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

Answer

$\text{L.H.S.}=\text{A}\cup(\text{B}-\text{A})$
$=\text{A}\cup(\text{B}\cap\text{A}')\ \big[\because \text{A}-\text{B}=\text{A}\cap\text{B}'\big]$
$=(\text{A}\cup\text{B})\cap(\text{A}\cup\text{A}')$ [distributive law]
$=(\text{A}\cup\text{B})\cap\text{U}\ [\because \text{A}\cup\text{A}'=\text{U}]$
$=(\text{A}\cup\text{B})=\text{R.H.S.} \ \big[\because \text{A}\cap\text{U}=\text{A}\big]$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, the given statement is proved.

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