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 is the following property valid?
$${\sim}\bigwedge_{i\in I} x_i=\bigvee_{i\in I} {\sim} x_i$$
 A: 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.)
