Solution:

As per the question,

Let’s say $A$ and $B$ are two sets such that $A \cap X=B \cap X=\phi$ and $A \cup X=B \cup X$ for some set $X$

We need to show, $A=B$

Proof:

$\begin{array}{l}
A=A \cap(A \cup X)=A \cap(B \cup X)[A \cup X=B \cup X] \\
=(A \cap B) \cup(A \cap X) \dots \dots \dots[\text {Using the distributive law} ] \\
=(A \cap B) \cup \Phi[A \cap X=\Phi] \\
=A \cap B \text { (i) }
\end{array}$

So now, $~B\text{ }=\text{ }B\text{ }\cap \text{ }\left( B\text{ }\cup \text{ }X \right)$

$\begin{aligned}
&=\text{ }B\text{ }\cap \text{ }\left( A\text{ }\cup \text{ }X \right)\text{ }\left[ A\text{ }\cup \text{ }X\text{ }=\text{ }B\text{ }\cup \text{ }X \right] \\
&=(B \cap A) \cup(B \cap X) \ldots \dots \dots[\text { Using the distributive law }] \\
&=(B \cap A) \cup \Phi[B \cap X=\Phi] \\
&=A \cap B \text { (i) }
\end{aligned}$

As a result, from eq. (i) and (ii), we get $\mathrm{A}=\mathrm{B}$.