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Question Number 22162 by j.masanja06@gmail.com last updated on 12/Oct/17
use the appropite set law to show   that  (A−B)∪(B−A)=(A∪B)−(A∩B)
$${use}\:{the}\:{appropite}\:{set}\:{law}\:{to}\:{show}\: \\ $$$${that} \\ $$$$\left({A}−{B}\right)\cup\left({B}−{A}\right)=\left({A}\cup{B}\right)−\left({A}\cap{B}\right) \\ $$
Answered by $@ty@m last updated on 12/Oct/17
(A∪B)−(A∩B)⇔(A∪B)∩(A∩B)^′   ⇔(A∪B)∩(A^′ ∪B^′ )  ⇔{(A∪B)∩A^′ }∪{(A∪B)∩B^′ }  ⇔{(A∩A^′ )∪(B∩A′)}∪{(A∩B^′ )∪(B∩B′)}  ⇔(B∩A′)∪(A∩B^′ )  ⇔(B−A)∪(A−B)
$$\left({A}\cup{B}\right)−\left({A}\cap{B}\right)\Leftrightarrow\left({A}\cup{B}\right)\cap\left({A}\cap{B}\right)^{'} \\ $$$$\Leftrightarrow\left({A}\cup{B}\right)\cap\left({A}^{'} \cup{B}^{'} \right) \\ $$$$\Leftrightarrow\left\{\left({A}\cup{B}\right)\cap{A}^{'} \right\}\cup\left\{\left({A}\cup{B}\right)\cap{B}^{'} \right\} \\ $$$$\Leftrightarrow\left\{\left({A}\cap{A}^{'} \right)\cup\left({B}\cap{A}'\right)\right\}\cup\left\{\left({A}\cap{B}^{'} \right)\cup\left({B}\cap{B}'\right)\right\} \\ $$$$\Leftrightarrow\left({B}\cap{A}'\right)\cup\left({A}\cap{B}^{'} \right) \\ $$$$\Leftrightarrow\left({B}−{A}\right)\cup\left({A}−{B}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 13/Oct/17
Nice!
$$\mathcal{N}{ice}! \\ $$

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