# 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]

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 defined coordinate-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)

-
The subalgebras of images of products refers to the kind of constructions arising in Birkhoff's HSP theorem in universal algebra. See en.wikipedia.org/wiki/Variety_(universal_algebra). –  Joel David Hamkins Jan 14 '10 at 15:03
The question is really asking whether the variety of algebras produced by closing that particular de Morgan algebra under HSP, is the same as the variety of all de Morgan algebras. This would be interesting, since I think he means Diamond to be some kind of prototypical de Morgan algebra that is not a Boolean algebra. But I'm not clear on exactly what his algebra is, though, since to my understanding, one should have a bounded distributive lattice with an involution satisfying the de Morgan equations. He instead has a binary operation a/b, but he doesn't seem to mean a-b = a meet not-b. –  Joel David Hamkins Jan 14 '10 at 18:05
I guess there are only two de Morgan algebras with 4 elements. One is the familiar Boolean algebra, and the other has - defined properly on 0 and 1, but has -i = i and -j = j in the diagram above. So clearly he means the second one, and if we have this understanding, we can ignore the operation a/b. – Joel –  Joel David Hamkins Jan 14 '10 at 18:18
@Joel: Excellent rewrite. –  Reid Barton Jan 14 '10 at 21:07
@Joel: thanks, it is now clear to me what the idea of "subalgebra of images of" algebras I asked about are. Observation: Relevance logic is another four-valued logic, where t and f have joins and meets, as the contradictory truth value, and the undertdetermined truth value. This is kind of a Boolean algebra of truth values: might it be interesting to consider alongside the Boolean/DeMorgan quasi-dichotomy? –  Charles Stewart Jan 15 '10 at 6:32