Question
For any two sets A and B, prove the following:
$\text{A}-\text{(A}-\text{B})=\text{A}\cap\text{B}$

Answer

For any sets A and B we have by De-morgan's laws
$\text{(A}\cup\text{B)}' = \text{A}'\cap\text{B}',\text{(A}\cap\text{B)}' = \text{A}'\cup\text{B}'$
Also
$\text{LHS}= \text{A} - \text{(A} - \text{B)}$
$=\text{A}\cap\text{(A}- \text{B)}'$
$=\text{A}\cap\text{(A}\cap\text{B}')'$
$=\text{A}\cap\text{(A}\cap\text{(B})')'$ [By De-morgan's law]
$=\text{A}\cap\text{(A}'\cup\text{B})$ $[\because\text{(B}')'=\text{B}]$
$=\text{(A}\cap\text{A}')\cup\text{(A}\cap\text{B})$
$=\oint\cup\text{(A}\cap\text{B})$ $[\because\text{A}\cap\text{A}'=\oint]$
$=\text{A}\cap\text{B}$ $[\because\oint\cup$ x = x, for and set x$]$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.

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