I'm working on some assigned problems for my Discrete Math class, and I have some questions about a couple of the last ones...

First, the question talks about finding equivalent wffs for AvB and A-->B.
Then, it asks me to show that AvB is equivalent to (A'^B')' and show A-->B is equivalent to (A^B')'.

I don't quite understand how to do this...
Initially I did something like "AvB <--> (A'^B')' => (AvB)' <-->A'^B' => A'^B' <--> A'^B'
Then I thought I was supposed to do a truth table, so I did.
Now I'm thinking that I wasn't supposed to do either of those....



The next question says "Show that every compound wff is equivalent to a wff using only the connectives of (v and ') and (--> and ').

I guess the problem is I don't really know what a wff is, unless it's just a string of variables and connectives including a --> ?

Thanks for any help =/

We did truth tables to show both are equivalent. I'm not sure if your instructor gave special instructions on how to show this, but I would assume Truth tables

If that's the case, then the next question really confuses me.

"Show that every compound wff is equivalent to a wff using only the connectives of (v and ') and (--> and ')"

What is every compound wff?
This is literally all the information given in the problem....

Up.

I still need help with this by Wednesday =/
I appreciate any help.

Hard to understand what exactly you're asked for...
A truth table is generally the best way to answer these kind of things.

Maybe it was just a question of formulation :

AvB is false when and only when neither A and B are false, hence : (AvB) <-> (A'^B')'

A->B is false when and only when A is true and B is false, hence : (A->B) <-> (A^B')'

??

The second part is more tricky. My understanding is that a wff uses v ,^ ,non , -> and <->.
In this case, maybe the question is to show than any of these connectives can be rewritten using only (v and ') or (--> and ').

You've already shown that v and -> can be respectively replaced by a combination of ^ and non...
<-> is of course (->) ^ (<-)
Now try to figure out why : (A^B) <-> (A' v B')'

Hope it helps...

Okay, so thinking some more about it, if I can show that every connective can be replaced with a combination of (v and ') or (--> and '), then in effect every possible wff can be rewritten using only these connectives....

I guess I'll be able to do that, but it seems a little more difficult than the previous problems =/

Thanks for the help.

/e Just re-read feanur's post and realized that's exactly what he suggested xD
Thanks again, that's probably what they're looking for =)

This post was edited by furbyjs on Aug 26 2013 11:16am