## Complete De Morgan algebra

Recall that an algebra $(A,\sim)$ is a De Morgan algebra if $A$ is a bounded distributive lattice and $\sim$ is a unary operation which satisfies:

${\sim} (x\vee y)={\sim} x\wedge {\sim} y$ and ${\sim\sim} x=x$.

When the following property is valid?

$${\sim}\bigwedge_{i\in I} x_i=\bigvee_{i\in I} {\sim} x_i$$

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 Negation ($\lnot x≡(x\Rightarrow\bot)$ in a Heyting algebra, or indeed where it exists in a distributive lattice, always satisfies your first property. I would call a structure with your second property a Boolean algebra. By a de Morgan algebra I would undersand one that satisfies the displayed equation; Peter Johnstone has written several papers about them in topos theory. – Paul Taylor May 4 at 14:56 Some of the users of this site have a habit of closing or deleting questions that they consider to be undergraduate material. I disapprove of this habit, but would point out that this question is definitely a very easy one whose answer is likely to be found in any undergraduate textbook on logic. – Paul Taylor May 4 at 14:59

I claim that the property is true in every de Morgan algebra, whenever the expressions in it make sense (on either side). The issue about making sense is that when $I$ is infinite, the expression $\bigwedge_{i\in I}x_i$ refers to the greatest lower bound of the set of $x_i$ for $i\in I$, and in general there may be no such element of the algebra that is such a greatest lower bound. It is a kind of completeness property to assert that there is such an element as $\bigwedge_{i\in I}x_i$.

But I claim that in any de Morgan algebra in which $\bigwedge_{i\in I}x_i$ exists, then your equation is satisfied. $${\sim}\Bigl(\bigwedge_{i\in I}x_i\Bigr)=\bigvee_{i\in I}{\sim} x_i$$ To see this, observe first that $\sim$ must be order-reversing: if $x\leq y$ in the lattice, then this means $x\wedge y=x$, which implies ${\sim}(x\wedge y)={\sim} x\vee{\sim} y={\sim} x$, which means ${\sim} y\leq {\sim} x$.

Now, if $x=\bigwedge_{i\in I} x_i$ exists, then $x$ is the greatest lower bound of the $x_i$. In particular, $x\leq x_i$ and so ${\sim} x_i\leq {\sim} x$ and so $\bigvee_i {\sim} x_i\leq {\sim} x$ for every $i\in I$. But also, any other upper bound $y$ of the ${\sim} x_i$ would have ${\sim} y$ as a lower bound of the $x_i$, which would lead by the definition of $x$ to ${\sim} y\leq x$ and so ${\sim} x\leq {\sim\sim} y=y$. Thus, ${\sim} x$ is a least upper bound of ${\sim} x_i$ for $i\in I$ and so your equation $${\sim}\Bigl(\bigwedge_{i\in I}x_i\Bigr)=\bigvee_{i\in I}{\sim} x_i$$ is true whenever it makes sense. (A similar argument works when the other side is defined.)

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An executive summary: for any De Morgan algebra $A$, $\sim$ is an isomorphism of $A$ and its dual. – Emil Jeřábek Jan 11 2012 at 11:36

I need some example of infinite De Morgan algebra

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 You should try googling. The Wikipedia article can help. – Todd Trimble May 4 at 10:45 (In classical set theory) any powerset (of any infinite set) is a Boolean algebra and hence an excample of this question. – Paul Taylor May 4 at 15:01