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(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

Rewrite Strategy: FULL

(1) CpxTrsToCpxRelTrsProof (BOTH BOUNDS(ID, ID) transformation)

Transformed TRS to relative TRS where S is empty.

(2) Obligation:

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

f(empty, l) → l
f(cons(x, k), l) → g(k, l, cons(x, k))
g(a, b, c) → f(a, cons(b, c))

S is empty.
Rewrite Strategy: FULL

(3) SlicingProof (LOWER BOUND(ID) transformation)

Sliced the following arguments:
cons/0
g/1

(4) Obligation:

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

f(empty, l) → l
f(cons(k), l) → g(k, cons(k))
g(a, c) → f(a, cons(c))

S is empty.
Rewrite Strategy: FULL

(5) DecreasingLoopProof (EQUIVALENT transformation)

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

(6) BOUNDS(n^1, INF)