Question
For all sets A, and B, $\text{A}-(\text{A}\cap\text{B})=\text{A} - \text{B}.$

Answer

$\text{L.H.S.}=\text{A}-(\text{A}\cap\text{B})$
$=\text{A}\cap(\text{A}\cap\text{B})'\ \big[\because \text{A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=\text{A}\cap(\text{A}'\cup\text{B}')\ \big[\because(\text{A}\cap\text{B})'=\text{A}'\cup\text{B}'\big]$
$=(\text{A}\cap\text{A}')\cup(\text{A}\cap\text{B}')\ \big[\text{distributive law}\big]$
$=\phi\cup(\text{A}-\text{B})\big[\because \text{A}\cap\text{A}'=\phi\text{ and A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=(\text{A}-\text{B})=\text{R.H.S.}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence proved.

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