Date: 02/27/2002 at 18:36:02 From: Beth Subject: Logic Proofs I have a logic proof that I'm trying to solve. I'm up to the point after I've written down all my givens. One of the givens is p-->q. I want to say ~p-->~q, with my reason being inverse. Am I allowed to do this?
Date: 02/27/2002 at 21:42:27 From: Doctor Twe Subject: Re: Logic Proofs Hi Beth - thanks for writing to Dr. Math. No. Although the statement ~p --> ~q is called the inverse of p --> q, it does not necessarily follow. Let's look at an example. Suppose that: p = "X is 2" and q = "X is an even number" Clearly, p --> q is true ("If X is 2 then X is an even number."). But is the inverse, ~p --> ~q, also true? This statement reads, "If X is NOT 2 then X is NOT an even number." Suppose X = 4. Then the "if" part, X is NOT 2, is true, but the "then" part, X is NOT an even number, is false. So the statement as a whole is false. What you *are* allowed to use in a logic proof is the contrapositive. The contrapositive of p --> q is ~q --> ~p. It turns out that any conditional proposition ("if-then" statement) and its contrapositive are logically equivalent. In our example, the contrapositive of "If X is 2 then X is an even number" would read, "If X is NOT an even number then X is NOT 2." We can see that this is also true. A third possible "switching" of the statement p --> q is q --> p. This is called the converse, but like the inverse, it does not follow logically from the original statement. The converse of our original statement would read, "If X is an even number then X is 2." Clearly, not all even numbers are 2. So the converse statement is false. (It turns out that the inverse and converse statements are logically equivalent to each other - but not logically equivalent to the original statement.) To summarize, given the statement p --> q: The inverse is q --> p, NOT equivalent to p --> q The converse is ~p --> ~q, NOT equivalent to p --> q The contrapositive is ~q --> ~p, IS equivalent to p --> q I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.com/dr.math/
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