Page 177 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 177
V: Further Topics in Skew Lattices
Theorem 5.3.8.
1) The following are equivalent for a skew lattice S.
i) S is left cancellative.
ii) None of M3, N5, NCL5 , NCR5 , NSR7 ,0 nor NSL7 ,1 are subalgebras of S.
iii) S is simply cancellative, S/L is upper symmetric, and S/R is lower symmetric.
2) The following are equivalent for a skew lattice S.
i) S is right cancellative.
ii) None of M3, N5, NCL5 , NCR5 , NSR7 ,1 nor NSL7 ,0 are subalgebras of S.
iii) S is simply cancellative, S/L is lower symmetric, and S/R is upper symmetric.
3) The following are equivalent for a skew lattice S.
i) S is cancellative.
ii) None of the above 5 or 7-element algebras occur as subalgebras of S.
iii) S is simply cancellative and symmetric.
4) Left [right, fully] cancellative skew lattices form a variety.
Proof. We begin with (1). If S is left cancellative, then S cannot contain copies of any of the
algebras listed in (ii) since none of them are left cancellative. Thus (i) implies (ii). The
equivalence of (ii) and (iii) follows from Theorems 5.3.7 and 5.1.2. Now assume S satisfies (iii),
and suppose a, b, c ∈ S satisfy a ∨ b = a ∨ c and a ∧ b = a ∧ c. Since S/L is upper symmetric
and S/R is lower symmetric, Theorem 5.1.2 gives
b ∨ a ∨ b = (a ∧ b) ∨ a ∨ b = (a ∧ c) ∨ a ∨ c = c ∨ a ∨ c
and
b ∧ a ∧ b = (a ∨ b) ∧ a ∧ b = (a ∨ c) ∧ a ∧ c = c ∧ a ∧ c.
By simple cancellativity, b = c. Therefore S is left cancellative and (i) holds. The proof of (2) is
similar. Indeed (2) follows from (1) via horizontal duality (new x ∨ y = old y ∨ x; new
x ∧ y = old y ∧ x). (3) is a consequence of (1) and (2) combined, together with Theorems 5.3.7
and 5.1.1. Finally (4) now follows from (1) – (3) and Theorems 5.1.1, 5.1.2 and 5.3.7. £
Equational bases. A basis for left cancellative skew lattices is thus as follows.
1) [x∧(y∨z)] ∧ [(x∧y) ∨ (x∧z)] ∧ [x∧(y∨z)] = x∧(y∨z).
2) f∨(e∧y∧z∧y∧e)∨f = f∨(e∧z∧y∧z∧e)∨f where e = x∨y∨z∨x and f = x∧y∧z∧y∧x.
3) x∨y∨x = (y∧x)∨y∨x and x∧y∧x = (y∨x)∧y∧x.
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Theorem 5.3.8.
1) The following are equivalent for a skew lattice S.
i) S is left cancellative.
ii) None of M3, N5, NCL5 , NCR5 , NSR7 ,0 nor NSL7 ,1 are subalgebras of S.
iii) S is simply cancellative, S/L is upper symmetric, and S/R is lower symmetric.
2) The following are equivalent for a skew lattice S.
i) S is right cancellative.
ii) None of M3, N5, NCL5 , NCR5 , NSR7 ,1 nor NSL7 ,0 are subalgebras of S.
iii) S is simply cancellative, S/L is lower symmetric, and S/R is upper symmetric.
3) The following are equivalent for a skew lattice S.
i) S is cancellative.
ii) None of the above 5 or 7-element algebras occur as subalgebras of S.
iii) S is simply cancellative and symmetric.
4) Left [right, fully] cancellative skew lattices form a variety.
Proof. We begin with (1). If S is left cancellative, then S cannot contain copies of any of the
algebras listed in (ii) since none of them are left cancellative. Thus (i) implies (ii). The
equivalence of (ii) and (iii) follows from Theorems 5.3.7 and 5.1.2. Now assume S satisfies (iii),
and suppose a, b, c ∈ S satisfy a ∨ b = a ∨ c and a ∧ b = a ∧ c. Since S/L is upper symmetric
and S/R is lower symmetric, Theorem 5.1.2 gives
b ∨ a ∨ b = (a ∧ b) ∨ a ∨ b = (a ∧ c) ∨ a ∨ c = c ∨ a ∨ c
and
b ∧ a ∧ b = (a ∨ b) ∧ a ∧ b = (a ∨ c) ∧ a ∧ c = c ∧ a ∧ c.
By simple cancellativity, b = c. Therefore S is left cancellative and (i) holds. The proof of (2) is
similar. Indeed (2) follows from (1) via horizontal duality (new x ∨ y = old y ∨ x; new
x ∧ y = old y ∧ x). (3) is a consequence of (1) and (2) combined, together with Theorems 5.3.7
and 5.1.1. Finally (4) now follows from (1) – (3) and Theorems 5.1.1, 5.1.2 and 5.3.7. £
Equational bases. A basis for left cancellative skew lattices is thus as follows.
1) [x∧(y∨z)] ∧ [(x∧y) ∨ (x∧z)] ∧ [x∧(y∨z)] = x∧(y∨z).
2) f∨(e∧y∧z∧y∧e)∨f = f∨(e∧z∧y∧z∧e)∨f where e = x∨y∨z∨x and f = x∧y∧z∧y∧x.
3) x∨y∨x = (y∧x)∨y∨x and x∧y∧x = (y∨x)∧y∧x.
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