# How do I find the equivalent statement to the statement ~(p -> ~q)

1. May 30, 2007

### MiffedMFG1106St

~(p -> ~q) <=> ?

I am having a heck of a time figuring this out, and I keep basically wanting to distribute the negation symbol to each individual simple statement in the conditional statement arriving at...

~[~(p -> ~q)] <=> ?
~(~p -> q) <=> ?
p -> ~q <=> ~p V ~q <=> ~(p ^ q)

But the book says I should come to p ^ q.

What am I doing wrong here.

2. May 30, 2007

### honestrosewater

Did you try changing the implication to a disjunction first, before distributing the negation?

1] ~(p -> ~q)
2] ~(~p v ~q)
3] ?

3. May 30, 2007

### MiffedMFG1106St

Okay if I play that way

~(~p v ~q) <=> ~[~(p ^ q)] <=> p ^ q

Is that the correct process to acquire that answer then. First negate the antecedent and change the conditional to a disjunction. Then does the ~ distribute to both as in my answer above? Or am I going about this the wrong way?

4. May 30, 2007

### honestrosewater

Sure, that is one way, if you have the rules to do that. You could also distribute the outer negation first and then drop the double negations.

1] ~(p -> ~q)
2] ~(~p v ~q)
3] ~~p ^ ~~q
4] p ^ q

Why does it not seem right to you?

5. May 30, 2007

### MiffedMFG1106St

It does now, I wasn't paying attention to the rule of starting with the statement in parentheses first, the least dominant statement. Is that correct?

BTW, thanks for the help so far.

Last edited: May 30, 2007
6. May 30, 2007

### honestrosewater

Well, people have different rules about parentheses, so I might not be following your rules. But they are just different ways of writing the same thing. I'm not sure what rule you are talking about.

If I added back in parentheses around every single proposition, it would look like this:

1] (~((p) -> (~(q))))
2] (~((~(p)) v (~(q)))
3] ((~(~(p))) ^ (~(~(q))))
4] ((p) ^ (q))

Unless I missed some, haha.

There is often more than one way to prove something. In fact, there are often infinitely many ways, since, for one thing, you can just add in extra steps that don't do anything. So I wouldn't worry about finding the way or even the shortest or best way. If you find one way, it usually suggests other ways.

Why are you unsure about your equivalences? You can just check that each one is an instance of one of your rules, right?

You're very welcome. Welcome to PF, by the bye.

7. May 31, 2007

### fopc

There's an important eqivalence relation regarding negation of a conditional
(assuming a propositional logic).

~(p->q) <=> (p ^ ~q)

It's important because it's the first step in a proof by contradiction.
(In a predicate logic equipped with quantifiers (and most are)
life can be more difficult.)
In your problem, the result p ^ q is immediate from the above identity.