Let A and B be two sets such that A X = B X = ϕ and A X = B X for...

Given, that, $A \cup X = B \cup X$

$⇒ A \cap ( A \cup X ) = A \cap ( B \cup X )$

$⇒ ( A \cap A ) \cup ( A \cap X ) = ( A \cap B ) \cup ( A \cap X )$

[Using distributive law]

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

$⇒ A = A \cap B... ( i )$

Again, consider $A \cup X = B \cup X$

$⇒ B \cap ( A \cup X ) = B \cap ( B \cup X )$

$⇒ ( B \cap A ) \cup ( B \cap X ) = ( B \cap B ) \cup ( B \cap X )$

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

$⇒ A \cap B = B... ( ii )$

Thus from Eq. $( i )$ and $( ii )$, we get

$A = B$