(0) Obligation:

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

f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(2) Obligation:

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

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

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

Infered types.

(4) Obligation:

TRS:
Rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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 :: b:a:c → b:a:c → f
a :: b:a:c → b:a:c
b :: b:a:c → b:a:c
c :: b:a:c → b:a:c
hole_f1_0 :: f
hole_b:a:c2_0 :: b:a:c
gen_b:a:c3_0 :: Nat → b:a:c

(5) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
f

(6) Obligation:

TRS:
Rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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 :: b:a:c → b:a:c → f
a :: b:a:c → b:a:c
b :: b:a:c → b:a:c
c :: b:a:c → b:a:c
hole_f1_0 :: f
hole_b:a:c2_0 :: b:a:c
gen_b:a:c3_0 :: Nat → b:a:c

Generator Equations:
gen_b:a:c3_0(0) ⇔ hole_b:a:c2_0
gen_b:a:c3_0(+(x, 1)) ⇔ a(gen_b:a:c3_0(x))

The following defined symbols remain to be analysed:
f

(7) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
f(gen_b:a:c3_0(+(1, n5_0)), gen_b:a:c3_0(b)) → *4_0, rt ∈ Ω(n50)

Induction Base:
f(gen_b:a:c3_0(+(1, 0)), gen_b:a:c3_0(b))

Induction Step:
f(gen_b:a:c3_0(+(1, +(n5_0, 1))), gen_b:a:c3_0(b)) →RΩ(1)
f(gen_b:a:c3_0(+(1, n5_0)), a(gen_b:a:c3_0(b))) →IH
*4_0

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

(8) Complex Obligation (BEST)

(9) Obligation:

TRS:
Rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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 :: b:a:c → b:a:c → f
a :: b:a:c → b:a:c
b :: b:a:c → b:a:c
c :: b:a:c → b:a:c
hole_f1_0 :: f
hole_b:a:c2_0 :: b:a:c
gen_b:a:c3_0 :: Nat → b:a:c

Lemmas:
f(gen_b:a:c3_0(+(1, n5_0)), gen_b:a:c3_0(b)) → *4_0, rt ∈ Ω(n50)

Generator Equations:
gen_b:a:c3_0(0) ⇔ hole_b:a:c2_0
gen_b:a:c3_0(+(x, 1)) ⇔ a(gen_b:a:c3_0(x))

No more defined symbols left to analyse.

(10) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
f(gen_b:a:c3_0(+(1, n5_0)), gen_b:a:c3_0(b)) → *4_0, rt ∈ Ω(n50)

(11) BOUNDS(n^1, INF)

(12) Obligation:

TRS:
Rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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 :: b:a:c → b:a:c → f
a :: b:a:c → b:a:c
b :: b:a:c → b:a:c
c :: b:a:c → b:a:c
hole_f1_0 :: f
hole_b:a:c2_0 :: b:a:c
gen_b:a:c3_0 :: Nat → b:a:c

Lemmas:
f(gen_b:a:c3_0(+(1, n5_0)), gen_b:a:c3_0(b)) → *4_0, rt ∈ Ω(n50)

Generator Equations:
gen_b:a:c3_0(0) ⇔ hole_b:a:c2_0
gen_b:a:c3_0(+(x, 1)) ⇔ a(gen_b:a:c3_0(x))

No more defined symbols left to analyse.

(13) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n1) was proven with the following lemma:
f(gen_b:a:c3_0(+(1, n5_0)), gen_b:a:c3_0(b)) → *4_0, rt ∈ Ω(n50)

(14) BOUNDS(n^1, INF)