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:

sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)

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:

sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)

S is empty.
Rewrite Strategy: FULL

(3) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
sort(cons(x, y)) →⁺ insert(x, sort(y))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1].
The pumping substitution is [y / cons(x, y)].
The result substitution is [ ].

(4) BOUNDS(n^1, INF)