Page 33 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 33
I: Preliminaries
Theorem 1.3.3. An algebra (N, ∨, ∧) where ∨ and ∧ are both associative and
idempotent is a skew lattice (B1, B2, C1 and C2) if and only if ∧≺L dualizes ∨≺R and ∧≺R dualizes
∨≺L, that is:
x ∨ y = x iff x ∧ y = y and x ∨ y = y iff x ∧ y = x.
Likewise, N is a skew* lattice (B3, B4, C3 and C4) if and only if ∧≺L dualizes ∨≺L and ∧≺R
dualizes ∨≺R, that is: x ∨ y = x iff y ∧ x = y and x ∨ y = y iff y ∧ x = x. £
Laslo has shown that no lesser combination than those given in Theorem 1.3.2 suffice to
force (N, ∨, ∧) to be a lattice. Moreover, the effect of Theorems 1.3.2 and 1.3.3 is that the
combination of any three of B1-B4 with any three of C1-C4 guarantees that (N, ∨, ∧) is at least a
skew* lattice, if not a lattice.
In the first section on lattices we saw that B1 and C1 together were sufficient to force both
∧ and ∨ to be idempotent operations. In general, we have:
Theorem 1.3.4. (Laslo and Cozac [1998]) Let Bi and Cj where 1 ≤ i, j ≤ 4 be a pair of
absorption identities and let (N, ∨, ∧) be an algebra with binary operations ∨ and ∧ \on N that
satisfy both Bi and Cj. Unless Bi and Cj form a converse pair in the above sense, then both ∨ and
∧ are idempotent. When Bi and Cj form a converse pair, then even assuming that both operations
are associative is not enough to force them to also be idempotent.
Proof. We have already seen that B1 and C1 imply that both operations are idempotent. Given
B1 and C3 we have a ∧ a = a ∧ [a ∨ (b ∧ a)] = a and a ∨ a = a ∨ (a ∧ a) = a. Next, given B1 and
C4 we have a ∨ a = (a ∧ (a ∨ a)) ∨ a = a and a ∧ a = a ∧ (a ∨ a) = a. Lastly, both B1 and C2 are
satisfied by the three-element algebra with binary operations given by tables:
∨abc ∧abc
a a b c and a a a a .
bccc baab
c ccc c aac
Both operations are associative, but clearly not idempotent.
We have settled the case for B1 and any of the Cj. What about B2 through B4? First,
suppose we are given algebra (N, ∨, ∧) where ∨ and ∧ are binary operations. For (N, ∨, ∧) to
satisfy B4 it is equivalent that (N, ∨, ∧*) satisfy B1. (Here x ∧* y = y ∧ x as before.) But
replacing ∧ by ∧* has the effect of permuting the Cj. In particular, (N, ∨, ∧*) satisfying C2,
corresponds to (N, ∨, ∧) satisfy C4. Thus, given B4, all Cj except C4 induce idempotent
operations. Continuing, the B1-C2 pairing for (N, ∨*, ∧) corresponds to the B3-C3 pairing for
(N, ∨, ∧) and the B1-C2 pairing for (N, ∨*, ∧*) corresponds to the B2-C1 pairing for (N, ∨, ∧).
Thus all Bi-Cj pairings insure idempotent operations except for the converse pairings. £
Comment. The proof shows that once one of the operations is known to be idempotent,
then the nonconverse pairing is enough to force the other operation to be idempotent also.
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Theorem 1.3.3. An algebra (N, ∨, ∧) where ∨ and ∧ are both associative and
idempotent is a skew lattice (B1, B2, C1 and C2) if and only if ∧≺L dualizes ∨≺R and ∧≺R dualizes
∨≺L, that is:
x ∨ y = x iff x ∧ y = y and x ∨ y = y iff x ∧ y = x.
Likewise, N is a skew* lattice (B3, B4, C3 and C4) if and only if ∧≺L dualizes ∨≺L and ∧≺R
dualizes ∨≺R, that is: x ∨ y = x iff y ∧ x = y and x ∨ y = y iff y ∧ x = x. £
Laslo has shown that no lesser combination than those given in Theorem 1.3.2 suffice to
force (N, ∨, ∧) to be a lattice. Moreover, the effect of Theorems 1.3.2 and 1.3.3 is that the
combination of any three of B1-B4 with any three of C1-C4 guarantees that (N, ∨, ∧) is at least a
skew* lattice, if not a lattice.
In the first section on lattices we saw that B1 and C1 together were sufficient to force both
∧ and ∨ to be idempotent operations. In general, we have:
Theorem 1.3.4. (Laslo and Cozac [1998]) Let Bi and Cj where 1 ≤ i, j ≤ 4 be a pair of
absorption identities and let (N, ∨, ∧) be an algebra with binary operations ∨ and ∧ \on N that
satisfy both Bi and Cj. Unless Bi and Cj form a converse pair in the above sense, then both ∨ and
∧ are idempotent. When Bi and Cj form a converse pair, then even assuming that both operations
are associative is not enough to force them to also be idempotent.
Proof. We have already seen that B1 and C1 imply that both operations are idempotent. Given
B1 and C3 we have a ∧ a = a ∧ [a ∨ (b ∧ a)] = a and a ∨ a = a ∨ (a ∧ a) = a. Next, given B1 and
C4 we have a ∨ a = (a ∧ (a ∨ a)) ∨ a = a and a ∧ a = a ∧ (a ∨ a) = a. Lastly, both B1 and C2 are
satisfied by the three-element algebra with binary operations given by tables:
∨abc ∧abc
a a b c and a a a a .
bccc baab
c ccc c aac
Both operations are associative, but clearly not idempotent.
We have settled the case for B1 and any of the Cj. What about B2 through B4? First,
suppose we are given algebra (N, ∨, ∧) where ∨ and ∧ are binary operations. For (N, ∨, ∧) to
satisfy B4 it is equivalent that (N, ∨, ∧*) satisfy B1. (Here x ∧* y = y ∧ x as before.) But
replacing ∧ by ∧* has the effect of permuting the Cj. In particular, (N, ∨, ∧*) satisfying C2,
corresponds to (N, ∨, ∧) satisfy C4. Thus, given B4, all Cj except C4 induce idempotent
operations. Continuing, the B1-C2 pairing for (N, ∨*, ∧) corresponds to the B3-C3 pairing for
(N, ∨, ∧) and the B1-C2 pairing for (N, ∨*, ∧*) corresponds to the B2-C1 pairing for (N, ∨, ∧).
Thus all Bi-Cj pairings insure idempotent operations except for the converse pairings. £
Comment. The proof shows that once one of the operations is known to be idempotent,
then the nonconverse pairing is enough to force the other operation to be idempotent also.
31
