A sort of generalization of De Morgan's law for Boolean algebras

Question

I've proved the De Morgan's law $\neg (x \vee y) = \neg x \wedge \neg y$.

How can I prove that in every complete Boolean algebra $\neg \bigvee_{i \in I} x_i = \bigwedge_{i \in I} \neg x_i$?

Answer 1

To show that $\neg \bigvee_{i \in I} x_i = \bigwedge_{i \in I} \neg x_i$, you need to show that $\neg \bigvee_{i \in I} x_i$ is the greatest lower bound of $\{\neg x_i\mid i\in I\}$.

One half is the argument you give in the comments: Since $x_j\leq \bigvee_{i\in I} x_i$, $\lnot \bigvee_{i\in I} x_i \leq \lnot x_j$, so $\lnot \bigvee_{i\in I} x_i$ is a lower bound of $\{\neg x_i\mid i\in I\}$.

To show it is the greatest lower bound, suppose $y\leq \lnot x_j$ for all $j\in I$. Then $x_j\leq \lnot y$ for all $j\in I$, so $\bigvee_{i\in I} x_i\leq \lnot y$, so $y\leq \lnot \bigvee_{i\in I} x_i$.