Page 171 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 171
V: Further Topics in Skew Lattices
Proof. Notice that S/L is upper symmetric if and only if S/L itself contains no copy of NSR7 ,0 .
Thus the equivalence of (i)(a) with the upper symmetry of S/L follows from Theorem 2.2.9. To
show the equivalence with (i)(b), we begin with x, y ∈ S, set u = u(x, y) = (y∧x)∨y. Since
x∧y∧x L y∧x, we have
(x∧y∧x) ∨ y ∨ (x∧y∧x) L (y∧x) ∨ y ∨ (y∧x) = (y∧x)∨y = u.
Setting w = (x∧y∧x) ∨ y ∨ (x∧y∧x), since x∧w = w∧x = x∧y∧x we get
(x∧u)∧(u∧x) = x∧u∧x = x∧u∧w∧x = (x∧u)∧(x∧w) =x∧u
and
(u∧x)∧(x∧u) = u∧x∧u = u∧w∧x∧u = u∧x∧w∧u = u∧x∧w = u∧w∧x = u∧x.
Thus, x∧u L u∧x. Moreover, if x∧y L y∧x, then u(x, y) must reduce to y. S/L is thus upper
symmetric if and only if for all x, y ∈ S, x ∨ u(x, y) L u(x, y) ∨ x. Using the x∧y = y∨x identity on
D-classes we get, (y∧x) ∨ y ∨ x ∨ (y∧x) ∨ y = x ∨ (y∧x) ∨ y and
x ∨ (y∧x) ∨ y ∨ (y∧x) ∨ y ∨ x = (y∧x) ∨ y ∨ x
which reduce to (y∧x) ∨ y ∨ x ∨ y = x ∨ y and x ∨ y ∨ x = (y∧x) ∨ y ∨ x by Lemma 2.1.4, with the
left identity being redundant in the presence of the right identity. (The lemma states that in any
skew lattice, a, c ≻ b ⇒ a∨b∨c = a∨c while a, c ≺ b ⇒ a∧b∧c = a∧c.)
The three other cases are similar. The final assertion is now clear. £
5.2 Distributive identities in the symmetric case
We give Cvetko-Vah’s 2006 proof of Spinks’ Theorem (Spinks [1998]) stating that for
symmetric skew lattices, the following identities are equivalent:
x ∧ (y ∨ z) ∧ x = (x ∧ y ∧ x) ∨ (x ∧ z ∧ x) (5.2.1)
and (5.2.2)
x ∨ (y ∧ z) ∨ x = (x ∨ y ∨ x) ∧ (x ∨ z ∨ x).
Recall also the following results that are used in further computations often without reference
Lemma 5.2.1 A band S is regular if and only if axb = ab holds for all a, b ≺ x ∈ S. £
Lemma 5.2.2 A skew lattice S satisfies any identity or equational implication satisfied
by both its left factor S/R and its right factor S/L. £
169
Proof. Notice that S/L is upper symmetric if and only if S/L itself contains no copy of NSR7 ,0 .
Thus the equivalence of (i)(a) with the upper symmetry of S/L follows from Theorem 2.2.9. To
show the equivalence with (i)(b), we begin with x, y ∈ S, set u = u(x, y) = (y∧x)∨y. Since
x∧y∧x L y∧x, we have
(x∧y∧x) ∨ y ∨ (x∧y∧x) L (y∧x) ∨ y ∨ (y∧x) = (y∧x)∨y = u.
Setting w = (x∧y∧x) ∨ y ∨ (x∧y∧x), since x∧w = w∧x = x∧y∧x we get
(x∧u)∧(u∧x) = x∧u∧x = x∧u∧w∧x = (x∧u)∧(x∧w) =x∧u
and
(u∧x)∧(x∧u) = u∧x∧u = u∧w∧x∧u = u∧x∧w∧u = u∧x∧w = u∧w∧x = u∧x.
Thus, x∧u L u∧x. Moreover, if x∧y L y∧x, then u(x, y) must reduce to y. S/L is thus upper
symmetric if and only if for all x, y ∈ S, x ∨ u(x, y) L u(x, y) ∨ x. Using the x∧y = y∨x identity on
D-classes we get, (y∧x) ∨ y ∨ x ∨ (y∧x) ∨ y = x ∨ (y∧x) ∨ y and
x ∨ (y∧x) ∨ y ∨ (y∧x) ∨ y ∨ x = (y∧x) ∨ y ∨ x
which reduce to (y∧x) ∨ y ∨ x ∨ y = x ∨ y and x ∨ y ∨ x = (y∧x) ∨ y ∨ x by Lemma 2.1.4, with the
left identity being redundant in the presence of the right identity. (The lemma states that in any
skew lattice, a, c ≻ b ⇒ a∨b∨c = a∨c while a, c ≺ b ⇒ a∧b∧c = a∧c.)
The three other cases are similar. The final assertion is now clear. £
5.2 Distributive identities in the symmetric case
We give Cvetko-Vah’s 2006 proof of Spinks’ Theorem (Spinks [1998]) stating that for
symmetric skew lattices, the following identities are equivalent:
x ∧ (y ∨ z) ∧ x = (x ∧ y ∧ x) ∨ (x ∧ z ∧ x) (5.2.1)
and (5.2.2)
x ∨ (y ∧ z) ∨ x = (x ∨ y ∨ x) ∧ (x ∨ z ∨ x).
Recall also the following results that are used in further computations often without reference
Lemma 5.2.1 A band S is regular if and only if axb = ab holds for all a, b ≺ x ∈ S. £
Lemma 5.2.2 A skew lattice S satisfies any identity or equational implication satisfied
by both its left factor S/R and its right factor S/L. £
169
