\[(A \cap B)'\]
: means 'not in both A and B' - \[A'\]
means 'not in A).Theorem
\[(A \cap B)'=A' \cup B'\]
Proof
If
\[x \notin (A \cap B)\]
then \[x \notin A\]
or \[x \notin B\]
hence \[x \in A'\]
or \[x \in B'\]
so \[x \in A' \cup B'\]
.Conversely if
\[x \notin A' \cup B'\]
then \[x \notin A'\]
or \[x \notin B'\]
so \[x \in (A \cap B) \rightarrow x \notin (A \cap B)'\]
Theorem
\[(A \cup B)'=A' \cap B'\]
Proof
If
\[x \notin (A \cup B)\]
then \[x \notin A\]
and \[x \notin B\]
hence \[x \in A'\]
and \[x \in B'\]
so \[x \in A' \cap B'\]
.Conversely if
\[x \notin \notin A' \cup B'\]
then \[x \notin A; \rightarrow x \in A\]
and \[x \notin B' \rightarrow x \in B\]
so \[x \notin (A \cup B)'\]