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Hint: Here we need to verify whether $A - B$, $A \cap B$ and $B - A$ are pairwise disjoint i.e.., intersection of two sets is a null set.
Complete step-by-step answer:
Note: Pairwise disjoint means the intersection of any two sets is a null set. A venn diagram can also be drawn to solve this problem with a visual representation.
Complete step-by-step answer:
A & B are two sets. Pairwise disjoint means no two elements of any two sets are equal. So we have to proof:
$(A - B) \cap (B - A) = (A - B) \cap (A \cap B) = (B - A) \cap (A \cap B) = \emptyset $
Where $\emptyset $ is null set $ \Rightarrow \emptyset = \{ \} $
Let,
$x \in (A - B) \cap (B - A)$
$ \therefore x \in (A - B){\text{ & }}x \in (B - A) \\$
$ \therefore x \in A{\text{ }}or{\text{ }}x \notin B\ & {\text{ }}x \in B\or x \notin A \\ $
From this it is clear that we are at contradiction $(\because x \in A{\text{ or }}x \notin A)$
$ \Rightarrow (A - B) \cap (B - A) = \emptyset \to (1)$
Now let $x \in (A - B) \cap (A \cap B)$
$\therefore x \in (A - B) x \in (A \cap B) \\$
$\therefore x \in A{\text{ }}or{\text{ }}x \notin B {\text{ }}x \in A\ x \in B \\$
From this it is clear that we are at contradiction $(\because x \in B{\text{ or }}x \notin B)$
$ \Rightarrow (A - B) \cap (A \cap B) = \emptyset \to (2)$
Similarly, $(B - A) \cap (A \cap B) = \emptyset \to (3)$
From equation 1, 2 and 3 it is clear that it is pairwise disjoint.
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