Let $A$ and $B$ be two sets such that $A \cap X=B \cap x=f$ and $A \cup X=B \cup X$ for some

To show: $A=B$

It can be seen that

$A=A \cap(A \cup X)=A \cap(B \cup X)[A \cup X=B \cup X]$

$=(A \cap B) \cup(A \cap X)$               [Distributive law]

$=(A \cap B) \cup \varnothing[A \cap X=\varnothing]$

$=A \cap B$          ………$(1)$

Now, $B=B \cap(B \cup X)$

$=B \cap(A \cup X)[A \cup X=B \cup X]$

$=(B \cap A) \cup(B \cap X)$            [Distributive law]

$=(B \cap A) \cup \varnothing[B \cap X=\varnothing]$

$=B \cap A$

$=A \cap B$        ………..$(2)$

Hence, from $(1)$ and $(2),$ we obtain $A = B$