3 fix typography (\sim is a \mathrel, but logician’s usage needs it to be a \mathord)

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. $$\sim(\bigwedge_{i\in I}x_i)=\bigvee_{i\in I}\sim {\sim}\Bigl(\bigwedge_{i\in I}x_i\Bigr)=\bigvee_{i\in I}{\sim} x_i$$ 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 $\sim(x\wedge y)=\sim x\vee\sim y=\sim {\sim}(x\wedge y)={\sim} x\vee{\sim} y={\sim} x$, which means $\sim {\sim} y\leq \sim {\sim} x$.

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

2 added 57 characters in body

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 $\bigwedge_{i\in I}a_i$ I}x_i$refers to the greatest lower bound of the set of$a_i$x_i$ 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 $\bigwedge_{i\in I}a_i$I}x_i$. But I claim that in any de Morgan algebra in which$\bigwedge_{i\in I}a_i$I}x_i$ exists, then your equation is satisfied. $$\sim(\bigwedge_{i\in I}a_i)=\bigvee_{i\in I}x_i)=\bigvee_{i\in I}\sim a_i$$ x_i$$To see this, observe first that \sim must be order-reversing: if a\leq b x\leq y in the lattice, then this means a\wedge b=ax\wedge y=x, which implies \sim(a\wedge b)=\sim a\vee\sim b=\sim a\sim(x\wedge y)=\sim x\vee\sim y=\sim x, which means \sim b\leq y\leq \sim ax. Now, if a=\bigwedge_{i\in x=\bigwedge_{i\in I} a_i x_i exists, then a x is the greatest lower bound of the a_i. x_i. In particular, a\leq a_i x\leq x_i and so \sim a_i\leq x_i\leq \sim a x and so \bigvee_i \sim a_i\leq x_i\leq \sim a x for every i\in I. But also, any other upper bound y of the \sim a_i x_i would have \sim y as a lower bound of the a_i, x_i, which would lead by the definition of a x to \sim y\leq a x and so \sim a\leq x\leq \sim\sim y=y. Thus, \sim a x is a least upper bound of \sim a_i x_i for i\in I and so your equation$$\sim(\bigwedge_{i\in I}a_i)=\bigvee_{i\in I}x_i)=\bigvee_{i\in I}\sim a_i$$x_i$$ is true whenever it makes sense. (A similar argument works when the other side is defined.)

1

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 $\bigwedge_{i\in I}a_i$ refers to the greatest lower bound of the set of $a_i$ 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 $\bigwedge_{i\in I}a_i$.

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

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