# proof involving unions, intersections and complements

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• October 6th 2009, 07:37 AM
leinadwerdna
proof involving unions, intersections and complements
The complement of (A intersect B) = A complement union B complement

Any help on this would be much appreciated
• October 6th 2009, 07:47 AM
Plato
Quote:

Originally Posted by leinadwerdna
The complement of (A intersect B) = A complement union B complement
Any help on this would be much appreciated

Do you know how to write the negation of the logical statement "A and B"?
That is all there is to it.
• October 6th 2009, 09:37 AM
leinadwerdna
i am still unclear as to how to write a formal proof for this statement.
• October 6th 2009, 09:54 AM
Danneedshelp
Quote:

Originally Posted by leinadwerdna
The complement of (A intersect B) = A complement union B complement

Any help on this would be much appreciated

$x\in{(A\cap\\B)^{c}}$
$\Leftrightarrow$ $x\notin{A\cap\\B}$
$\Leftrightarrow$ it is not the case that $x\in{A}$ and $x\in{B}$
$\Leftrightarrow$ $x\notin{A}$ or $x\notin{B}$
$\Leftrightarrow$ $x\in{A^{c}}$ or $x\in{B^{c}}$
$\Leftrightarrow$ $x\in{A^{c}\cup\\B^{c}}$
• October 6th 2009, 09:54 AM
Plato
Quote:

Originally Posted by leinadwerdna
i am still unclear as to how to write a formal proof for this statement.

Danneedshelp handed you the solution.
I hope you will learn something from the gift.
• October 6th 2009, 12:45 PM
leinadwerdna
thank you both for helping me out
• October 6th 2009, 01:06 PM
Danneedshelp
Quote:

Originally Posted by leinadwerdna
i am still unclear as to how to write a formal proof for this statement.

All I did was use the definition of union / intersection and complement, which is $A^{c}=\{x\in{U}|x\notin{A}\}$. 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, $P\Leftrightarrow\\Q$ is equivalent to $(P\Rightarrow\\Q)$ and $(Q\Rightarrow\\P)$. 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.
• October 16th 2009, 05:09 PM
leinadwerdna
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