(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)

[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

Let ⋄ be 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.

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

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

[Ed: I can't quite make sense of this question. Could people please edit it into shape?]

Define ⋄ as

    τ=τ/τ

i=τ/f    j=f/τ

    f=f/f

⋄ is a De Morgan algebra. Is every De Morgan algebra a subalgebra of images of products of ⋄?

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

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