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# Does ⋄ generate all De Morgan algebras?

(Asked by Nathaniel Hellerstein on the Q&A board at JMM)

This question is about De Morgan algebras (see also Wikipedia), which are something like Boolean algebras, but with a different weaker sense of the complement ∼. Namely, a De Morgan algebra is a bounded distributive lattice with an involution ∼ satisfying de Morgans laws.

Let ⋄ be the four element De Morgan algebra that is not a Boolean algebra, pictured below.

    1

i      j

0


where ∼0 = 1, ∼1 = 0, but ∼i = i and ∼j = j, so i and j and self-dual with respect to ∼. This algebra seems to express one of the fundamental differences between De Morgan algebras and Boolean algebras.

Question. Does the algebra ⋄ generate all De Morgan algebras, in the sense that every De Morgan algebra is a subalgebra of a homomorphic image of a product of ⋄?

Please see the related Birkhoff's HSP Theorem in universal algebra, concerning varieties of algebras closed under H, S, and P (homomorphic image, subalgebra and product).

(Edit: I edited the question to express the question as I understood it. I'm not sure whether the OP intended SHP as stated or HSP, which would conform with Birkhoff's theorem. Probably it was intended to take the variety generated by ⋄, that is, close {⋄} under H, S and P. The question then is whether this is equal to the class of all De Morgan algebras. Please revert if my edits are off-base.-JDH]

From the original question:

The ⋄ algebra can also be defined in terms of the usual 2 element Boolean algebra { f, τ } by using pairs denoted a/b, with the and operations on pairs defined belowcoordinate-wise, but where, as mentioned by Dorais, the operation exchanges coordinates in addition to negating them, making for a "twisted square".

     1 = τ/τ

i = τ/f    j = f/τ

0 = f/f


~(a/b)=(~b/~a)
(a/b)∧(c/d) = (a∧c)/(b∧d)
(a/b)∨(c/d) = (a∨c)/(b∨d)

2 Substantial edit, to explain what the question is asking

[Ed: I can't quite make

This question is about De Morgan algebras (see also Wikipedia), which are something like Boolean algebras, but with a different weaker sense of this questionthe complement ∼. Namely, a De Morgan algebra is a bounded distributive lattice with an involution satisfying de Morgans laws. Could people please edit it into shape?]

Define as

i=τ/f j=f/τ

Letbe the four element De Morgan algebra that is not a Boolean algebra, pictured below.

where ∼0 = 1, ∼1 = 0, but ∼i = i and ∼j = j, so i and j and self-dual with respect to ∼. This algebra seems to express one of the fundamental differences between De Morgan algebraalgebras and Boolean algebras. Is

Question. Does the algebra generate all De Morgan algebras, in the sense that every De Morgan algebra is a subalgebra of images a homomorphic image of products a product of ⋄?

Please see the related Birkhoff's HSP Theorem in universal algebra, concerning varieties of algebras closed under H, S, and P.

(Edit: I edited the question to express the question as I understood it. I'm not sure whether the OP intended SHP as stated or HSP, which would conform with Birkhoff's theorem. Probably it was intended to take the variety generated by ⋄, that is, close {⋄} under H, S and P. The question then is whether this is equal to the class of all De Morgan algebras. Please revert if my edits are off-base.-JDH]

From the original question: The algebra can also be defined in terms of the usual 2 element Boolean algebra { f, τ } by using pairs a/b, with the operations on pairs defined below.

1 = τ/τi = τ/f j = f/τ 0 = f/f

[Ed: the above box has to do with the definition of a De Morgan algebra; see above link]