Raising the bar

De Morgan's laws explained

Source: onlinemathlearning.com

Simple form: $latex \overline{A\cap B}=\bar{A}\cup \bar{B},$ (1) $latex \overline{A\cup B}=\bar{A}\cap \bar{B}.$ (2)

Note that the usage of the complement assumes that there is a universal set $latex U$ of which $latex A,B$ are subsets, so that $latex \bar{A}=U\backslash A.$

These laws can be applied for help with logical reasoning. For example, let $latex A$ be the set of girls, $latex B$ be the set of blondes. For simplicity, assume that $latex \bar{A}$ is the set of boys and $latex \bar{B}$ is the set of brunettes. Then $latex A\cap B$ means "blonde girls" and the left side of equation (1) means NOT "blonde girls". It is not good to have a negated statement, so instead we can have a statement in positive terms from the right side of equation (1): this is the set of boys (any hair color) or brunettes (any sex) or brunette boys.

General form. Let $latex \left\{ A_{i}:i\in I\right\} $ be any family of sets (can be infinite). Then

$latex \overline{\cap _{i}A_{i}}=\cup _{i}\bar{A}_{i},$ (1')

$latex \overline{\cup _{i}A_{i}}=\cap _{i}\bar{A}_{i}.$ (2')

Proof. To see the essence behind formal arguments, verbalize the definitions for yourself. Verbalizations should be concise (easy to grasp) and precise.
For example:

$latex A\cap B$ means the set of common elements of $latex A$ and $latex B.$ Hence, $latex \overline{A\cap B}$ means those elements that are not in at least one of them.

$latex A\cup B$ is the set of elements that belong to at least one of $latex A$ or $latex B.$ Hence, $latex \overline{A\cup B}$ is the set of elements that are not in any of $latex A$ or  $latex B.$

Equation (1') is proved as follows. Denote $latex X=$ $latex \overline{\cap _{i}A_{i}}$ the left side and $latex Y=\cup _{i}\bar{A}_{i}$ the right side. If we show that $latex X\subseteq Y$ and $latex X\supseteq Y$ at the same time, then it will follow that $latex X=Y$. This is shown element by element.

Suppose $latex x\in X.$ Then $latex x$ does not belong to $latex \cap _{i}A_{i}.$ That is, it is not a common element of $latex A_{i}.$ Hence there exists $latex A_{i}$ which does not contain $latex x$ and therefore $latex x\in \bar{A}_{i}.$ It follows that $latex x\in Y.$ Since $latex x\in X$ is arbitrary, we have $latex X\subseteq Y.$

Conversely, let $latex y\in Y.$ Then $latex y$ belongs to at least one of $latex \bar{A}_{i}.$ This means $latex x$ does not belong to $latex A_{i}.$ So it cannot be an element common to all $latex A_{i.}.$ Hence, it belongs to $latex \overline{\cap _{i}A_{i}}=X.$ Again, since $latex y$ was an arbitrary element of $latex Y,$ we have $latex X\supseteq Y.$