The complement of (A intersect B) = A complement union B complement

Any help on this would be much appreciated

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- Oct 6th 2009, 07:37 AMleinadwerdnaproof involving unions, intersections and complements
The complement of (A intersect B) = A complement union B complement

Any help on this would be much appreciated - Oct 6th 2009, 07:47 AMPlato
- Oct 6th 2009, 09:37 AMleinadwerdna
i am still unclear as to how to write a formal proof for this statement.

- Oct 6th 2009, 09:54 AMDanneedshelp
$\displaystyle x\in{(A\cap\\B)^{c}}$

$\displaystyle \Leftrightarrow$$\displaystyle x\notin{A\cap\\B}$

$\displaystyle \Leftrightarrow$ it is not the case that $\displaystyle x\in{A}$ and $\displaystyle x\in{B}$

$\displaystyle \Leftrightarrow$$\displaystyle x\notin{A}$ or $\displaystyle x\notin{B}$

$\displaystyle \Leftrightarrow$$\displaystyle x\in{A^{c}}$ or $\displaystyle x\in{B^{c}}$

$\displaystyle \Leftrightarrow$$\displaystyle x\in{A^{c}\cup\\B^{c}}$ - Oct 6th 2009, 09:54 AMPlato
- Oct 6th 2009, 12:45 PMleinadwerdna
thank you both for helping me out

- Oct 6th 2009, 01:06 PMDanneedshelp
All I did was use the definition of union / intersection and complement, which is $\displaystyle A^{c}=\{x\in{U}|x\notin{A}\}$. I made a list of biconditional statements. Simply, this just means whatever is on the left side of the arrow implies whats on the right and whatever is on the right side of the arrow implies whatever is on the left side. This can be veiwed symbolically as, $\displaystyle P\Leftrightarrow\\Q$ is equivalent to $\displaystyle (P\Rightarrow\\Q)$ and $\displaystyle (Q\Rightarrow\\P)$. So, to begin the proof, just choose an arbitrary element x (since we are trying to prove the general case). From there, begin looking for ways you can sart connecting definitions using logical conectives. I usally make a list of what I am given and then write down my ultimate goal. From there start breaking down your givens and goal using definitions and start looking for ways to make connections between your givens and goal.

Easier said than done, but keep pluggin' away at it. - Oct 16th 2009, 05:09 PMleinadwerdna
wait when you go from x is not an element of a and b to x is not an element of a or x is not an element dont you need to prove a lemma