Contrapositive

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/