Page 56 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 56
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Proof. First observe that each of these conditions holds on S if and only if it holds on both its left
factor S/R and its right factor S/L. Thus, we may prove the theorem by proving it for all right
[left]-handed algebras. So suppose that (1) holds for a right-handed skew lattice, S. We show
that if a ≥ b, c in S with b R c, then b = c. Indeed, the following instance of (1)
b ∧ (a ∨ c) ∧ a = (b ∧ a∧ a) ∨ (b ∧ c ∧ a)
reduces first to b ∧ c ∧ a = b ∨ c, and then to c = b. Hence S is right normal and thus normal.
Next, consider the D-equivalent expressions, a∨(b∧c)∨a and (a∨b∨a)∧(a∨c∨a). Assuming right
normality. They are actually the R-equivalent expressions, a∨(b∧c) and (a∨b)∧(a∨c). Since
[(a∨b)∧(a∨c)] ∨ (a∨c) = a∨c by absorption, and
[a∨(b∧c)] ∨ (a∨c) = a ∨ (b∧c) ∨ c] = a∨c.
Hence a∨c ≥ both a∨(b∧c) and (a∨b)∧(a∨c). S being normal with a∨(b∧c) R (a∨b)∧(a∨c) we
get a∨(b∧c) = (a∨b)∧(a∨c). Thus (1) implies (2). Next, (2) clearly implies (3). Suppose that (3)
holds. Then by normality,
a ∧ d ≥ a ∧ (b ∨ c) ∧ d, a ∧ b∧ d, a ∧ c ∧ d and hence a ∧ d ≥ (a ∧ b∧ d) ∨ (a ∧ c ∧ d)
with a∧(b ∨ c)∧d D (a∧ b∧ d) ∨ (a∧c∧d) since S/D is distributive. Whence,
a ∧ (b ∨ c) ∧ d = (a ∧ b ∧ d) ∨ (a ∧ c ∧ d) ,
again by normality. Thus (3) implies (1). £
Lemma 2.3.3. In any normal skew lattice, a∨b = b∨a implies a∧b = b∧a.
Proof. Indeed, a∨b = b∨a implies a, b ∈(a∨b)∧S∧(a∨b), a sublattice, so that a∧b = b∧a. £
A skew lattice is strongly distributive if it satisfies both (4) and (5) below:
4) a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c).
5) (b ∨ c) ∧ d = (b ∧ d) ∨ (c ∧ d).
Theorem 2.3.4. A skew lattice S is strongly distributive if and only if it is distributive,
normal and symmetric.
Proof. A skew lattice satisfying (4) and (5) also satisfies (1), and thus is distributive and normal.
Being normal, a∨b = b∨a implies a∧b = b∧a holds. So let a∧b = b∧a. If a∨b ≠ b∨a, then either
a ∧ (b ∨ a) ≠ a or (b ∨ a) ∧ b ≠ b, for otherwise absorption gives a ∨ b ∨ a = b ∨ a = b ∨ a ∨ b so
that a∨b = b∨a. Suppose that a ∧ (b ∨ a) ≠ a. But then (4) fails in general since a ∧ (b ∨ a) ≠ a
while
(a∧b) ∨ (a∧a) = (a∧b) ∨ a = (b∧a) ∨ a = a.
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Proof. First observe that each of these conditions holds on S if and only if it holds on both its left
factor S/R and its right factor S/L. Thus, we may prove the theorem by proving it for all right
[left]-handed algebras. So suppose that (1) holds for a right-handed skew lattice, S. We show
that if a ≥ b, c in S with b R c, then b = c. Indeed, the following instance of (1)
b ∧ (a ∨ c) ∧ a = (b ∧ a∧ a) ∨ (b ∧ c ∧ a)
reduces first to b ∧ c ∧ a = b ∨ c, and then to c = b. Hence S is right normal and thus normal.
Next, consider the D-equivalent expressions, a∨(b∧c)∨a and (a∨b∨a)∧(a∨c∨a). Assuming right
normality. They are actually the R-equivalent expressions, a∨(b∧c) and (a∨b)∧(a∨c). Since
[(a∨b)∧(a∨c)] ∨ (a∨c) = a∨c by absorption, and
[a∨(b∧c)] ∨ (a∨c) = a ∨ (b∧c) ∨ c] = a∨c.
Hence a∨c ≥ both a∨(b∧c) and (a∨b)∧(a∨c). S being normal with a∨(b∧c) R (a∨b)∧(a∨c) we
get a∨(b∧c) = (a∨b)∧(a∨c). Thus (1) implies (2). Next, (2) clearly implies (3). Suppose that (3)
holds. Then by normality,
a ∧ d ≥ a ∧ (b ∨ c) ∧ d, a ∧ b∧ d, a ∧ c ∧ d and hence a ∧ d ≥ (a ∧ b∧ d) ∨ (a ∧ c ∧ d)
with a∧(b ∨ c)∧d D (a∧ b∧ d) ∨ (a∧c∧d) since S/D is distributive. Whence,
a ∧ (b ∨ c) ∧ d = (a ∧ b ∧ d) ∨ (a ∧ c ∧ d) ,
again by normality. Thus (3) implies (1). £
Lemma 2.3.3. In any normal skew lattice, a∨b = b∨a implies a∧b = b∧a.
Proof. Indeed, a∨b = b∨a implies a, b ∈(a∨b)∧S∧(a∨b), a sublattice, so that a∧b = b∧a. £
A skew lattice is strongly distributive if it satisfies both (4) and (5) below:
4) a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c).
5) (b ∨ c) ∧ d = (b ∧ d) ∨ (c ∧ d).
Theorem 2.3.4. A skew lattice S is strongly distributive if and only if it is distributive,
normal and symmetric.
Proof. A skew lattice satisfying (4) and (5) also satisfies (1), and thus is distributive and normal.
Being normal, a∨b = b∨a implies a∧b = b∧a holds. So let a∧b = b∧a. If a∨b ≠ b∨a, then either
a ∧ (b ∨ a) ≠ a or (b ∨ a) ∧ b ≠ b, for otherwise absorption gives a ∨ b ∨ a = b ∨ a = b ∨ a ∨ b so
that a∨b = b∨a. Suppose that a ∧ (b ∨ a) ≠ a. But then (4) fails in general since a ∧ (b ∨ a) ≠ a
while
(a∧b) ∨ (a∧a) = (a∧b) ∨ a = (b∧a) ∨ a = a.
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