Hi,

I have been trying to work around the following question, but can't seem to get it right.

(P ^ ¬Q) V (¬P ^ Q) <=> ¬(P ^ Q) ^ (P VQ)

Any help is highly appreciated.

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- Feb 4th 2013, 02:06 PMaprilrocks92Logical Equivalences and Boolean Algebra
Hi,

I have been trying to work around the following question, but can't seem to get it right.

**(P ^ ¬Q) V (¬P ^ Q) <=> ¬(P ^ Q) ^ (P VQ)**

Any help is highly appreciated. - Feb 4th 2013, 02:44 PMemakarovRe: Logical Equivalences and Boolean Algebra
Use De Morgan's law on the right-hand side, then apply distributivity to the result, i.e., transform it into a disjunction of conjunctions. Simplify, and you'll get the left-hand side.

- Feb 4th 2013, 02:55 PMPlatoRe: Logical Equivalences and Boolean Algebra

$\displaystyle \begin{gathered} (P \wedge \neg Q) \vee (\neg P \wedge Q) \Leftrightarrow \hfill \\ \left[ {(P \wedge \neg Q) \vee \neg P} \right] \wedge \left[ {(P \wedge \neg Q) \vee Q} \right] \Leftrightarrow \hfill \\ \left[ {\left( {P \vee \neg P} \right) \wedge \left( {\neg Q \vee \neg P} \right)} \right] \wedge \left[ {\left( {P \vee Q} \right) \wedge \left( {\neg Q \vee Q} \right)} \right] \Leftrightarrow \hfill \\ \neg \left( {P \wedge Q} \right) \wedge (P \vee Q) \hfill \\\end{gathered}$ - Feb 4th 2013, 03:52 PMaprilrocks92Re: Logical Equivalences and Boolean Algebra
I have tried doing this:

**(P ^ ¬Q) V (¬P ^ Q) <=> ¬(P ^ Q) ^ (P VQ)**

Swap sides

**¬(P ^ Q) ^ (P VQ) <=> (P ^ ¬Q) V (¬P ^ Q)**

Hence, since**¬(P ^Q) <=> ¬P V ¬Q**(using De Morgan's Laws)

This gives us:**(¬P V ¬Q) ^ (P V Q)**

And in turn,**(¬P V ¬Q) ^ (P V Q)<=> (P ^ ¬Q) V (¬P ^ Q)**

However, I can't seem to figure out where to go from here. - Feb 4th 2013, 03:56 PMemakarovRe: Logical Equivalences and Boolean Algebra