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, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
f(cons(a, k), cons(a358_0, k359_0)) →⁺ f(k, k359_0)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [k / cons(a, k), k359_0 / cons(a358_0, k359_0)].
The result substitution is [ ].

(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, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

S is empty.
Rewrite Strategy: FULL

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

Infered types.

(6) Obligation:

TRS:
Rules:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

Types:
f :: empty:cons → empty:cons → empty:cons
empty :: empty:cons
cons :: a → empty:cons → empty:cons
hole_empty:cons1_0 :: empty:cons
hole_a2_0 :: a
gen_empty:cons3_0 :: Nat → empty:cons

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

Heuristically decided to analyse the following defined symbols:
f

(8) Obligation:

TRS:
Rules:
f(x, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

Types:
f :: empty:cons → empty:cons → empty:cons
empty :: empty:cons
cons :: a → empty:cons → empty:cons
hole_empty:cons1_0 :: empty:cons
hole_a2_0 :: a
gen_empty:cons3_0 :: Nat → empty:cons

Generator Equations:
gen_empty:cons3_0(0) ⇔ empty
gen_empty:cons3_0(+(x, 1)) ⇔ cons(hole_a2_0, gen_empty:cons3_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, empty) → x
f(empty, cons(a, k)) → f(cons(a, k), k)
f(cons(a, k), y) → f(y, k)

Types:
f :: empty:cons → empty:cons → empty:cons
empty :: empty:cons
cons :: a → empty:cons → empty:cons
hole_empty:cons1_0 :: empty:cons
hole_a2_0 :: a
gen_empty:cons3_0 :: Nat → empty:cons

Generator Equations:
gen_empty:cons3_0(0) ⇔ empty
gen_empty:cons3_0(+(x, 1)) ⇔ cons(hole_a2_0, gen_empty:cons3_0(x))

No more defined symbols left to analyse.