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Math Help - Logical Equivalences and Boolean Algebra

  1. #1
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    Logical 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.
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  2. #2
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    Re: 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.
    Thanks from aprilrocks92
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  3. #3
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    Re: Logical Equivalences and Boolean Algebra

    Quote Originally Posted by aprilrocks92 View Post
    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)

    \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}
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  4. #4
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    Re: 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.
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  5. #5
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    Re: Logical Equivalences and Boolean Algebra

    Quote Originally Posted by aprilrocks92 View Post
    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.
    As I said, use distributivity on the left-hand side.
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