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Thread: Determine whether he following logic argument is valid.

  1. #1
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    Determine whether he following logic argument is valid.

    I am stuck in this question. Please help.

    Determine whether he following logic argument is valid.-untitled.jpg
    Last edited by mr fantastic; Aug 21st 2011 at 12:30 PM. Reason: Re-titled.
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  2. #2
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    Re: Need help in solving this question

    We have to deduce $\displaystyle t\rightarrow w$, which is equivalent to $\displaystyle \neg w \rightarrow \neg t$, from:

    $\displaystyle \\ \neg p \rightarrow (r \wedge \neg s) \quad (1) \\ t \rightarrow s \quad (2) \\ u \rightarrow \neg p \quad (3) \\ \neg w \quad (4) \\ u \vee w \quad (5) \\ u \quad (6) \mbox{ From 5 and 4} \\ \neg p \quad (7) \mbox{ From 3 and 6 } \\ r \wedge \neg s \quad (8) \mbox{ From 1 and 7 } \\ \neg s \quad (9)\mbox{ From 8 } \\ \neg t \quad (10) \mbox{ From 2 and 9 } \\ \neg w \rightarrow \neg t \quad \mbox{ From 4 and 10 }$

    Hence, it's a valid argument. Hope this helps you.
    Last edited by MATHNEM; Aug 21st 2011 at 12:02 PM. Reason: Corrected negation symbol
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  3. #3
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    Re: Need help in solving this question

    How do you get (9) from 8? Truth table gives F T F F which is something different.
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  4. #4
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    Re: Need help in solving this question

    Quote Originally Posted by terrorsquid View Post
    How do you get (9) from 8? Truth table gives F T F F which is something different.
    $\displaystyle \frac{\alpha \land \beta}{\alpha } $

    and,

    $\displaystyle \frac{\alpha \land \beta}{\beta }$

    Edit:

    Proof:

    $\displaystyle \alpha \land \beta$ true only when $\displaystyle \alpha$ and $\displaystyle \beta $ are both true. Therefor, if $\displaystyle \alpha \land \beta$ true at particular reference, so are $\displaystyle \alpha$ and $\displaystyle \beta$ are true in that particular reference.

    And we can write, $\displaystyle \alpha \land \beta \models \alpha$ and $\displaystyle \alpha \land \beta \models \beta$
    Last edited by Also sprach Zarathustra; Aug 21st 2011 at 02:15 AM.
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  5. #5
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    Re: Need help in solving this question

    Quote Originally Posted by Also sprach Zarathustra View Post
    $\displaystyle \frac{\alpha \land \beta}{\alpha } $

    and,

    $\displaystyle \frac{\alpha \land \beta}{\beta }$
    Is there somewhere I can read more about that? is it a law of logic? I don't understand.
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  6. #6
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    Re: Need help in solving this question

    Quote Originally Posted by Also sprach Zarathustra View Post
    Proof:

    $\displaystyle \alpha \land \beta$ true only when $\displaystyle \alpha$ and $\displaystyle \beta $ are both true. Therefor, if $\displaystyle \alpha \land \beta$ true at particular reference, so are $\displaystyle \alpha$ and $\displaystyle \beta$ are true in that particular reference.

    And we can write, $\displaystyle \alpha \land \beta \models \alpha$ and $\displaystyle \alpha \land \beta \models \beta$
    How do we know that r is true and that s is false though? such that $\displaystyle (r\wedge -s) \equiv~$ $\displaystyle (T\wedge -F)$

    Not following sry :S
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  7. #7
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    Re: Need help in solving this question

    We have correctly inferred that $\displaystyle r \wedge \sim s$ is true, but this implies that $\displaystyle r$ is true and so is $\displaystyle \sim s$.

    Simplification - Wikipedia, the free encyclopedia

    I'm sorry I couldn't find something else.
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  8. #8
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    Re: Need help in solving this question

    A word about not and LaTeX.
    [tex]\neg P[/tex] gives $\displaystyle \neg P $
    Last edited by mr fantastic; Aug 21st 2011 at 12:32 PM. Reason: Fixed [tex] and [noparse] tags.
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