Page 176 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 176
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Proof. Simply cancellative skew lattices are quasi-distributive since M3 and N5 cannot occur as
subalgebras. Assume S is at least quasi-distributive. If a, aʹ, b ∈ S exist such that
a∨b∨a = aʹ∨b∨aʹ and a∧b∧a = aʹ∧b∧aʹ, then a D aʹ since a and aʹ have the same image in the
distributive lattice S/D. If a and aʹ are comparable to b (a, aʹ ≻ b or b ≻ a, aʹ), then either
a∨b∨a = aʹ∨b∨aʹ or a∧b∧a = aʹ∧b∧aʹ will reduce to a = aʹ. Thus we need only consider the case
where b is incomparable to a and aʹ. In any case, the first assertion is by now clear. As for the
equivalence of (i) through (iv) for a skew diamond T, (ii) and (iii) are trivially equivalent by
Theorem 2.2.1. Clearly these conditions imply (iv). Conversely, suppose (iv) holds. Then given
any c ∈ A and d ∈ B, setting a = f∨(e∧c∧e)∨f and b = f∨(e∧d∧e)∨f gives a pair a, b satisfying
these inequalities. Suppose e > a, aʹ > f where a D aʹ. Then a R a∧aʹ L aʹ. Thus if a ≠ aʹ then
either a ≠ a∧aʹ or a∧aʹ ≠ aʹ. Moreover e > a∧aʹ > f so that either {a, a∧aʹ, b, e, f} or
{ aʹ, a∧aʹ, b, e, f} is a copy of NC5L or NCR5 in T contrary to (iv). Thus a is unique and
similarly so is b. Hence (iv) implies (ii).
Suppose that (i) holds and let e ∈ J and f ∈ M be as in (ii) and (iii). As we have seen,
a ∈ A and b ∈ B exist such that e > a, b > f. If aʹ ∈ A is such that e > aʹ > f also, then
a∨b∨a = e = aʹ∨b∨aʹ and a∧b∧a = f = aʹ∧b∧aʹ. Simple cancellation yields a = aʹ. Similarly b
is unique and (ii) follows. Conversely suppose that (ii) – (iv) hold and let a∨b∨a = aʹ∨b∨aʹ and
a∧b∧a = aʹ∧b∧aʹ in T. Since T is quasi-distributive, at least a D aʹ. Again, if a and b are
comparable, then a = aʹ, as seen above. So we may assume they are incomparable and hence, say
a ∈ A and b ∈ B. Setting e = a∨b∨a = aʹ∨b∨aʹ in J and f = a∧b∧a = aʹ∧b∧aʹ in M, we have
e ≥ a, aʹ > f. By (ii), a = aʹ and (i) follows. £
Theorem 5.3.7. A skew lattice S is simply cancellative if and only if it contains no copy
of M3, N5, NCL5 or NCR5 . Collectively, these skew lattices form a variety.
Proof. The first assertion is clear from the preceding discussion. An equational base for this
class of skew lattices is given by the identity for quasi-distributivity and the identity
f ∨ (e∧y∧z∧y∧e) ∨ f = f ∨ (e∧z∧y∧z∧e) ∨ f
where e = e(x, y, z) = x∨y∨z∨x and f = f(x, y, z) = x∧e∧y∧z∧y∧e∧x = x∧y∧z∧y∧x. On either side
are two typical D-related elements having common commuting joins and meets with a third
element (here x). The implication defining simple cancellativity (where x, y and z assume
different roles!) equates these elements. £
Returning to left [right, full cancellativity] we have:
174
Proof. Simply cancellative skew lattices are quasi-distributive since M3 and N5 cannot occur as
subalgebras. Assume S is at least quasi-distributive. If a, aʹ, b ∈ S exist such that
a∨b∨a = aʹ∨b∨aʹ and a∧b∧a = aʹ∧b∧aʹ, then a D aʹ since a and aʹ have the same image in the
distributive lattice S/D. If a and aʹ are comparable to b (a, aʹ ≻ b or b ≻ a, aʹ), then either
a∨b∨a = aʹ∨b∨aʹ or a∧b∧a = aʹ∧b∧aʹ will reduce to a = aʹ. Thus we need only consider the case
where b is incomparable to a and aʹ. In any case, the first assertion is by now clear. As for the
equivalence of (i) through (iv) for a skew diamond T, (ii) and (iii) are trivially equivalent by
Theorem 2.2.1. Clearly these conditions imply (iv). Conversely, suppose (iv) holds. Then given
any c ∈ A and d ∈ B, setting a = f∨(e∧c∧e)∨f and b = f∨(e∧d∧e)∨f gives a pair a, b satisfying
these inequalities. Suppose e > a, aʹ > f where a D aʹ. Then a R a∧aʹ L aʹ. Thus if a ≠ aʹ then
either a ≠ a∧aʹ or a∧aʹ ≠ aʹ. Moreover e > a∧aʹ > f so that either {a, a∧aʹ, b, e, f} or
{ aʹ, a∧aʹ, b, e, f} is a copy of NC5L or NCR5 in T contrary to (iv). Thus a is unique and
similarly so is b. Hence (iv) implies (ii).
Suppose that (i) holds and let e ∈ J and f ∈ M be as in (ii) and (iii). As we have seen,
a ∈ A and b ∈ B exist such that e > a, b > f. If aʹ ∈ A is such that e > aʹ > f also, then
a∨b∨a = e = aʹ∨b∨aʹ and a∧b∧a = f = aʹ∧b∧aʹ. Simple cancellation yields a = aʹ. Similarly b
is unique and (ii) follows. Conversely suppose that (ii) – (iv) hold and let a∨b∨a = aʹ∨b∨aʹ and
a∧b∧a = aʹ∧b∧aʹ in T. Since T is quasi-distributive, at least a D aʹ. Again, if a and b are
comparable, then a = aʹ, as seen above. So we may assume they are incomparable and hence, say
a ∈ A and b ∈ B. Setting e = a∨b∨a = aʹ∨b∨aʹ in J and f = a∧b∧a = aʹ∧b∧aʹ in M, we have
e ≥ a, aʹ > f. By (ii), a = aʹ and (i) follows. £
Theorem 5.3.7. A skew lattice S is simply cancellative if and only if it contains no copy
of M3, N5, NCL5 or NCR5 . Collectively, these skew lattices form a variety.
Proof. The first assertion is clear from the preceding discussion. An equational base for this
class of skew lattices is given by the identity for quasi-distributivity and the identity
f ∨ (e∧y∧z∧y∧e) ∨ f = f ∨ (e∧z∧y∧z∧e) ∨ f
where e = e(x, y, z) = x∨y∨z∨x and f = f(x, y, z) = x∧e∧y∧z∧y∧e∧x = x∧y∧z∧y∧x. On either side
are two typical D-related elements having common commuting joins and meets with a third
element (here x). The implication defining simple cancellativity (where x, y and z assume
different roles!) equates these elements. £
Returning to left [right, full cancellativity] we have:
174
