### (0) Obligation:

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

f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

Rewrite Strategy: FULL

### (1) DecreasingLoopProof (EQUIVALENT transformation)

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

### (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(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

S is empty.
Rewrite Strategy: FULL

Infered types.

### (6) Obligation:

TRS:
Rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

Types:
f :: c:b:a → c:b:a → f
a :: c:b:a → c:b:a
b :: c:b:a → c:b:a
c :: c:b:a → c:b:a
hole_f1_0 :: f
hole_c:b:a2_0 :: c:b:a
gen_c:b:a3_0 :: Nat → c:b:a

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

Heuristically decided to analyse the following defined symbols:
f

### (8) Obligation:

TRS:
Rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

Types:
f :: c:b:a → c:b:a → f
a :: c:b:a → c:b:a
b :: c:b:a → c:b:a
c :: c:b:a → c:b:a
hole_f1_0 :: f
hole_c:b:a2_0 :: c:b:a
gen_c:b:a3_0 :: Nat → c:b:a

Generator Equations:
gen_c:b:a3_0(0) ⇔ hole_c:b:a2_0
gen_c:b:a3_0(+(x, 1)) ⇔ a(gen_c:b:a3_0(x))

The following defined symbols remain to be analysed:
f

### (9) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
f(gen_c:b:a3_0(+(1, n5_0)), gen_c:b:a3_0(b)) → *4_0, rt ∈ Ω(n50)

Induction Base:
f(gen_c:b:a3_0(+(1, 0)), gen_c:b:a3_0(b))

Induction Step:
f(gen_c:b:a3_0(+(1, +(n5_0, 1))), gen_c:b:a3_0(b)) →RΩ(1)
f(gen_c:b:a3_0(+(1, n5_0)), a(gen_c:b:a3_0(b))) →IH
*4_0

We have rt ∈ Ω(n1) and sz ∈ O(n). Thus, we have ircR ∈ Ω(n).

### (11) Obligation:

TRS:
Rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

Types:
f :: c:b:a → c:b:a → f
a :: c:b:a → c:b:a
b :: c:b:a → c:b:a
c :: c:b:a → c:b:a
hole_f1_0 :: f
hole_c:b:a2_0 :: c:b:a
gen_c:b:a3_0 :: Nat → c:b:a

Lemmas:
f(gen_c:b:a3_0(+(1, n5_0)), gen_c:b:a3_0(b)) → *4_0, rt ∈ Ω(n50)

Generator Equations:
gen_c:b:a3_0(0) ⇔ hole_c:b:a2_0
gen_c:b:a3_0(+(x, 1)) ⇔ a(gen_c:b:a3_0(x))

No more defined symbols left to analyse.

### (12) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
f(gen_c:b:a3_0(+(1, n5_0)), gen_c:b:a3_0(b)) → *4_0, rt ∈ Ω(n50)

### (14) Obligation:

TRS:
Rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))

Types:
f :: c:b:a → c:b:a → f
a :: c:b:a → c:b:a
b :: c:b:a → c:b:a
c :: c:b:a → c:b:a
hole_f1_0 :: f
hole_c:b:a2_0 :: c:b:a
gen_c:b:a3_0 :: Nat → c:b:a

Lemmas:
f(gen_c:b:a3_0(+(1, n5_0)), gen_c:b:a3_0(b)) → *4_0, rt ∈ Ω(n50)

Generator Equations:
gen_c:b:a3_0(0) ⇔ hole_c:b:a2_0
gen_c:b:a3_0(+(x, 1)) ⇔ a(gen_c:b:a3_0(x))

No more defined symbols left to analyse.

### (15) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
f(gen_c:b:a3_0(+(1, n5_0)), gen_c:b:a3_0(b)) → *4_0, rt ∈ Ω(n50)