We shall look at some worked examples that require the application of key results you came across in Section 1.
Worked example 1: Show that .
Let , we show that
.
Now implies
and
.
So . Hence
.
Note: It is also true that .
Worked example 2: Show that .
First, we note that .
By the complement of an intersection, we obtain .
Now LHS is simplified to .
By the distributive law of intersection over union, we obtain .
By the Note in example 1, we arrive at the desired result.