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Thread: Logic -- have no idea how to finish this!

  1. #1
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    Logic -- have no idea how to finish this!

    Demonstrate that p ->q is equivalent to (p^q')'.

    This is what i got so far: T = true, F= false

    p q p -> q q'
    T T T F
    T F F T
    F T T F
    F F T T

    now what?? grrr
    any help is appreciated thanks!
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  2. #2
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    $\displaystyle \begin{array}{ccccc}
    p & q &\vline & {p \to q} & {\neg \left( {p \wedge \neg q} \right)} \\
    \hline
    t & t &\vline & t & t \\
    t & f &\vline & f & f \\
    f & t &\vline & t & t \\
    f & f &\vline & t & t \\
    \end{array}
    $
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