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

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• Mar 19th 2013, 07:33 PM
Elusive1324
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 :|?
• Mar 19th 2013, 07:43 PM
Shakarri
Re: How do you generally prove (and disprove) If-then theorems?
Quote:

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.
• Mar 30th 2013, 12:38 PM
HallsofIvy
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.
• Mar 30th 2013, 02:41 PM
SworD
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.
• Apr 6th 2013, 09:55 PM
Mathhead200
Re: How do you generally prove (and disprove) If-then theorems?
You can also go the direct route and create a logical chain:
$\displaystyle P \rightarrow A$
$\displaystyle A \rightarrow B$
$\displaystyle B \rightarrow C$
$\displaystyle C \rightarrow Q$
$\displaystyle \therefore P \rightarrow Q$