVALENT transformation)

Proved the following rewrite lemma:
f(gen_g:h2_0(n4_0)) → gen_g:h2_0(n4_0), rt ∈ Ω(2ⁿ)

Induction Base:
f(gen_g:h2_0(0)) →_R^Ω(1)
h

Induction Step:
f(gen_g:h2_0(+(n4_0, 1))) →_R^Ω(1)
g(f(f(gen_g:h2_0(n4_0)))) →_IH
g(f(gen_g:h2_0(c5_0))) →_IH
g(gen_g:h2_0(c5_0))

We have rt ∈ Ω(2ⁿ) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(2ⁿ)

(12) BOUNDS(2^n, INF)