Page 172 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press

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

If S is a right-handed skew lattice then the two symmetry identities of (5.1.1) and (5.1.2)

above simplify as

x∧y∧(x∨y) = y∧x (5.2.3)

and

x∨y = (y∧x)∨y∨x. (5.2.4).

Likewise for right-handed skew lattices, identities (5.2.1) and (5.2.2) above simplify as

(y ∨ z) ∧ x = (y ∧ x) ∨ (z ∧ x) (5.2.5)

and

x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). (5.2.6)

We are ready to prove Spinks’ Theorem, doing so first for the right-handed case.

Theorem 5.2.3. For any skew lattice S the identities (5.2.1) and (5.1.1) imply (5.2.2),
and dually, the identities (5.2.2) and (5.1.2) imply (5.2.1).

Proof. Assume S is a right-handed skew lattice satisfying (5.2.3) and (5.2.5). Set α = x ∨ (y ∧ z)
and β = (x ∨ y) ∧ (x ∨ z). By (5.2.5) and absorption one also has

β = x ∨ (y ∧ (x ∨ z)). (5.2.7)

Equality (5.2.3) yields y ∧ (x ∨ z) = (x ∨ z) ∧ y ∧ ( x ∨ z ∨ y). Thus by (5.2.7),

β = x ∨ ((x ∨ z) ∧ y ∧ (x ∨ z ∨ y)) by (5.2.5)
= x ∨ (x ∧ y ∧ (x ∨ z ∨ y)) ∨ (z ∧ y ∧ (x ∨ z ∨ y)) (5.2.8)
= x ∨ (z ∧ y ∧ (x ∨ z ∨ y))

by absorption. Apply ∧ (x ∨ y ∨ z) on the right of (5.2.8). By (5.2.5) and absorption,

β ∧ (x ∨ y ∨ z) = x ∨ (z ∧ y ∧ (x ∨ y ∨ z)) (5.2.9)

On the other hand,

α ∨ (x ∨ z ∨ y) = x ∨ (y∧z) ∨ x ∨ z ∨ y = x ∨ (y∧z) ∨ z ∨ y = x ∨ z ∨ y.

Hence α ≤ x ∨ z ∨ y and thus α ∧ (x ∨ z ∨ y) = α. Thus by (5.2.5) and absorption again,

x ∨ (y ∧ z) = x ∨ ((y ∧ z) ∧ (x ∨ z ∨ y)). (5.2.10)

Switch y and z in (5.2.10) to get x ∨ (z ∧ y) = x ∨ ((z ∧ y) ∧ (x ∨ y ∨ z)) and use (5.2.9) to obtain

β ∧ (x ∨ y ∨ z) = x ∨ (z ∧ y). (5.2.11)

Apply ∧ (x ∨ y) on the right to obtain

β ∧ (x ∨ y ∨ z) ∧ (x ∨ y) = (x ∨ (z ∧ y)) ∧ (x ∨ y) = x ∨ (z ∧ y ∧ (x ∨ y)). (5.2.12)

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