# proof that (p=>q) is the same as (not p => not q)

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Aug 2nd 2009, 11:52 AM
jasonfranklin
proof that (p=>q) is the same as (not p => not q)
I cannot come up with the proof that (p => q) is the same as (not p => not q).

If anyone can help - that would be great.

As I understand it: (p => q) is always true except when p is true and q is false.
If that definition is true, then (not p => not q) cannot possibly be the same.

http://jasoninclass.wordpress.com/fi.../picture-1.png

This cannot be. What am I missing?

Thanks a million for any constructive help.

Jason
• Aug 2nd 2009, 12:10 PM
Danneedshelp
Are you sure you are not trying to show $P\Rightarrow{Q}$ is equivalent to $\sim{Q}\Rightarrow{\sim{P}}$ ?
• Aug 2nd 2009, 12:30 PM
Bruno J.
You're usually going to have a hard time proving false things; that's the beauty of mathematics. It never lets us fool ourselves.

I suggest always thinking of a small example to see if your "theorem" is true before you start trying to prove it.

If I have a cold, I have a sore throat.
Does that imply that if I don't have a cold, I don't have a sore throat?
• Aug 2nd 2009, 01:49 PM
jasonfranklin
This is not a question of perspectives.

I am talking about the contrapositive of p=>q.

is p=>q the same as (not p => not q).

can this be proved in a truth table.
• Aug 2nd 2009, 02:09 PM
o_O
The contrapositive of $p \Rightarrow q$ is : ${\color{white}.} \sim q \Rightarrow \ \sim p$

Contrapositive
• Aug 2nd 2009, 02:11 PM
Plato
Quote:

Originally Posted by jasonfranklin
This is not a question of perspectives.
I am talking about the contrapositive of p=>q.
is p=>q the same as (not p => not q).
can this be proved in a truth table.

The contrapositive is: $P\, \Rightarrow Q \equiv \neg Q\, \Rightarrow \neg P$.
Not what you wrote.

$\neg P\, \Rightarrow \neg Q$ is the inverese of $P\, \Rightarrow Q$

The inverse is not the same as the contrapositive.
• Aug 2nd 2009, 02:14 PM
jasonfranklin
this is not my information.

p=>q
q=>p is the converse
not p => not q is the contrapositive

what is then the contrapositive in your book???
• Aug 2nd 2009, 02:18 PM
o_O
Perhaps you wrote it wrong in your notes. Three members have told you what the contrapositive is as well as providing a link to the wikipedia article on what it is.

Try a google search and tell us what you find: What is contrapositive

You even told us that you couldn't show by truth tables that the two statements you have aren't equivalent. And I assure you that you won't be able to simply because they aren't.
• Aug 2nd 2009, 02:22 PM
Plato
Quote:

Originally Posted by jasonfranklin
this is not my information.
p=>q
q=>p is the converse
not p => not q is the contrapositive

You information is incorrect.
It is at odds with the whole history of logic.
I assure that it must be a typo or missprint in your textbook.
Reread my post to you. It is the correct information.
• Aug 2nd 2009, 02:30 PM
Danneedshelp
Quote:

Originally Posted by jasonfranklin
this is not my information.

p=>q
q=>p is the converse
not p => not q is the contrapositive

what is then the contrapositive in your book???

At least three posts already contain the definition of the contrapositive.
• Aug 2nd 2009, 02:59 PM
jasonfranklin
thanks for the help - but the wiki article is telling me this:

http://jasoninclass.wordpress.com/fi.../picture-2.png

is the above truth table correct?

or this one?

http://jasoninclass.wordpress.com/fi.../picture-1.png
• Aug 2nd 2009, 03:08 PM
Plato
Using your notation this is correct.
$
\begin{array}{c|c|c|c}
P & Q & {P\, \Rightarrow Q} & {\neg Q \Rightarrow \neg P} \\
\hline
0 & 0 & 1 & 1 \\
0 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 \\

\end{array}$
• Aug 2nd 2009, 03:21 PM
jasonfranklin
Plato . thanks a million billion - i thought i was going mad.

the only thing is i dont get why....

if the definition for p=>q is that it is false only when p is true and q is false then surely this table should be correct. (even though it is obviously not).

http://jasoninclass.wordpress.com/fi.../picture-1.png

if p=>q is false only when p is true and q is false,

is also (not p=> not q) false only when not p is true and not q is false?
• Aug 2nd 2009, 03:29 PM
Plato
That table is correct. But it proves nothing.
If one wants to show two statement forms are equivalent, then there truth columns must be identical.
That is not the case in the above.
You have compared the statement “If P then Q.” with its inverse “If not P then not Q’.
You have shown that they are not equivalent. Their columns are not the same.
• Aug 3rd 2009, 12:05 AM
jasonfranklin
Hi Plato,

This is a nightmare.

How can both tables be true?

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

http://www.mathhelpforum.com/math-he...001022a0-1.gif

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?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last