Page 269 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 269
VII: Further Topics in Skew Boolean Algebras
Proof. First, note that for left normal bands ≤ is described by b ≤ a if a ∧ b = b and ⎡a⎤ is the set
{a ∧ s⎮s ∈S}. Let b, c lie in ⎡a⎤. By Lemma 4.4.9, b \ c = (a ∧ b) \ (a ∧ c) = a ∧ (b \ c) ∈ ⎡a⎤
also. In addition 0 = a ∧ 0 ∈ ⎡a⎤ so that ⎡a⎤ is seen to be an iBCS-subalgebra of S. Since ∧ is
commutative on (⎡a⎤; \, 0), the latter satisfies the iBCK identities (i) – (iv) above and in particular
(ii) which implies that ∧ is commutative on ⎡a⎤. Since (⎡a⎤; \, 0) is an iBCK-algebra with
maximal element a, it forms a Boolean lattice (⎡a⎤; ∨, ∧, \, a, 0) with ∧ and \ as already given, and
x ∨ y defined as a \ [(a \ x) ∧ (a \ y)] for all x, y ≤ a.
Conversely, let (S; ∧, 0) be a left normal band with zero such that for each a ∈ S the set
⎡a⎤ is a Boolean lattice under ≤. Let a \ b denote the usual relative complement a \ a∧b in ⎡a⎤.
The clearly B6 holds. B7 reduces to
(a \ a∧b) \ [(a \ a∧b) ∧ c] ≈ (a \ a∧c) \ [(a \ a∧c) ∧ b].
This holds in the Boolean case of ⎡a⎤ where both sides reduce to [a \ (a∧b ∨ a∧c)] ∨ (a∧b∧c). In
similar fashion B8 and B9 are seen to hold. That both processes are reciprocal follows from B14
which implies a \ (a ∧ b) = a \ b and the Boolean identity a \ (a \ a∧b) = a∧b. £
Observe that this theorem implies the converse of the main statement of Theorem 7.1.1 in
the case of left-handed skew Boolean algebras. That is, we have:
Theorem 7.1.1L. An algebra (S; ∨, ∧, \, 0) of type 〈2, 2, 2, 0〉 forms a left-handed skew
Boolean algebra if and only if:
i) (S; ∨, ∧) is a left-handed strongly distributive skew lattice.
ii) (S; \, 0) is an iBCS-algebra.
iii) The identity e \ (e \ f) ≈ e∧f holds.
Proof. (⇒) is clear. For (⇐), (ii) and (iii) along with the previous theorem imply that for all a in
S, ⎡a⎤ = {b ∈S⎮b ≤ a} is a Boolean lattice on which \ is the relative complement. (S; ∨, ∧, 0) is
thus at least a skew Boolean algebra reduct. The \ of its skew Boolean structure agrees with the
given \ within all ⎡a⎤. But then they agree in general since a\b = a\(a∧b) holds for the algebra. £
Given a skew lattice (S; ∨, ∧), set x∧Ly = x∧y∧x and x∨Ry = y∨x∨y. Then (S; ∧L, ∨R) is a
left-handed skew lattice. (See Section 3.4.) Observe that:
i) x∧Ly = y∧Lx iff x∧y = y∧x, in which case all four expressions are equal.
ii) x∨Ry = y∨Rx iff x∨y = y∨x, in which case all four expressions are equal.
iii) Thus (S; ∧L, ∨R) is symmetric iff (S; ∧, ∨) is symmetric.
iv) (S; ∧L, ∨R) and (S; ∧, ∨) share the same natural partial order ≥ and natural quasi-order ≻.
v) Thus (S; ∧L, ∨R) is normal iff (S; ∧, ∨) is normal
vi) In particular, (S; ∧L, ∨R) forms skew Boolean algebra iff (S; ∧, ∨) does.
vii) In which case, (S; ∧L, ∨R, 0) and (S; ∧, ∨, 0) share a common difference \.
267
Proof. First, note that for left normal bands ≤ is described by b ≤ a if a ∧ b = b and ⎡a⎤ is the set
{a ∧ s⎮s ∈S}. Let b, c lie in ⎡a⎤. By Lemma 4.4.9, b \ c = (a ∧ b) \ (a ∧ c) = a ∧ (b \ c) ∈ ⎡a⎤
also. In addition 0 = a ∧ 0 ∈ ⎡a⎤ so that ⎡a⎤ is seen to be an iBCS-subalgebra of S. Since ∧ is
commutative on (⎡a⎤; \, 0), the latter satisfies the iBCK identities (i) – (iv) above and in particular
(ii) which implies that ∧ is commutative on ⎡a⎤. Since (⎡a⎤; \, 0) is an iBCK-algebra with
maximal element a, it forms a Boolean lattice (⎡a⎤; ∨, ∧, \, a, 0) with ∧ and \ as already given, and
x ∨ y defined as a \ [(a \ x) ∧ (a \ y)] for all x, y ≤ a.
Conversely, let (S; ∧, 0) be a left normal band with zero such that for each a ∈ S the set
⎡a⎤ is a Boolean lattice under ≤. Let a \ b denote the usual relative complement a \ a∧b in ⎡a⎤.
The clearly B6 holds. B7 reduces to
(a \ a∧b) \ [(a \ a∧b) ∧ c] ≈ (a \ a∧c) \ [(a \ a∧c) ∧ b].
This holds in the Boolean case of ⎡a⎤ where both sides reduce to [a \ (a∧b ∨ a∧c)] ∨ (a∧b∧c). In
similar fashion B8 and B9 are seen to hold. That both processes are reciprocal follows from B14
which implies a \ (a ∧ b) = a \ b and the Boolean identity a \ (a \ a∧b) = a∧b. £
Observe that this theorem implies the converse of the main statement of Theorem 7.1.1 in
the case of left-handed skew Boolean algebras. That is, we have:
Theorem 7.1.1L. An algebra (S; ∨, ∧, \, 0) of type 〈2, 2, 2, 0〉 forms a left-handed skew
Boolean algebra if and only if:
i) (S; ∨, ∧) is a left-handed strongly distributive skew lattice.
ii) (S; \, 0) is an iBCS-algebra.
iii) The identity e \ (e \ f) ≈ e∧f holds.
Proof. (⇒) is clear. For (⇐), (ii) and (iii) along with the previous theorem imply that for all a in
S, ⎡a⎤ = {b ∈S⎮b ≤ a} is a Boolean lattice on which \ is the relative complement. (S; ∨, ∧, 0) is
thus at least a skew Boolean algebra reduct. The \ of its skew Boolean structure agrees with the
given \ within all ⎡a⎤. But then they agree in general since a\b = a\(a∧b) holds for the algebra. £
Given a skew lattice (S; ∨, ∧), set x∧Ly = x∧y∧x and x∨Ry = y∨x∨y. Then (S; ∧L, ∨R) is a
left-handed skew lattice. (See Section 3.4.) Observe that:
i) x∧Ly = y∧Lx iff x∧y = y∧x, in which case all four expressions are equal.
ii) x∨Ry = y∨Rx iff x∨y = y∨x, in which case all four expressions are equal.
iii) Thus (S; ∧L, ∨R) is symmetric iff (S; ∧, ∨) is symmetric.
iv) (S; ∧L, ∨R) and (S; ∧, ∨) share the same natural partial order ≥ and natural quasi-order ≻.
v) Thus (S; ∧L, ∨R) is normal iff (S; ∧, ∨) is normal
vi) In particular, (S; ∧L, ∨R) forms skew Boolean algebra iff (S; ∧, ∨) does.
vii) In which case, (S; ∧L, ∨R, 0) and (S; ∧, ∨, 0) share a common difference \.
267
