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

Answer

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

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