Question:
For any two sets $A$ and $B$, prove that : $A^{\prime}-B^{\prime}=B-A$
Solution:
$\mathrm{LHS}=A^{\prime}-B^{\prime}$
$=A^{\prime} \cap\left(B^{\prime}\right)^{\prime} \quad\left[\because C-D=C \cap D^{\prime}\right]$
$=A^{\prime} \cap B$
$=B \cap A^{\prime}$
$=B-A \quad\left[\because C \cap D^{\prime}=C-D\right]$
$\mathrm{RHS}=B-A$So, LHS = RHS
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