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Math Help - proof that (p=>q) is the same as (not p => not q)

  1. #16
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    ok - now I got it!!!

    My mistrake was that I assumed the headings in both the tables were the same!!





    And of course they are not!

    You all saw the difference in the headings - I did not - I just assumed that I had the same headings! A very very stupid mistake.

    Anyway just to conclude this thread - re-reading the definitions helped be realize my mistake.

    converse: q -> p
    ===========
    the hypothesis and the conclusion switch places -- the conclusion becomes the hypothesis, the hypothesis becomes the conclusion

    inverse: not p -> not q
    ===============
    negate both the hypothesis and the conclusion

    contrapositive
    =========
    (a combination of the converse and the inverse): not q -> not p
    negate and switch the hypothesis and the conclusion

    Thanks for your help - I got there eventually.

    Jason
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  2. #17
    Super Member Gamma's Avatar
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    Another exercise for you

    can you prove the converse and inverse are logically equivalent?
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  3. #18
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    For what its worth.....


    assuming i got this right.

    (not p => not q) <=> (q => p)
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  4. #19
    Super Member Gamma's Avatar
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    Spot on! I think you have it figured out now.
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  5. #20
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    yep, i think so too.

    I have actually started to blog everything for dummies (i.e me) while I go along Jasoninclass&#039;s Blog
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  6. #21
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    Quote Originally Posted by jasonfranklin View Post
    Hi Plato,

    This is a nightmare.

    How can both tables be true?

    I want to be able to show that this table is true.



    It is clear why the p=>q colums gets those values. (p=>q) is only false when p is true and q is false. But.....Why on earth does the (not q => not p) column get those values? based on what?
    Based on the definition of =>, of course. a=> b is true in all cases except when a is true and b is false. That means that ~Q=> ~P is true in all cases except the on where ~Q is true (i.e. Q is false) and ~P is false (i.e. P is true). If you look at the last column you will see "1" ("true") in all rows except the one where P= 1 (P is true) and Q= 0 (Q is false).

    By the way, have you realized that you started by talking about "not p => not q" which is certainly NOT the contrapositive you are now talking about "not q=> not p".
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  7. #22
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    Hallsofivy, many thanks for your feedback at this.

    I got there eventually - I realized the mistake I was making at the end of the day and documented it earlier on in the thread.

    Thanks!
    Last edited by jasonfranklin; August 5th 2009 at 10:54 PM. Reason: not to plato - but to hallsofivy
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