JAMB 1994 · UME · Q8

Given that for sets $A$ and $B$ in a universal set $E$ , $A \subseteq B$ , then $A \cap (A \cap B)'$ is

Worked solution (try it first)

$A \subseteq B$ means every element of $A$ is in $B$ , so $A \cap B = A$ .
So the expression is $A \cap A'$ .
Nothing is in a set and in its complement at once, so $A \cap A' = \varnothing$ , option B.