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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"?

Marti M.
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  • What is wrong with my question? I would be happy to amend it... – Marti M. Jun 27 '18 at 14:27
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    I didn't downvote, but a possible reason for downvoting would be the vagueness of your definition of commutative Lawvere theory. Add a precise definition and you'll get a +1 from me. – Eran Jun 28 '18 at 18:46

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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).

Pece
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