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How do you prove this proposition from set theory, where $\overline X$ denotes the complement of $X$?

$\overline{\overline{A}\cap{\overline{B}}}\cup\overline{\overline{A}\cap{B}}=A$

My gut feeling tells me to apply De Morgan Law but I don't know how to go about this. Thank you!

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  • $\begingroup$ Does the overline denote complement or closure? $\endgroup$ – mxnoqwerty Oct 8 '18 at 0:13
  • $\begingroup$ @mrnoqwerty - I rather suspect complement $\endgroup$ – Henry Oct 8 '18 at 0:14
  • $\begingroup$ It denotes complement. $\endgroup$ – James Warthington Oct 8 '18 at 0:15
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    $\begingroup$ This is not true. The left-hand side is the entire universe. $\endgroup$ – David Peterson Oct 8 '18 at 0:25
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This is false if your space is $\{0\}$ and $A=B=\emptyset$. Because $\overline{A}=\overline{B}=\{0\}$, the left hand side becomes $\overline{\{0\}}\cup\overline{\emptyset}=\emptyset\cup\{0\}=\{0\}$ which is not $A$.

What is true, is that $\overline{\overline{A}\cup\overline{B}}\cup\overline{\overline{A}\cup B}=A$.

Proof in words:

If we have $\overline{V\cup W}$, this is everything not in $V\cup W$, so everything neither in $V$ nor $W$. Hence $\overline{V\cup W}=\overline{V}\cap\overline{W}$. So we get $\overline{\overline{A}\cup\overline{B}}\cup\overline{\overline{A}\cup B}=(A\cap B)\cup(A\cap\overline{B})=A\cap(B\cup\overline{B})=A$.

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