### (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 [ ].

### (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

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) RewriteLemmaProof (EQUIVALENT transformation)

Proved the following rewrite lemma:
f(gen_g:a2_0(+(1, n4_0))) → *3_0, rt ∈ Ω(2n)

Induction Base:
f(gen_g:a2_0(+(1, 0)))

Induction Step:
f(gen_g:a2_0(+(1, +(n4_0, 1)))) →RΩ(1)
f(a(g(g(f(gen_g:a2_0(+(1, n4_0))))), g(f(gen_g:a2_0(+(1, n4_0)))))) →IH
f(a(g(g(*3_0)), g(f(gen_g:a2_0(+(1, n4_0)))))) →IH
f(a(g(g(*3_0)), g(*3_0)))

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