The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

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

f(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(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(y))) →⁺ f(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(y))].
The result substitution is [x / a(b(x))].

(2) BOUNDS(n^1, 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(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

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

Types:
f :: b:a → b:a → f
a :: b:a → b:a
b :: b:a → b:a
hole_f1_0 :: f
hole_b:a2_0 :: b:a
gen_b:a3_0 :: Nat → 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(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

Types:
f :: b:a → b:a → f
a :: b:a → b:a
b :: b:a → b:a
hole_f1_0 :: f
hole_b:a2_0 :: b:a
gen_b:a3_0 :: Nat → b:a

Generator Equations:
gen_b:a3_0(0) ⇔ hole_b:a2_0
gen_b:a3_0(+(x, 1)) ⇔ a(gen_b:a3_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(x, a(b(y))) → f(a(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))

Types:
f :: b:a → b:a → f
a :: b:a → b:a
b :: b:a → b:a
hole_f1_0 :: f
hole_b:a2_0 :: b:a
gen_b:a3_0 :: Nat → b:a

Generator Equations:
gen_b:a3_0(0) ⇔ hole_b:a2_0
gen_b:a3_0(+(x, 1)) ⇔ a(gen_b:a3_0(x))

No more defined symbols left to analyse.