### (0) Obligation:

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

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

Rewrite Strategy: FULL

### (1) DecreasingLoopProof (EQUIVALENT transformation)

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

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

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

S is empty.
Rewrite Strategy: FULL

Infered types.

### (6) Obligation:

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

Types:
g :: f → f
f :: f → f → f
hole_f1_0 :: f
gen_f2_0 :: Nat → f

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

Heuristically decided to analyse the following defined symbols:
g

### (8) Obligation:

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

Types:
g :: f → f
f :: f → f → f
hole_f1_0 :: f
gen_f2_0 :: Nat → f

Generator Equations:
gen_f2_0(0) ⇔ hole_f1_0
gen_f2_0(+(x, 1)) ⇔ f(hole_f1_0, gen_f2_0(x))

The following defined symbols remain to be analysed:
g

### (9) RewriteLemmaProof (EQUIVALENT transformation)

Proved the following rewrite lemma:
g(gen_f2_0(+(1, n4_0))) → *3_0, rt ∈ Ω(2n)

Induction Base:
g(gen_f2_0(+(1, 0)))

Induction Step:
g(gen_f2_0(+(1, +(n4_0, 1)))) →RΩ(1)
f(f(g(g(hole_f1_0)), g(g(gen_f2_0(+(1, n4_0))))), f(g(g(hole_f1_0)), g(g(gen_f2_0(+(1, n4_0)))))) →IH
f(f(g(g(hole_f1_0)), g(*3_0)), f(g(g(hole_f1_0)), g(g(gen_f2_0(+(1, n4_0)))))) →IH
f(f(g(g(hole_f1_0)), g(*3_0)), f(g(g(hole_f1_0)), g(*3_0)))

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