In Section 2, after the account of various properties of the cozero map, it is claimed thatfor r > 0 in Q and arbitrary α, β ≥ 0 in RL. This is clearly false (although it does hold in the particular situation considered later on, where coz(α) ∨ coz(β) = e), and hence the proof of Lemma 2 as given is incomplete. It should be replaced by the following.Proof For any α, β ≥ 0 ∈ RL such that coz(α) ∨ coz(β) = e, we show (i) coz((α − β) + ) ∨ coz(β) = e, and (ii) coz((α − β) + ) ≺ ≺ coz(α).The following calculations specifically use the rules for ∨ and ∧ involving coz listed above as well as the fact that coz(sγ ) = coz(tγ ) for any γ ∈ RL and s, t > 0 in Q. Now, regarding (i)The online version of the original article can be found under