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Math Help - valid arguement?

  1. #1
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    valid arguement?

    p goes to q
    not p
    therefore not q

    I think this is invalid because while not q could be the case q could also be true since p is false. Is this sound logic?

    p-->(q-->r)

    q--> (p-->r)

    therefore (p or q) --> r

    I know this is invalid because I have the answer but I do not understand why??? Could someone please explan this? Thank you
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  2. #2
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    Lexington, MA (USA)
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    Hello, frostking2!

    Are you supposed to talk your way through these?
    What methods have you been taught?
    These can be solved with truth tables . . .


    \begin{array}{c}p \to q\\<br />
\sim p \\ \hline<br />
\therefore\;\sim q\end{array}
    This is reasoning by the Inverse . . . which is invalid.

    Or we can run this through a truth table: . \left[(p \to q)\:\wedge \sim p\right] \to \:\sim q



    \begin{array}{cc} & p \to (q\to r) \\<br />
& q \to (p \to r) \\ \hline<br />
\therefore & (p \vee q) \to r \end{array}
    Make a truth table for: . \bigg(\left[p\to (q\to r)\right] \wedge \left[q\to(p\to r)\right]\bigg) \:\to\:\left[(p \vee r) \to r\right]

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  3. #3
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    for frostking2's 2nd question, is there an easier way to do this? cause i stink at making truth tables, and constantly make mistakes on them...
    Have a great one, and thanks in advance...
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