using set algebra laws prove the following:

1) (A-B) U (B-A) = (A U B) - (A ∩ B)

2) ((A ∩ B)' U B)' = 0

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- Nov 3rd 2013, 06:42 AMbeastset algebra
using set algebra laws prove the following:

1) (A-B) U (B-A) = (A U B) - (A ∩ B)

2) ((A ∩ B)' U B)' = 0 - Nov 3rd 2013, 08:15 AMemakarovRe: set algebra
Start by replacing X - Y by X ∩ Y', then use De Morgan's laws and distributivity of ∩ over U.

- Nov 3rd 2013, 10:50 AMSorobanRe: set algebra
Hello, beast!

Here is the second one.

I'll let you supply the reasons.

Quote:

$\displaystyle [2]\;\;\big[(A \cap B)' \cup B\big]' \:=\:\emptyset$

$\displaystyle \big[(A \cap B)' \cup B\big]' $

. . $\displaystyle =\;\big[(A' \cup B') \cup B\big]'$

. . $\displaystyle =\;\big[A' \cup (B' \cup B)\big]'$

. . $\displaystyle =\; \big[A' \cup U\big]'$

. . $\displaystyle =\;U'$

. . $\displaystyle =\; \emptyset$

- Nov 3rd 2013, 05:59 PMbeastRe: set algebra
1st line - deMorgans?

2nd line - complement law?

3rd line - ???

4th line - ???

please help