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Math Help - Showing a conclusion is correct (with deduction/reduction)

  1. #1
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    Showing a conclusion is correct (with deduction/reduction)

    I'll use some replacement characters since I don't know how to make the correct mathematical ones.

    A = Logic symbol for and
    --> = symbol for implies
    - (it's that hook thing normally) =
    negative

    Here's the problem:

    Show that the conclusion

    (p \rightarrow q)  \wedge (-p  \rightarrow r)  \wedge ((-p \wedge r) \rightarrow s) \wedge -q ==> s

    is correct. Do this using both deduction and reduction methods.

    I'm just stranded here, I really don't know where to begin and how to proceed. Thanks for any help.
    Last edited by JacobSkylar; September 27th 2009 at 12:37 PM.
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  2. #2
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    Quote Originally Posted by JacobSkylar View Post
    I'll use some replacement characters since I don't know how to make the correct mathematical ones.

    A = Logic symbol for and
    --> = symbol for implies
    - (it's that hook thing normally) =
    negative

    Here's the problem:

    Show that the conclusion

    (p --> q) A (-p --> r) A ((-p A r) --> s) A -q ==> s

    is correct. Do this using both deduction and reduction methods.

    I'm just stranded here, I really don't know where to begin and how to proceed. Thanks for any help.
    You can learn to use symbols
    [tex] \wedge [/tex] gives  \wedge is and
    [tex] \Rightarrow [/tex] gives  \Rightarrow is imply
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  3. #3
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    Quote Originally Posted by Plato View Post
    You can learn to use symbols
    [tex] \wedge [/tex] gives  \wedge is and
    [tex] \Rightarrow [/tex] gives  \Rightarrow is imply
    Fixed.
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  4. #4
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    Quote Originally Posted by JacobSkylar View Post
    (p \rightarrow q)  \wedge (-p  \rightarrow r)  \wedge ((-p \wedge r) \rightarrow s) \wedge -q ==> s
    \neg q from the first gives \neg p.
    \neg p gives r which gives \neg p\wedge r.
    That in turn gives s
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  5. #5
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    Quote Originally Posted by Plato View Post
    \neg q from the first gives \neg p.
    \neg p gives r which gives \neg p\wedge r.
    That in turn gives s
    Which of the two methods was that, and would you mind showing the other one too? Feel free to add in why things do what they do, what rules you apply.

    Thanks alot.
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  6. #6
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    Quote Originally Posted by JacobSkylar View Post
    Which of the two methods was that, and would you mind showing the other one too? Feel free to add in why things do what they do, what rules you apply.
    I don't know what you mean.
    I used simplification to get \neg q.
    Then modus tollens to get \neg p.
    Then modus ponens to get  r.
    Then modus conjunction to get \neg p\wedge r.
    Then modus ponens to get s.
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  7. #7
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    Quote Originally Posted by Plato View Post
    I don't know what you mean.
    I used simplification to get \neg q.
    Then modus tollens to get \neg p.
    Then modus ponens to get  r.
    Then modus conjunction to get \neg p\wedge r.
    Then modus ponens to get s.
    Okay, thanks alot.

    Nvm, I realised you used the deduction method. Would you know how to solve it using the reduction method too?
    Last edited by JacobSkylar; September 30th 2009 at 12:33 PM.
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  8. #8
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    So, do you know how to solve it using the reduction method?

    1. Assume that the conclusion is false
    2. The premises are true but the conclusion is false
    3. This would mean is true etc etc

    (note I don't know the mathematical terms in english for everything)

    Would be really appreciated.
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