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Thread: Proofs

  1. #1
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    Proofs

    Prove by direct proof, using only equivalence and inference rules. Remember to name each rule used.
    1. (p ^ q) → r

    2. p ^ r

    Therefore q.
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  2. #2
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    $\displaystyle \begin{gathered}
    \left( {p \wedge q} \right) \to r \hfill \\
    p \wedge \neg r \hfill \\
    \------- \hfill \\
    p \hfill \\
    \neg r \hfill \\
    \neg \left( {p \wedge q} \right) \hfill \\
    \neg p \vee \neg q \hfill \\
    \neg q \hfill \\
    \end{gathered} $
    You fill in the reasons.
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  3. #3
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    Hello, captainjapan!

    $\displaystyle (p \to q) \;\Leftrightarrow\;\sim p \vee q$ . Alternate definition of implication (ADI)


    Prove by direct proof, using only equivalence and inference rules.
    Remember to name each rule used.

    $\displaystyle \begin{array}{c}(p \wedge q) \to r \\ p \:\wedge \sim\! r \\ \hline
    \therefore\;\sim\! q \end{array}$

    . . $\displaystyle \begin{array}{ccc}
    (p \wedge q ) \to r & & \text{Given} \\
    \sim(p \wedge q) \vee r & & \text{ADI} \\
    (\sim\!p \:\vee \sim\!q) \vee r & & \text{DeMorgan} \\
    (\sim\! p \vee r) \:\vee \sim\! q & & \text{comm/assoc} \\
    \sim(\sim\!p \vee r) \to \:\sim\!q & & \text{ADI} \\
    (p \:\wedge \sim\!r) \to \:\sim\! q & & \text{DeMorgan} \\
    p \:\wedge \sim r & & \text{Given} \\
    \therefore\;\;\sim\!q & & \text{Modus Ponens}
    \end{array}$

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  4. #4
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    Quote Originally Posted by Soroban View Post
    $\displaystyle (p \to q) \;\Leftrightarrow\;\sim p \vee q$ . Alternate definition of implication (ADI)
    In almost all standard logic textbooks the property
    $\displaystyle (p \to q) \;\Leftrightarrow\;\sim p \vee q$ is known as Material Implication (Impl.).
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