The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(2^n, INF).

(0) Obligation:

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

bsort(nil) → nil
bsort(.(x, y)) → last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) → nil
bubble(.(x, nil)) → .(x, nil)
bubble(.(x, .(y, z))) → if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) → 0
last(.(x, nil)) → x
last(.(x, .(y, z))) → last(.(y, z))
butlast(nil) → nil
butlast(.(x, nil)) → nil
butlast(.(x, .(y, z))) → .(x, butlast(.(y, z)))

Rewrite Strategy: INNERMOST

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(2ⁿ):
The rewrite sequence
bubble(.(x, .(y, z))) →⁺ if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [1,1].
The pumping substitution is [z / .(y, z)].
The result substitution is [ ].

The rewrite sequence
bubble(.(x, .(y, z))) →⁺ if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
gives rise to a decreasing loop by considering the right hand sides subterm at position [2,1].
The pumping substitution is [z / .(y, z)].
The result substitution is [x / y].

(2) BOUNDS(2^n, INF)