On the algebraic theory of Boolean algebras

Question

I have a question which (I think) should be easy for the experts:

Is the Lawvere theory of Boolean algebras commutative, i.e. are its operations "algebra homomorphisms under any interpretation"?

Answer 1

I don't think so. If it were true, for any boolean algebra $B$, we should have $$ B\times B \overset \land \to B $$ to respect joins, that is for all $a,b,c,d \in B$ $$ (a\lor b) \land (c \lor d) = (a\land c) \lor (b\land d) $$

Take now $a = d = \bot$ and $b=c=\top$. The left hand side is $\top$ while the right hand side is $\bot$.

So actually, no boolean algebra is commutative except if $\top= \bot$ (which only occurs in the trivial boolean algebra).