I see from this answer "if you want an algebraic definition, you have to stick with equations.", and this comment, that equations are preferred for defining lattice properties. But it seems to me there's plenty of (useful) properties can't be expressed as equations.

  • Every element has a complement. all x. exists z. (x ^ z = 0 & x v z = 1);

    (How to eliminate the existential quant?)

  • Every element has a unique pseudo-complement x* = max{ y ∈ L | x ∧ y = 0 };

    (Now to eliminate the nested quantifying over 'candidate' complements?)

The Burris and Sankappanavar textbook just doesn't give formulae if it can't give an equation. That's not helpful. (For example Definition 8.6/page 156 for 'relatively complemented'.)

In my specific example, I'm working with a family of lattices which (I think) are 'relatively pseudo-complemented'. That is, within every convex sublattice, every element has a unique max relative complement. How do I express that equationally? (Currently I have a bunch of clunky formulae with existential quant, nested foralls and implications.)

  • $\begingroup$ You eliminate the quantifiers for complemented lattices just like you do it for groups, for example. In the book in mention, there is a definition of group that doesn't include quantifiers. In fact, all variables are implicitly universally quantified. So for complemented lattices, define a unary operation, say $'$, and add the equations $x \wedge x' = 0$ and $x \vee x' = 1$. $\endgroup$
    – amrsa
    Jun 25 at 7:45
  • $\begingroup$ In my example, complements are not nec. unique. So I can't express that as a operation/function(?). I want to express that all 'candidate' complements have a property in common (which I can express as an equation); do I set up two functions, one to be the max (i.e. pseudocomplement), one to be the min? Ah, except the min definitely isn't unique/the 'candidates' form a semilattice. $\endgroup$
    – AntC
    Jun 25 at 11:02
  • $\begingroup$ If complements are not necessarily unique, there is no way to assert their existence by way of equations. But that is not such a rare event. For example, in my answer to the question which has a comment you link to above, there is a class of lattices which are definable by an implication, but I also prove there that it's impossible to define it by equations. That's also true for complemented lattices: every lattice is embeddable into a complemented lattice, so the least class of lattices which has all the complemented lattices is the class of all lattices. $\endgroup$
    – amrsa
    Jun 25 at 11:18

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