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Math Help - Need help solving this argument....PLEASE

  1. #1
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    Post Need help solving this argument....PLEASE

    1. A v (B -> C)
    2. B / A v C
    3.
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  2. #2
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    Quote Originally Posted by ambergarrett955 View Post
    1. A v (B -> C)
    2. B
    3./ A v C
    Do by indirect argument.
    Assume ~(AvC) then ~A ^ ~C.
    ~A by simplification.
    (B->C) by disjunctive syllogism.
    ~C by simplification.
    ~B by modus tollens.
    That is a contradiction.
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  3. #3
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    Reply to answer

    I'm trying to get A v C by itselfs, though. So, is that still the answer? Thank you
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  4. #4
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    Quote Originally Posted by ambergarrett955 View Post
    I'm trying to get A v C by itselfs, though. So, is that still the answer?
    The answer is yes only if you know how to use indirect argument.
    If you do not have that tool, then I do not how to help you with this.
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  5. #5
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    In class we always came out with the answer that was behind this /xx as our last #. I'm just confused. This stuff is so confusing to me. I'm sorry.
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  6. #6
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    Hello, Amber!

    \begin{array}{c}A \vee (B \to C)\\<br />
B\qquad\qquad\quad \\ \hline<br />
A \vee C\qquad\quad\end{array}

    I vaguely recall a theorem . . . the Law of Detachment (?)

    . . \begin{array}{c} p \vee q \\ \sim q\quad \\ \hline p\quad \end{array}


    The first line is: . A \vee (B \to C) \quad\Rightarrow\quad A \vee (\sim\!B \vee C)\quad\Rightarrow\quad (A \vee C)\: \vee \sim\!B

    So the argument becomes: . \begin{array}{c}(A \vee C)\:\vee\: \sim\!B \\ B\qquad\qquad \\ \hline \end{array}


    By the Law of Detachment, the conclusion is: .  A \vee C

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