Page 196 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press

athan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond

To begin, set u = [x ∨ (y∧z)] ∧ [z ∨ (y∧z)]. Since [z ∨ (y∧z)] ∧ y = y ∧ z by (5.5.8),

u∧y = [x ∨ (y ∧ z)] ∧ (y ∧ z) = y ∧ z.

with the second identity due to absorption. But then,

y∧u=y∧u∧y=y∧y∧z=y∧z=u∧y

so that y ∨ u = u ∨ y by upper symmetry. Thus:

x∨y∨u = x ∨ (y ∧ z) ∨ y ∨ u
= x ∨ (y ∧ z) ∨ u ∨ y
= x ∨ (y ∧ z) ∨ ([x ∨ (y ∧ z)] ∧ [z ∨ (y ∧ z)]) ∨ y
= x ∨ (y ∧ z) ∨ y
= x ∨ y,

where the first and fifth equalities are due to (5.5.6), the second equality is the established case of
commutation, the third is replacing u by its full expression and the fourth is due to absorption.
But since S is left-handed, x ∨ y ∨ u = x ∨ y establishes x ∨ y ≥ u.

ii) For all x, y, z ∈ S, z ∧ [x ∨ (y ∧ z)] = z ∧ (x ∨ y) ∧ [x ∨ (y ∧ z)].

By duality u = (x ∨ y) ∧ u, and so

z ∧ [x ∨ (y ∧ z)] = z ∧ [z ∨ (y ∧ z)] ∧ [x ∨ (y ∧ z)]
= z ∧ [z ∨ (y ∧ z)] ∧ u
= z ∧ [z ∨ (y ∧ z)] ∧ (x ∨ y) ∧ u
= z ∧ [z ∨ (y ∧ z)] ∧ (x ∨ y) ∧ [x ∨ (y ∧ z)]
= z ∧ (x ∨ y) ∧ [x ∨ (y ∧ z)],

using reverse absorption in the first equality, left-handedness in the second and fourth equalities,
part (i) in the middle equality and absorption in the final equality.

iii) For all x, y, z ∈ S, z ∧ [x ∨ (y ∧ x ∧ z)] = z ∧ [x ∨ (y ∧ x)].

Replace y with y∧x in (ii) to get

z ∧ [x ∨ (y ∧ x ∧ z)] = z ∧ [x ∨ (y ∧ x)] ∧ [x ∨ (y ∧ x ∧ z)]
= z ∧ [x ∨ (y ∧ x)] ∧ x ∧ [x ∨ (y ∧ x ∧ z)]
= z ∧ [x ∨ (y ∧ x)] ∧ x
= z ∧ [x ∨ (y ∧ x)],

using (5.5.7) in the second and fourth equalities and absorption in the third.

iv) To conclude the left-handed case, replace x by y ∨ x in (iii). On the left side, absorption

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