Erratum to "On the reflection invariance of residuated chains" [Ann. Pure Appl. Logic 161 (2009) 220-227]

In Section 2 replace the definition of ∗◦Q in Definition 1 by x∗◦Q y = inf{u ∗◦ v | u > x, v > y}. It is defined only if the infimum exists. Proposition 1 remains unchanged. Theorem 0. Let (X, ∗◦,→∗◦,≤) be a commutative residuated semigroup on a complete chain equipped with the order topology. Let a, b, c ∈ X be such that a = b→∗◦ c. Let (x, y) ∈ X × X be such that 1. neither x nor y equals the top element of the chain (if any) and we have x ∗◦ y = x∗◦Q y, 2. x is a-closed, and y is b-closed, 3. either we have sup{t→∗◦ c | t > x ∗◦ y} = x ∗◦ y→∗◦ c (-1) or we have sup{t→∗◦ c | t > (x→∗◦ a) ∗◦ (y→∗◦ b)} = (x→∗◦ a) ∗◦ (y→∗◦ b)→∗◦ c. (0) Then we have (x→∗◦ a) ∗◦ (y→∗◦ b) = x ∗◦ y→∗◦ c. (1) Proof. First assume (-1) holds. Note that ∗◦Q is an operation on X since the chain is complete. We have [x ∗◦ y] ∗◦ [(x→∗◦ a) ∗◦ (y→∗◦ b)] = [y ∗◦ (y→∗◦ b)] ∗◦ [x ∗◦ (x→∗◦ a)] ≤ b ∗◦ a = b ∗◦ (b→∗◦ c) ≤ c, and hence (x→∗◦ a) ∗◦ (y→∗◦ b) ≤ x ∗◦ y→∗◦ c . In order to prove (x→∗◦ a) ∗◦ (y→∗◦ b) ≥ x ∗◦ y→∗◦ c it suffices to show that (x→∗◦ a) ∗◦ (y→∗◦ b) > x1 ∗◦ y1→∗◦ c for x1 > x, y1 > y. Indeed, if it holds, then we have (x→∗◦ a) ∗◦ (y→∗◦ b) ≥ sup{x1 ∗◦ y1→∗◦ c | x1 > x, y1 > y}.We have inf{x1 ∗◦ y1 | x1 > x, y1 > y} = x∗◦Q y = x ∗◦ y by definition of ∗◦Q and condition 1 and hence, by (-1) we have sup{x1 ∗◦ y1→∗◦ c | x1 > x, y1 > y} = x ∗◦ y→∗◦ c , as stated. To this end, it suffices to verify [x1 ∗◦ y1] ∗◦ [(x→∗◦ a) ∗◦ (y→∗◦ b)] > c for x1 > x, y1 > y since X is a chain. Since x is a-closedwe have x→∗◦a > x1→∗◦a by Proposition 1/2. Analogouslywe obtain y→∗◦b > y1→∗◦b. Therefore x1∗◦(x→∗◦ a) > a