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)) → f(a(g(g(f(x))), g(f(x))))

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
f(g(x)) →⁺ f(a(g(g(f(x))), g(f(x))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,0,0,0].
The pumping substitution is [x / g(x)].
The result substitution is [ ].

The rewrite sequence
f(g(x)) →⁺ f(a(g(g(f(x))), g(f(x))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [0,1,0].
The pumping substitution is [x / g(x)].
The result substitution is [ ].

(2) BOUNDS(2^n, 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)) → f(a(g(g(f(x))), g(f(x))))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
f(g(x)) → f(a(g(g(f(x))), g(f(x))))

Types:
f :: g:a → g:a
g :: g:a → g:a
a :: g:a → g:a → g:a
hole_g:a1_0 :: g:a
gen_g:a2_0 :: Nat → g:a

(7) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
f

(8) Obligation:

TRS:
Rules:
f(g(x)) → f(a(g(g(f(x))), g(f(x))))

Types:
f :: g:a → g:a
g :: g:a → g:a
a :: g:a → g:a → g:a
hole_g:a1_0 :: g:a
gen_g:a2_0 :: Nat → g:a

Generator Equations:
gen_g:a2_0(0) ⇔ hole_g:a1_0
gen_g:a2_0(+(x, 1)) ⇔ g(gen_g:a2_0(x))

The following defined symbols remain to be analysed:
f

(9) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

Could not prove a rewrite lemma for the defined symbol f.

(10) Obligation:

TRS:
Rules:
f(g(x)) → f(a(g(g(f(x))), g(f(x))))

Types:
f :: g:a → g:a
g :: g:a → g:a
a :: g:a → g:a → g:a
hole_g:a1_0 :: g:a
gen_g:a2_0 :: Nat → g:a

Generator Equations:
gen_g:a2_0(0) ⇔ hole_g:a1_0
gen_g:a2_0(+(x, 1)) ⇔ g(gen_g:a2_0(x))

No more defined symbols left to analyse.