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Math Help - How do you generally prove (and disprove) If-then theorems?

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    How do you generally prove (and disprove) If-then theorems?

    Suppose a theorem proposes If P then Q.

    Are the only two ways to prove that the If-then statement is true is to show that for any P, Q must be the consequent?
    I've heard that another way to prove that the statement is true is to show that for any ~Q, the consequent must be ~P. That is, if we show that If ~Q then ~P is true, then we can conclude If P then Q is true.


    Disproving an If-then statement would be: Show that for at least one P, ~Q or for one ~Q, not ~P.

    That all there is to this logic stuff :|?
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    Re: How do you generally prove (and disprove) If-then theorems?

    Are the only two ways to prove that the If-then statement is true is to show that for any P, Q must be the consequent?
    I've heard that another way to prove that the statement is true is to show that for any ~Q, the consequent must be ~P. That is, if we show that If ~Q then ~P is true, then we can conclude If P then Q is true.
    Yes that line of logic works. Proof by contradiction uses this idea.
    Last edited by Shakarri; March 19th 2013 at 07:49 PM.
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    Re: How do you generally prove (and disprove) If-then theorems?

    That is proving the "contrapositive". "if P then Q" is equivalent to "if not Q then not P". "Proof by contradiction" is a little more general than that. You use "P" to prove statement "X", then use "not Q" to prove statement "Y" and show that "X" and "Y" contradict each other.
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    Re: How do you generally prove (and disprove) If-then theorems?

    Yes. But it is very important that you see why this is the case. It is not simply a formal thing: its logical. If you prove that ~Q implies ~P, you are showing that if P is true, ~Q COULDNT be the case, because then ~P. Therefore Q. Sorta what Plato said.
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    Re: How do you generally prove (and disprove) If-then theorems?

    You can also go the direct route and create a logical chain:
    P \rightarrow A
    A \rightarrow B
    B \rightarrow C
    C \rightarrow Q
    \therefore P \rightarrow Q
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