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Math Help - Prove validity of an argument by Indirect derivation

  1. #1
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    Arrow Prove validity of an argument by Indirect derivation

    Establish the validity of this argument by means of an unmixed indirect derivation.

    ~(P-->Q)-->Q
    \therefore P-->Q

    Attempt:

    1. Show P --> Q
    2. ~ (P -->Q) Assumption( Indirect Derivation)
    3. ~ (P -->Q) --> Q
    4. Q

    Remark: I cannot find the contradiction, and am not sure whether line 4 can be used for Indirect Derivation.
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  2. #2
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    \neg \left( {p \to q} \right) \equiv p \wedge \neg q
    Simplification gives \neg q.
    But \neg \left( {p \to q} \right) \to q\;\& \,\neg q gives p\to g.
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  3. #3
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    Quote Originally Posted by Plato View Post
    \neg \left( {p \to q} \right) \equiv p \wedge \neg q
    Simplification gives \neg q.
    But \neg \left( {p \to q} \right) \to q\;\& \,\neg q gives p\to g.
    I see you did it by Modus tollens.
    Last edited by novice; November 13th 2010 at 07:37 PM.
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