Page 17 - Textos de Matemática Vol. 41

1.4. Pseudocomplements and complements 3 (H9) for every a,b∈L, b=(b∨a)∧(a→b).
Proof. (H1) is a special case of (-distr).
(H2): x ≤ a→a iff x ∧ a ≤ a, that is, always; x ≤ 1→a iff x = x ∧ 1 ≤ a. (H3): 1 ≤ a→b iff a = 1 ∧ a ≤ b.
(H4): since a ∧ b ≤ a.
(H5): by (H1) and (H2).
(H6): a ∧ (a→b) ≤ b since a→b ≤ a→b, and a ∧ b ≤ a ∧ (a→b) by (H4). (H7) follows from (H1) and (H6).
(H8): x ≤ (a ∧ b)→c iff x ∧ a ∧ b ≤ c iff x ∧ a ≤ b→c iff x ≤ a→(b→c). (H9): b ≤ (b ∨ a) ∧ (a → b) by (H4); by distributivity and (H6),
(b ∨ a) ∧ (a → b) = (b ∧ (a → b)) ∨ (a ∧ (a → b)) ≤ b.

1.4. Pseudocomplements and complements. An element b is a pseudocomple- ment of a in a lattice (with bottom 0) if
a∧x=0 iff x≤b. (Psc)
It does not necessarily exist; if it does it is obviously uniquely determined. It will be denoted by a∗.
In a Heyting algebra pseudocomplements exist, namely a∗ =a→0.
1.4.1. Proposition. We have (1) a≤a∗∗,
(2) a≤b ⇒ a∗ ≥b∗,
(3) a∗∗∗ = a∗,
(4) 0∗ is the top 1 of the lattice in question and 1∗ = 0, (5) in any Heyting algebra, (i∈J ai)∗ = i∈J a∗i .