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:

rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))

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:

rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(x, k), a) → r1(k, cons(x, a))

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:

rev(ls) → r1(ls, empty)
r1(empty, a) → a
r1(cons(k), a) → r1(k, cons(a))

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
r1(cons(k), a) →⁺ r1(k, cons(a))
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)