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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

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(a, empty) → g(a, empty)
f(a, cons(x, k)) → f(cons(x, a), k)
g(empty, d) → d
g(cons(x, k), d) → g(k, cons(x, d))

S is empty.
Rewrite Strategy: FULL

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

Sliced the following arguments:
cons/0

(4) Obligation:

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

f(a, empty) → g(a, empty)
f(a, cons(k)) → f(cons(a), k)
g(empty, d) → d
g(cons(k), d) → g(k, cons(d))

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(a, cons(k)) →⁺ f(cons(a), 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 [a / cons(a)].

(6) BOUNDS(n^1, INF)