Page 155 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 155
IV: Skew Boolean Algebras
(aʹ \ b) ∨ bʹ = (aʹ \ (aʹ∧b)) ∨ bʹ = (aʹ \ a) ∨ (aʹ \ bʹ)∨bʹ = (aʹ \ bʹ)∨bʹ = aʹ∨bʹ.
Similarly (bʹ \ a) ∨ aʹ = aʹ∨bʹ so that (aʹ \ b) ∨ bʹ = (bʹ \ a) ∨ aʹ follows. Conversely given the
latter, one has aʹ∧[(aʹ \ b) ∨ bʹ]∧b = aʹ∧[(bʹ \ a) ∨ aʹ]∧b which yields aʹ∧bʹ = aʹ∧b. Similarly,
a∧[(aʹ \ b) ∨ bʹ]∧bʹ = a∧[(bʹ \ a) ∨ aʹ]∧bʹ yields a∧bʹ = aʹ∧b. £
Lemma 4.5.4. In ω(B):
i) (a, aʹ) ≻ (b, bʹ) if and only if a ≥ b in B.
ii) (a, aʹ) D (b, bʹ) if and only if a = b in B.
iii) (a, aʹ) ≥ (b, bʹ) if and only if a ≥ b in B with bʹ = aʹ ∧ b.
Proof. Since (b, bʹ) ∧ (a, aʹ) ∧ (b, bʹ) = (a∧b, bʹ∧a∧b), we have (b, bʹ) ∧ (a, aʹ) ∧ (b, bʹ) = (b, bʹ)
if and only if a∧b = b in B. Thus (i) holds and (ii) follows immediately. As for (iii),
(a, aʹ) ∧ (b, bʹ) = (b, bʹ) = (b, bʹ) ∧ (a, aʹ)
becomes aʹ∧b = bʹ = bʹ∧a in the second coordinate, and b = a∧b in the first coordinate. But these
conditions, plus b ≥ bʹ are equivalent to both a ≥ b and bʹ = aʹ∧b holding in B. £
Theorem 4.5.5. Upon letting (a, aʹ) \ (b, bʹ) be the complement of (a, aʹ) ∧ (b, bʹ) in the
Boolean sublattice ⎡(a, aʹ)⎤, (ω(B); ∨, ∧, \, (0, 0)) becomes a skew Boolean algebra for which
ω(B)/D ≅ B. Moreover, ω(B) has finite intersections. Thus, as a skew Boolean ∩-algebra
ω(B) has a distributive congruence lattice.
Proof. ω(B) is symmetric by Lemma 4.5.3. Since ⎡(a, aʹ)⎤ = {(b, aʹ∧b)⎮a ≥ b} ≅ ⎡a⎤ by Lemma
4.5.4, we see that ⎡(a, aʹ)⎤ is a Boolean lattice, making ω(B) also normal and hence a skew
Boolean lattice. The first assertion is now clear.
Next, (a, aʹ) ∩ (b, bʹ) is given by ([aʹ∧bʹ] + [(a∧b) \ (aʹ∨bʹ)], aʹ∧bʹ) where + is the
symmetric difference in B: x + y = (x\y) ∨ (y\x). That
(a, aʹ), (b, bʹ) ≥ ([aʹ∧bʹ] + [(a∧b) \ (aʹ∨bʹ)], aʹ∧bʹ)
follows from Lemma 4.5.4 and basic Boolean algebra. So let (a, aʹ), (b, bʹ) ≥ (c, cʹ). Then
a∧b ≥ c and aʹ∧bʹ ≥ cʹ with aʹ∧c = cʹ and bʹ∧c = cʹ imply
c ∧ {[aʹ∧bʹ] + [a∧b \ (aʹ∨bʹ)]} = cʹ + (c \ cʹ) = c and c ∧ [aʹ∧bʹ] = cʹ∧cʹ = cʹ
so that ([aʹ∧bʹ] + [a∧b \ (aʹ∨bʹ)], aʹ∧bʹ) ≥ (c, cʹ). Thus (a, aʹ) ∩ (b, bʹ) is as stated. The final
assertion follows from Theorem 4.4.3. £
153
(aʹ \ b) ∨ bʹ = (aʹ \ (aʹ∧b)) ∨ bʹ = (aʹ \ a) ∨ (aʹ \ bʹ)∨bʹ = (aʹ \ bʹ)∨bʹ = aʹ∨bʹ.
Similarly (bʹ \ a) ∨ aʹ = aʹ∨bʹ so that (aʹ \ b) ∨ bʹ = (bʹ \ a) ∨ aʹ follows. Conversely given the
latter, one has aʹ∧[(aʹ \ b) ∨ bʹ]∧b = aʹ∧[(bʹ \ a) ∨ aʹ]∧b which yields aʹ∧bʹ = aʹ∧b. Similarly,
a∧[(aʹ \ b) ∨ bʹ]∧bʹ = a∧[(bʹ \ a) ∨ aʹ]∧bʹ yields a∧bʹ = aʹ∧b. £
Lemma 4.5.4. In ω(B):
i) (a, aʹ) ≻ (b, bʹ) if and only if a ≥ b in B.
ii) (a, aʹ) D (b, bʹ) if and only if a = b in B.
iii) (a, aʹ) ≥ (b, bʹ) if and only if a ≥ b in B with bʹ = aʹ ∧ b.
Proof. Since (b, bʹ) ∧ (a, aʹ) ∧ (b, bʹ) = (a∧b, bʹ∧a∧b), we have (b, bʹ) ∧ (a, aʹ) ∧ (b, bʹ) = (b, bʹ)
if and only if a∧b = b in B. Thus (i) holds and (ii) follows immediately. As for (iii),
(a, aʹ) ∧ (b, bʹ) = (b, bʹ) = (b, bʹ) ∧ (a, aʹ)
becomes aʹ∧b = bʹ = bʹ∧a in the second coordinate, and b = a∧b in the first coordinate. But these
conditions, plus b ≥ bʹ are equivalent to both a ≥ b and bʹ = aʹ∧b holding in B. £
Theorem 4.5.5. Upon letting (a, aʹ) \ (b, bʹ) be the complement of (a, aʹ) ∧ (b, bʹ) in the
Boolean sublattice ⎡(a, aʹ)⎤, (ω(B); ∨, ∧, \, (0, 0)) becomes a skew Boolean algebra for which
ω(B)/D ≅ B. Moreover, ω(B) has finite intersections. Thus, as a skew Boolean ∩-algebra
ω(B) has a distributive congruence lattice.
Proof. ω(B) is symmetric by Lemma 4.5.3. Since ⎡(a, aʹ)⎤ = {(b, aʹ∧b)⎮a ≥ b} ≅ ⎡a⎤ by Lemma
4.5.4, we see that ⎡(a, aʹ)⎤ is a Boolean lattice, making ω(B) also normal and hence a skew
Boolean lattice. The first assertion is now clear.
Next, (a, aʹ) ∩ (b, bʹ) is given by ([aʹ∧bʹ] + [(a∧b) \ (aʹ∨bʹ)], aʹ∧bʹ) where + is the
symmetric difference in B: x + y = (x\y) ∨ (y\x). That
(a, aʹ), (b, bʹ) ≥ ([aʹ∧bʹ] + [(a∧b) \ (aʹ∨bʹ)], aʹ∧bʹ)
follows from Lemma 4.5.4 and basic Boolean algebra. So let (a, aʹ), (b, bʹ) ≥ (c, cʹ). Then
a∧b ≥ c and aʹ∧bʹ ≥ cʹ with aʹ∧c = cʹ and bʹ∧c = cʹ imply
c ∧ {[aʹ∧bʹ] + [a∧b \ (aʹ∨bʹ)]} = cʹ + (c \ cʹ) = c and c ∧ [aʹ∧bʹ] = cʹ∧cʹ = cʹ
so that ([aʹ∧bʹ] + [a∧b \ (aʹ∨bʹ)], aʹ∧bʹ) ≥ (c, cʹ). Thus (a, aʹ) ∩ (b, bʹ) is as stated. The final
assertion follows from Theorem 4.4.3. £
153
