Page 173 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 173
V: Further Topics in Skew Lattices
On the other hand regularity and right-handedness with (5.2.5) imply
β ∧ (x ∨ y ∨ z) ∧ (x ∨ y) = β ∧ (x ∨ y)
= (x ∨ y) ∧ (x ∨ z) ∧ (x ∨ y) = (x ∨ z) ∧ (x ∨ y) = x ∨ (z ∧ (x ∨ y)). (5.2.13)
Together (5.2.12) and (5.2.13) give the identity
x ∨ (z ∧ y ∧ (x ∨ y)) = x ∨ (z ∧ (x ∨ y)). (5.2.14)
Notice that α ∨ x ∨ z = x ∨ (y ∧ z) ∨ z = x ∨ z. Hence α ≤ x ∨ z and thus α ∧ (x ∨ z) = α. Finally,
α = α ∧ (x ∨ z) = (x ∨ (y ∧ z)) ∧ (x ∨ z)
= x ∨ (y ∧ z ∧ (x ∨ z)) by (5.2.5)
= x ∨ (y ∧ (x ∨ z)) by (5.5.16) with y and z switched
= β by (5.2.7).
A similar argument shows that a right-handed skew lattice satisfying (5.2.4) and (5.2.6) must also
satisfy (5.2.5). Duality yields that the assertion follows for left handed-skew lattices. The theorem
now follows from Lemma 5.2.2. £
Corollary 5.2.4 (Spinks [1998] and [2000]) For symmetric skew lattices, the distributive
identities (5.2.1) and (5.2.2) are equivalent. £
5.3 Cancellation in skew lattices
A skew lattice is cancellative if both
x ∨ z = y ∨ z and x ∧ z = y ∧ z imply x = y, (5.3.1) 17
and
x ∨ y = x ∨ z and x ∧ y = x ∧ z imply y = z. (5.3.2) 18
A skew lattice satisfying (5.3.1) [or (5.3.2)] is right [left] cancellative. Lattices are cancellative
precisely when they are distributive. For skew lattices, we at least have:
Lemma 5.3.1. Left [right, fully] cancellative skew lattices are quasi-distributive.
Proof. All forms of cancellation prevent M3 or N5 from being subalgebras. £
Proposition 5.3.2 Any skew chain is cancellative.
Proof. Assume that S is a right-handed skew chain with x, y, z ∈ S satisfying x ∨ z = y ∨ z and
x ∧ z = y ∧ z. Then xDy since S/D is totally ordered and thus cancellative. Assume that
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On the other hand regularity and right-handedness with (5.2.5) imply
β ∧ (x ∨ y ∨ z) ∧ (x ∨ y) = β ∧ (x ∨ y)
= (x ∨ y) ∧ (x ∨ z) ∧ (x ∨ y) = (x ∨ z) ∧ (x ∨ y) = x ∨ (z ∧ (x ∨ y)). (5.2.13)
Together (5.2.12) and (5.2.13) give the identity
x ∨ (z ∧ y ∧ (x ∨ y)) = x ∨ (z ∧ (x ∨ y)). (5.2.14)
Notice that α ∨ x ∨ z = x ∨ (y ∧ z) ∨ z = x ∨ z. Hence α ≤ x ∨ z and thus α ∧ (x ∨ z) = α. Finally,
α = α ∧ (x ∨ z) = (x ∨ (y ∧ z)) ∧ (x ∨ z)
= x ∨ (y ∧ z ∧ (x ∨ z)) by (5.2.5)
= x ∨ (y ∧ (x ∨ z)) by (5.5.16) with y and z switched
= β by (5.2.7).
A similar argument shows that a right-handed skew lattice satisfying (5.2.4) and (5.2.6) must also
satisfy (5.2.5). Duality yields that the assertion follows for left handed-skew lattices. The theorem
now follows from Lemma 5.2.2. £
Corollary 5.2.4 (Spinks [1998] and [2000]) For symmetric skew lattices, the distributive
identities (5.2.1) and (5.2.2) are equivalent. £
5.3 Cancellation in skew lattices
A skew lattice is cancellative if both
x ∨ z = y ∨ z and x ∧ z = y ∧ z imply x = y, (5.3.1) 17
and
x ∨ y = x ∨ z and x ∧ y = x ∧ z imply y = z. (5.3.2) 18
A skew lattice satisfying (5.3.1) [or (5.3.2)] is right [left] cancellative. Lattices are cancellative
precisely when they are distributive. For skew lattices, we at least have:
Lemma 5.3.1. Left [right, fully] cancellative skew lattices are quasi-distributive.
Proof. All forms of cancellation prevent M3 or N5 from being subalgebras. £
Proposition 5.3.2 Any skew chain is cancellative.
Proof. Assume that S is a right-handed skew chain with x, y, z ∈ S satisfying x ∨ z = y ∨ z and
x ∧ z = y ∧ z. Then xDy since S/D is totally ordered and thus cancellative. Assume that
171
