Page 133 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 133
Skew Boolean Algebras
Example 4.2.2. For X = {x, y, z, w}, the 15 atomic classes and the 4(23) = 32 atoms are:
{x \ (y∨z∨w)} {y \ (x∨z∨w)} {z \ (x∨y∨w)} {w \ (x∨y∨z)}
{(x∧y)\(z∨w), (y∧x)\(z∨w)} {(x∧z)\(y∨w), (z∧x)\(y∨w)}
{(x∧w)\(y∨z), (w∧x)\(y∨z)} {(y∧z)\(x∨w), (z∧y)\(x∨w)}
{(y∧w)\(x∨z), (w∧y)\(x∨z)} {(z∧w)\(x∨y), (w∧z)\(x∨y)}
{(x∧y∧z)\w, (y∧z∧x)\w, (z∧x∧y)\w} {(x∧y∧w)\z, (y∧w∧x)\z, (w∧x∧y)\z}
{(x∧z∧w)\y, (z∧w∧x)\y, (w∧x∧z)\y} {(y∧z∧w)\x, (z∧w∧y)\x, (w∧y∧z)\x}
{x∧y∧z∧w, y∧z∧w∧x, z∧w∧x∧y, w∧x∧y∧z}. £
Theorem 4.2.5. Given the free left-handed skew Boolean algebra LSBAn on
{x1, … , xn}:
i) LSBAn is a finite algebra whose atoms are the terms (y1∧…∧yk) \ (yk+1∨…∨yn)
for k ≥ 1 and where (y1, … , yn) is a permutation of {x1, … , xn}.
ii) Atoms (y1∧…∧yk) \ (yk+1∨…∨yn) and (z1∧…∧zl) \ (zl+1∨…∨zn) lie in the same
atomic class if and only if k = l, (z1, … , zk) is a permutation of {y1, … , yk} and
thus (zl+1, … , zn) is a permutation of {yk+1. … , yn}.
iiiL) (y1∧…∧yk) \ (yk+1∨…∨yn) = (z1∧…∧zl) \ (zl+1∨…∨zn) if besides (ii), y1 = z1.
For the free right-handed dual algebra RSBAn, (i) and (ii) again hold along with:
iiiR) (y1∧…∧yk) \ (yk+1∨…∨yn) = (z1∧…∧zl) \ (zl+1∨…∨zn) if in addition to (ii), yk = zk.
Proof. We consider the left-handed case. The right-handed assertion is similar. To begin, given
a permutation (z1, … , zk) of {y1, … , yk}, (z1∧…∧zk)\(yk+1∨…∨yn) and (y1∧ …∧yk)\(yk+1∨…∨yn)
are L-related; they are not equal if z1 ≠ y1. Indeed y1∧…∧yk L z1∧…∧zk plus (a\c)∧(b\c) =
(a∧b)\c implies they are L-related; they are not equal if z1 ≠ y1 since they are not equal when
operating as functions on 3L. Give y1 and z1 values 1 and 2 respectively, the remaining front
variables 1, and all n–k back variables 0. The outcome for (y1∧ …∧yk)\ (yk+1∨…∨yn) is 1 and for
(z1∧…∧zk)\(yk+1∨…∨yn) is 2.
In general, given distinct partitions {L|M} and {Lʹ|Mʹ} of {x1, x2, …, xn} with L and Lʹ
nonempty, some element m lies in L∩Mʹ or in Lʹ∩M, say the former. Viewing m as a generator,
given any {L|M}-term u and any {Lʹ|Mʹ}-term v, we have u∧m = u but v∧m = 0 = m∧v. Thus
u∧v = u∧m∧v = u∧0 = 0 = v∧u. Thus all {L|M}-terms are orthogonal to all {Lʹ|Mʹ}-terms. Since
(x1∧ …∧xk) \ (xk+1∨…∨xn) = 0 is not an identity in 3L for k ≥ 1, all {L|M}-classes are non-0
classes and distinct {L|M}-classes are orthogonal. Returning to the example above,
{(x∧y)\(z∨w), (y∧x)\(z∨w)} is disjoint from {(x∧w)\(y∨z), (w∧x)\(y∨z)} with pairs from distinct
classes being orthogonal.
To see that they are full D-classes of LSBAn, and that they (all the) atomic D-classes,
observe first that they are the atomic D-classes in the subalgebra of LSBAn that they generate.
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Example 4.2.2. For X = {x, y, z, w}, the 15 atomic classes and the 4(23) = 32 atoms are:
{x \ (y∨z∨w)} {y \ (x∨z∨w)} {z \ (x∨y∨w)} {w \ (x∨y∨z)}
{(x∧y)\(z∨w), (y∧x)\(z∨w)} {(x∧z)\(y∨w), (z∧x)\(y∨w)}
{(x∧w)\(y∨z), (w∧x)\(y∨z)} {(y∧z)\(x∨w), (z∧y)\(x∨w)}
{(y∧w)\(x∨z), (w∧y)\(x∨z)} {(z∧w)\(x∨y), (w∧z)\(x∨y)}
{(x∧y∧z)\w, (y∧z∧x)\w, (z∧x∧y)\w} {(x∧y∧w)\z, (y∧w∧x)\z, (w∧x∧y)\z}
{(x∧z∧w)\y, (z∧w∧x)\y, (w∧x∧z)\y} {(y∧z∧w)\x, (z∧w∧y)\x, (w∧y∧z)\x}
{x∧y∧z∧w, y∧z∧w∧x, z∧w∧x∧y, w∧x∧y∧z}. £
Theorem 4.2.5. Given the free left-handed skew Boolean algebra LSBAn on
{x1, … , xn}:
i) LSBAn is a finite algebra whose atoms are the terms (y1∧…∧yk) \ (yk+1∨…∨yn)
for k ≥ 1 and where (y1, … , yn) is a permutation of {x1, … , xn}.
ii) Atoms (y1∧…∧yk) \ (yk+1∨…∨yn) and (z1∧…∧zl) \ (zl+1∨…∨zn) lie in the same
atomic class if and only if k = l, (z1, … , zk) is a permutation of {y1, … , yk} and
thus (zl+1, … , zn) is a permutation of {yk+1. … , yn}.
iiiL) (y1∧…∧yk) \ (yk+1∨…∨yn) = (z1∧…∧zl) \ (zl+1∨…∨zn) if besides (ii), y1 = z1.
For the free right-handed dual algebra RSBAn, (i) and (ii) again hold along with:
iiiR) (y1∧…∧yk) \ (yk+1∨…∨yn) = (z1∧…∧zl) \ (zl+1∨…∨zn) if in addition to (ii), yk = zk.
Proof. We consider the left-handed case. The right-handed assertion is similar. To begin, given
a permutation (z1, … , zk) of {y1, … , yk}, (z1∧…∧zk)\(yk+1∨…∨yn) and (y1∧ …∧yk)\(yk+1∨…∨yn)
are L-related; they are not equal if z1 ≠ y1. Indeed y1∧…∧yk L z1∧…∧zk plus (a\c)∧(b\c) =
(a∧b)\c implies they are L-related; they are not equal if z1 ≠ y1 since they are not equal when
operating as functions on 3L. Give y1 and z1 values 1 and 2 respectively, the remaining front
variables 1, and all n–k back variables 0. The outcome for (y1∧ …∧yk)\ (yk+1∨…∨yn) is 1 and for
(z1∧…∧zk)\(yk+1∨…∨yn) is 2.
In general, given distinct partitions {L|M} and {Lʹ|Mʹ} of {x1, x2, …, xn} with L and Lʹ
nonempty, some element m lies in L∩Mʹ or in Lʹ∩M, say the former. Viewing m as a generator,
given any {L|M}-term u and any {Lʹ|Mʹ}-term v, we have u∧m = u but v∧m = 0 = m∧v. Thus
u∧v = u∧m∧v = u∧0 = 0 = v∧u. Thus all {L|M}-terms are orthogonal to all {Lʹ|Mʹ}-terms. Since
(x1∧ …∧xk) \ (xk+1∨…∨xn) = 0 is not an identity in 3L for k ≥ 1, all {L|M}-classes are non-0
classes and distinct {L|M}-classes are orthogonal. Returning to the example above,
{(x∧y)\(z∨w), (y∧x)\(z∨w)} is disjoint from {(x∧w)\(y∨z), (w∧x)\(y∨z)} with pairs from distinct
classes being orthogonal.
To see that they are full D-classes of LSBAn, and that they (all the) atomic D-classes,
observe first that they are the atomic D-classes in the subalgebra of LSBAn that they generate.
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