Page 194 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 194
athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
We continue with a further characterization of quasi-distributivity in the left-handed case.
Lemma 5.5.8. A left-handed skew lattice is quasi-distributive if and only if the following
identity holds:
x ∧ [(y∧x) ∨ z] = x ∧ (y∨z). (5.5.9)
Proof. Sufficiency is clear since neither M3 nor N5 can satisfy (5.5.9). For necessity, observe
first that (5.5.9) holds in any distributive lattice. Hence in a quasi-distributive skew lattice one
has at least x ∧ [(y∧x) ∨ z] D x ∧ (y∨z). If we can show that x ∧ [(y∧x) ∨ z] ≤ x ∧ (y∨z) in the left-
handed case, then equality will hold, given the rectangular context of both expressions. Left-
handedness clearly gives y∧x ≤ y since y∧x = y∧x∧y. Thus (y∧x) ∨z ≤ y∨z follows since both
(y∧x) ∨ z ∨ y ∨ z = (y∧x∧y) ∨ y ∨ z = y ∨ z and y ∨ z ∨ (y∧x) ∨ z = y ∨ (y∧x) ∨ z = y ∨ z
by left-handedness and absorption. Left-handedness plus the inequality gives
x∧[(y∧x)∨z] ∧ x∧(y∨z) = x∧[(y∧x)∨z] ∧ (y∨z) = x∧[(y∧x)∨z]
and similarly x∧(y∨z) ∧ x∧[(y∧x)∨z] = x∧[(y∧x)∨z]. Thus indeed x∧[(y∧x)∨z] ≤ x∧(y∨z)
within a D-class context and equality follows. £
Besides quasi-distributivity and linear distributivity, a third consequence of distributivity is
the following pair of identities. Combined, all three consequences together guarantee that a skew
lattice is distributive.
Lemma 5.5.9. The following identities are respective consequence of (5.2.1) and (5.2.2):
x ∧ [(y∧x∧y) ∨ z ∨ y ∨ z ∨ (y∧x∧y)] ∧ x = x ∧ (y ∨ z ∨ y) ∧ x. (5.5.10).
x ∨ [(y∨x∨y) ∧ z ∧ y ∧ z ∧ (y∨x∨y)] ∨ x = x ∨ (y ∧ z ∧ y) ∨ x. (5.5.11).
Their left are right-handed variants are:
x∧[y ∨ z ∨ (y ∧ x)] = x∧(z ∨ y) and [(x ∧ y) ∨ z ∨ y]∧x = (y ∨ z)∧x. (5.5.10L and R)
[(x ∨ y) ∧ z ∧ y]∨x = (y ∧ z)∨x and x∨ [y ∧ z ∧ (y ∧ x)] = x∨(z ∧ y). (5.5.11L and R)
Proof. (5.2.1) plus regularity first gives
x ∧ [(y∧x∧y)∨z∨y∨z∨(y∧x∧y)] ∧ x = (x∧y∧x) ∨ (x∧z∧x) ∨ (x∧y∧x) ∨ (x∧z∧x) ∨ (x∧y∧x)
which reduces to (x∧y∧x) ∨ (x∧z∧x) ∨ (x∧y∧x) and then x ∧ (y∨z∨y) ∧ x by (5.2.1) again. The
derivation of (5.5.11) from (5.2.2) and regularity is similar. £
Both (5.5.10) and (5.5.11) can be given interpretations regarding the behavior of cosets and
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We continue with a further characterization of quasi-distributivity in the left-handed case.
Lemma 5.5.8. A left-handed skew lattice is quasi-distributive if and only if the following
identity holds:
x ∧ [(y∧x) ∨ z] = x ∧ (y∨z). (5.5.9)
Proof. Sufficiency is clear since neither M3 nor N5 can satisfy (5.5.9). For necessity, observe
first that (5.5.9) holds in any distributive lattice. Hence in a quasi-distributive skew lattice one
has at least x ∧ [(y∧x) ∨ z] D x ∧ (y∨z). If we can show that x ∧ [(y∧x) ∨ z] ≤ x ∧ (y∨z) in the left-
handed case, then equality will hold, given the rectangular context of both expressions. Left-
handedness clearly gives y∧x ≤ y since y∧x = y∧x∧y. Thus (y∧x) ∨z ≤ y∨z follows since both
(y∧x) ∨ z ∨ y ∨ z = (y∧x∧y) ∨ y ∨ z = y ∨ z and y ∨ z ∨ (y∧x) ∨ z = y ∨ (y∧x) ∨ z = y ∨ z
by left-handedness and absorption. Left-handedness plus the inequality gives
x∧[(y∧x)∨z] ∧ x∧(y∨z) = x∧[(y∧x)∨z] ∧ (y∨z) = x∧[(y∧x)∨z]
and similarly x∧(y∨z) ∧ x∧[(y∧x)∨z] = x∧[(y∧x)∨z]. Thus indeed x∧[(y∧x)∨z] ≤ x∧(y∨z)
within a D-class context and equality follows. £
Besides quasi-distributivity and linear distributivity, a third consequence of distributivity is
the following pair of identities. Combined, all three consequences together guarantee that a skew
lattice is distributive.
Lemma 5.5.9. The following identities are respective consequence of (5.2.1) and (5.2.2):
x ∧ [(y∧x∧y) ∨ z ∨ y ∨ z ∨ (y∧x∧y)] ∧ x = x ∧ (y ∨ z ∨ y) ∧ x. (5.5.10).
x ∨ [(y∨x∨y) ∧ z ∧ y ∧ z ∧ (y∨x∨y)] ∨ x = x ∨ (y ∧ z ∧ y) ∨ x. (5.5.11).
Their left are right-handed variants are:
x∧[y ∨ z ∨ (y ∧ x)] = x∧(z ∨ y) and [(x ∧ y) ∨ z ∨ y]∧x = (y ∨ z)∧x. (5.5.10L and R)
[(x ∨ y) ∧ z ∧ y]∨x = (y ∧ z)∨x and x∨ [y ∧ z ∧ (y ∧ x)] = x∨(z ∧ y). (5.5.11L and R)
Proof. (5.2.1) plus regularity first gives
x ∧ [(y∧x∧y)∨z∨y∨z∨(y∧x∧y)] ∧ x = (x∧y∧x) ∨ (x∧z∧x) ∨ (x∧y∧x) ∨ (x∧z∧x) ∨ (x∧y∧x)
which reduces to (x∧y∧x) ∨ (x∧z∧x) ∨ (x∧y∧x) and then x ∧ (y∨z∨y) ∧ x by (5.2.1) again. The
derivation of (5.5.11) from (5.2.2) and regularity is similar. £
Both (5.5.10) and (5.5.11) can be given interpretations regarding the behavior of cosets and
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