Page 108 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 108
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
The power of these varying forms of distribution is seen in the following results.
Theorem 3.5.1. A distributive, noncommutative lattice is a paralattice if and only if it is
a quasilattice.
Proof. Any paralattice satisfying D7 and D8 is a quasilattice. Just expand a∧(b∨a∨b)∧a and
a∨(b∧a∧b)∨a using D7 and D8 and then simplify using B6, B7, C6 and C7 to obtain respectively
B5 and C5. On the other hand, let N be a quasilattice satisfying both D7 and D8. To see that N is
also a paralattice, suppose to the contrary that there exist a, b ∈ N such that a∧b = b = b∧a, but
either a∨b ≠ a or b∨a ≠ a. Then we obtain either
a∧(a∨b)∧a = a ≠ a∨b = (a∧a∧a)∨(a∧b∧a)
or
a∧(b∨a)∧a = a ≠ b∨a = (a∧b∧a)∨(a∧a∧a)
and thus deny D7. Similarly, if a,b ∈ N exist such that a∨b = b = b∨a, but either a∧b ≠ a or
b∧a ≠ a then we obtain denials of D8. Hence the two natural partial orderings indeed dualize
each other and N is a paralattice. £
Passing to D5 and D6 we obtain a much stronger result.
Theorem 3.5.2. A bidistributive quasilattice (paralattice) factors into the product of a
distributive lattice and an antilattice. Conversely, every such product is a bidistributive
quasilattice (paralattice).
Proof. Let N be a (necessarily fine) bidistributive quasilattice. We show that if x ≥ y, z with
y D z, then y = z. Since both u ≥ u∧x∧y∧u, u∧y∧x∧u and u∧x∧y∧u D u∧y∧x∧u in general, the
implication yields the identity u∧x∧y∧u = u∧y∧x∧u. So let x ≥ y,z in N. Since y R(∧) y∧z L(∧) z,
with x ≥ y∧z also, we show that y = z under the added assumption that either y L(∧) z or y R(∧) z.
Let us assume that y L(∧) z. Consider y∨z and z∨y. D5 gives
y∧(y∨z) = y∧(y∨z)∧x = (y∧y∧x)∨(y∧z∧x) = y∨y = y
and
(y∨z)∧y = x∧(y∨z)∧y = (x∧y∧y)∨(x∧z∧y) = y∨z.
Hence y R(∨) y∨z L(∨) z, in a common L(∧)-class of N. Thus, we may assume further that in
addition to y L(∧) z either y L(∨) z or y R(∨) z. Supposing that y L(∨) z, then by D4,
y = y∨z = (x∧y∧z)∨(x∧x∧z) = x∧(y∨x)∧z = x∧x∧z = z.
If we take y R(∨) z instead, then D5 again yields
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The power of these varying forms of distribution is seen in the following results.
Theorem 3.5.1. A distributive, noncommutative lattice is a paralattice if and only if it is
a quasilattice.
Proof. Any paralattice satisfying D7 and D8 is a quasilattice. Just expand a∧(b∨a∨b)∧a and
a∨(b∧a∧b)∨a using D7 and D8 and then simplify using B6, B7, C6 and C7 to obtain respectively
B5 and C5. On the other hand, let N be a quasilattice satisfying both D7 and D8. To see that N is
also a paralattice, suppose to the contrary that there exist a, b ∈ N such that a∧b = b = b∧a, but
either a∨b ≠ a or b∨a ≠ a. Then we obtain either
a∧(a∨b)∧a = a ≠ a∨b = (a∧a∧a)∨(a∧b∧a)
or
a∧(b∨a)∧a = a ≠ b∨a = (a∧b∧a)∨(a∧a∧a)
and thus deny D7. Similarly, if a,b ∈ N exist such that a∨b = b = b∨a, but either a∧b ≠ a or
b∧a ≠ a then we obtain denials of D8. Hence the two natural partial orderings indeed dualize
each other and N is a paralattice. £
Passing to D5 and D6 we obtain a much stronger result.
Theorem 3.5.2. A bidistributive quasilattice (paralattice) factors into the product of a
distributive lattice and an antilattice. Conversely, every such product is a bidistributive
quasilattice (paralattice).
Proof. Let N be a (necessarily fine) bidistributive quasilattice. We show that if x ≥ y, z with
y D z, then y = z. Since both u ≥ u∧x∧y∧u, u∧y∧x∧u and u∧x∧y∧u D u∧y∧x∧u in general, the
implication yields the identity u∧x∧y∧u = u∧y∧x∧u. So let x ≥ y,z in N. Since y R(∧) y∧z L(∧) z,
with x ≥ y∧z also, we show that y = z under the added assumption that either y L(∧) z or y R(∧) z.
Let us assume that y L(∧) z. Consider y∨z and z∨y. D5 gives
y∧(y∨z) = y∧(y∨z)∧x = (y∧y∧x)∨(y∧z∧x) = y∨y = y
and
(y∨z)∧y = x∧(y∨z)∧y = (x∧y∧y)∨(x∧z∧y) = y∨z.
Hence y R(∨) y∨z L(∨) z, in a common L(∧)-class of N. Thus, we may assume further that in
addition to y L(∧) z either y L(∨) z or y R(∨) z. Supposing that y L(∨) z, then by D4,
y = y∨z = (x∧y∧z)∨(x∧x∧z) = x∧(y∨x)∧z = x∧x∧z = z.
If we take y R(∨) z instead, then D5 again yields
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