Need help with simple proof

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mzer
 
Reply Wed 25 Jun, 2008 08:16 pm
I haven't studied logic in 25 years, and while I seem to remember more than I would imagine I would, one simple proof is escaping me. In fact, it seems I remember the more complex issues better than the simple ones.

In trying to assist my nephew with a problem, I have become stumped. Obviously, the relationship is obvious intuitively, but I am having some odd problem with the conversion. If somebody could show me the proof for this, I would be most thankful:

A iff B is equivalent to (A and B) or (not A and not B).
 
VideCorSpoon
 
Reply Thu 26 Jun, 2008 08:13 am
@mzer,
http://i30.tinypic.com/m1o9h.jpg

If you have any questions or more explanation, I'll be happy to answer any of your questions.
 
mzer
 
Reply Thu 26 Jun, 2008 08:40 am
@VideCorSpoon,
Thanks. I can get there by truth table, but I was hoping to show something more like this...

(P -> R) ^ (Q->R) / (P v Q) -> R

(-P v R) ^ (-Q v R) ---(Conditional law)
(-P ^ -Q) v R ---(Distributive law)
-(P v Q) v R ---(de Morgan's law)
(P v Q) -> R ---(Conditional law)

I'm not sure what he really needs, perhaps a TT is enough, but the above is what I remember doing years ago, so that is what I was trying to help him do. Thanks again for your help.
 
VideCorSpoon
 
Reply Thu 26 Jun, 2008 08:56 am
@mzer,
No problem. But I'm not quite sure I answered your question though.

Are you looking for truth functional translations to the inference rules or examples in which they are used?

Do you need a literal proof for that equation?
 
mzer
 
Reply Thu 26 Jun, 2008 09:07 am
@VideCorSpoon,
I need a literal proof for that particular equation.
 
VideCorSpoon
 
Reply Thu 26 Jun, 2008 09:21 am
@mzer,
Ok! But first I need to know what the exact argument is (i.e. premises and conclusion.)

The equation you gave was an example of the different yet equivalent translation of A if and only if B. Normally, this is not proofable (if that ever was a word. LOL!) because it is not a well formed formula (WFF). I can only suppose that the word "equivalent" is the conclusion indicator because there isn't any other indicator.

Are any of these formulas what you need?

A <--> B / (A & B) v (~A & ~B)
(most likely candidate for a proof as it has a premise and conclusion)

(A <--> B) <--> [(A & B) v (~A & ~B)] (Though this one needs a conclusion to form a proof. If you just wanted the argument translated into logic, this is the translation you want. You will not be able to proof it the way it is now, only translate it.)

(BTW, the slash is how I do my conclusion indicator. But there are other ways to signify it, like tridot or turnstyle)
 
mzer
 
Reply Thu 26 Jun, 2008 09:38 am
@VideCorSpoon,
VideCorSpoon;17055 wrote:
Ok! But first I need to know what the exact argument is (i.e. premises and conclusion.)

The equation you gave was an example of the different yet equivalent translation of A if and only if B. Normally, this is not proofable (if that ever was a word. LOL!) because it is not a well formed formula (WFF). I can only suppose that the word "equivalent" is the conclusion indicator because there isn't any other indicator.

Are any of these formulas what you need?

A <--> B / (A & B) v (~A & ~B)
(most likely candidate for a proof as it has a premise and conclusion)

(A <--> B) <--> [(A & B) v (~A & ~B)] (Though this one needs a conclusion to form a proof. If you just wanted the argument translated into logic, this is the translation you want. You will not be able to proof it the way it is now, only translate it.)

(BTW, the slash is how I do my conclusion indicator. But there are other ways to signify it, like tridot or turnstyle)

The first one you mention. What I need, and I guess I don't have the correct technical vocabulary, is how you get from A<-->B to (A&B) v (~A&~B) using the basic rules.

Thanks again.
 
VideCorSpoon
 
Reply Thu 26 Jun, 2008 09:52 am
@mzer,
It's absolutely no problem at all. This is the only way you could successfully complete the proof with inference and replacement rules.
http://i30.tinypic.com/kykhv.jpg
Basically, we derive (A&B)v(~A&~B) from A<-->B by means of the equivalence rule.

However, equivalence is a replacement rule. Still part of the basic rules, but more difficult to work with in bigger proofs.

Equivalence basically states that whenever P<-->Q is given, (P&Q)v(~P&~Q) or (P-->Q) & (Q-->P) can replace it... and vice versa because they basically mean the same thing.
 
 

 
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