91Ó°ÊÓ

The De Morgan’s Laws states that for every two sets A and B:

\({\left( {A \cup B} \right)^c} = {A^c} \cap {B^c}\), and

\({\left( {A \cap B} \right)^c} = {A^c} \cup {B^c}.\)

Let x be an arbitrary element, such that \(x \in {\left( {A \cup B} \right)^c}.\)

\(\begin{aligned}{l} \Rightarrow x \notin \left( {A \cup B} \right)\\ \Rightarrow x \notin A\;{\rm{and}}\;x \notin B\\ \Rightarrow x \in {A^c}\;{\rm{and}}\;x \in {B^c}\\ \Rightarrow x \in \left( {{A^c} \cap {B^c}} \right)\end{aligned}\)

Therefore,\({\left( {A \cup B} \right)^c} \subseteq {A^c} \cap {B^c}\;\;\;\;\;............(1)\)

Similarly, let y be an arbitrary element, such that \(y \in {A^c} \cap {B^c}.\)

\(\begin{aligned}{l} \Rightarrow y \in {A^c}\;{\rm{and}}\;y \in {B^c}\\ \Rightarrow y \notin A\;{\rm{and}}\;y \notin B\\ \Rightarrow y \notin \left( {A \cup B} \right)\\ \Rightarrow y \in {\left( {A \cup B} \right)^c}\end{aligned}\)

Therefore,\({A^c} \cap {B^c} \subseteq {\left( {A \cup B} \right)^c}\;\;\;\;............(2)\)

By combining equations (1) and (2), we get:

\({\left( {A \cup B} \right)^c} = {A^c} \cap {B^c}.\)

Let x be an arbitrary element, such that \(x \in {\left( {A \cap B} \right)^c}.\)

\(\begin{aligned}{l} \Rightarrow x \notin \left( {A \cap B} \right)\\ \Rightarrow x \notin A\;{\rm{or}}\;x \notin B\\ \Rightarrow x \in {A^c}\;{\rm{or}}\;x \in {B^c}\\ \Rightarrow x \in \left( {{A^c} \cup {B^c}} \right)\end{aligned}\)

Therefore,\({\left( {A \cap B} \right)^c} \subseteq {A^c} \cup {B^c}\;\;\;\;\;............(3)\)

Similarly, let y be an arbitrary element, such that\(y \in {A^c} \cup {B^c}\)

\(\begin{aligned}{l} \Rightarrow y \in {A^c}\;{\rm{or}}\;y \in {B^c}\\ \Rightarrow y \notin A\;{\rm{or}}\;y \notin B\\ \Rightarrow y \notin \left( {A \cap B} \right)\\ \Rightarrow y \in {\left( {A \cap B} \right)^c}\end{aligned}\)

Therefore,\({A^c} \cup {B^c} \subseteq \;{\left( {A \cap B} \right)^c}\;\;............(4)\)

By combining equations (3) and (4), we get:

\({\left( {A \cap B} \right)^c} = {A^c} \cup {B^c}.\)