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

1. ok - now I got it!!!

My mistrake was that I assumed the headings in both the tables were the same!!

And of course they are not!

You all saw the difference in the headings - I did not - I just assumed that I had the same headings! A very very stupid mistake.

Anyway just to conclude this thread - re-reading the definitions helped be realize my mistake.

converse: q -> p
===========
the hypothesis and the conclusion switch places -- the conclusion becomes the hypothesis, the hypothesis becomes the conclusion

inverse: not p -> not q
===============
negate both the hypothesis and the conclusion

contrapositive
=========
(a combination of the converse and the inverse): not q -> not p
negate and switch the hypothesis and the conclusion

Thanks for your help - I got there eventually.

Jason

2. Another exercise for you

can you prove the converse and inverse are logically equivalent?

3. For what its worth.....

assuming i got this right.

(not p => not q) <=> (q => p)

4. Spot on! I think you have it figured out now.

5. yep, i think so too.

I have actually started to blog everything for dummies (i.e me) while I go along Jasoninclass&#039;s Blog

6. Originally Posted by 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.

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?
Based on the definition of =>, of course. a=> b is true in all cases except when a is true and b is false. That means that ~Q=> ~P is true in all cases except the on where ~Q is true (i.e. Q is false) and ~P is false (i.e. P is true). If you look at the last column you will see "1" ("true") in all rows except the one where P= 1 (P is true) and Q= 0 (Q is false).

By the way, have you realized that you started by talking about "not p => not q" which is certainly NOT the contrapositive you are now talking about "not q=> not p".

7. Hallsofivy, many thanks for your feedback at this.

I got there eventually - I realized the mistake I was making at the end of the day and documented it earlier on in the thread.

Thanks!

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