# Thread: Negation help

1. ## Negation help

Hey guys, the question is to negate:

$\sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q)$

- $\sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q)$
- $\sim ( \sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q))$
- $\sim ( \sim p \vee \sim q) \leftrightarrow \sim (p \wedge q)$
- $\sim ( \sim p \vee \sim q) \leftrightarrow ( \sim p \vee \sim q)$

is this correct? thanks.

2. To negate a formula F, it is sufficient to put ~ before it: ~F.

3. Originally Posted by emakarov
To negate a formula F, it is sufficient to put ~ before it: ~F.
dont I have to apply the negation laws to it?

4. $
\sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q)
$

To negate this you need to first simply the given, it makes it easier...
So ~(~p v ~q) = p AND q.
NOW, the negation of the above is ~p v ~q

The second part is p A q. so the negation of that is ~p or ~ q.

When you put the two together you get...

~p v ~q <-> ~p v ~q

5. Originally Posted by matthayzon89
$
\sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q)
$
?? That's the proposition I have to negate.......

6. Originally Posted by jvignacio
?? That's the proposition I have to negate.......
Sorry, I edited it... I hope it helps.

7. Originally Posted by matthayzon89
Sorry, I edited it... I hope it helps.
thanks but im a little confused!! When it says to negate a proposition doesnt it mean put brackets around the whole proposition and then negate that using the laws of negation..? thats what I did for my answer..but that first negation sign doesnt seem to be touched if the negation for $p \leftrightarrow q = p \leftrightarrow \sim q$ , leaving p = $\sim ( \sim p \vee \sim q)$ untouched..

8. To negate a formula F, it is sufficient to put ~ before it: ~F.
I said this with tongue in cheek. In fact, I still don't understand what you problem asks. To negate a formula F means to write the negation of F, which is ~F. I am not sure if is the whole formula you need to negate, or if you need to prove this formula by negating one part. Or maybe you need to negate the whole formula but transform the result in such a way that negations ~ occur only before propositional variables... Please state your problem more precisely.

9. I made a lil mistake, but i just fixed it, in my post above....

In order to negate, like you said, you CAN put brackets around the entire proposition and then apply the laws of negation.

Are you suppose to negate [ (~p v ~q) <-> p A q ] or [~(~p v ~q) <-> p A q] ????

If the second option is what you are suppose to NEGATE then my post above should work. I applied the laws of negation to that, like you said. Be careful when negating And and OR signs.

For example:
(p v q) <---- negation of= ~p AND ~q
(p A q) <--- negation of= ~p OR ~ q

10. Originally Posted by emakarov
I said this with tongue in cheek. In fact, I still don't understand what you problem asks. To negate a formula F means to write the negation of F, which is ~F. I am not sure if is the whole formula you need to negate, or if you need to prove this formula by negating one part. Or maybe you need to negate the whole formula but transform the result in such a way that negations ~ occur only before propositional variables... Please state your problem more precisely.
Sorry for any misunderstanding. I am suppose to negate the WHOLE PROPOSITION. so: $\sim \big{(} \sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q) \big{)}$ with the outta negation sign meaning to negate the whole proposition..

11. Originally Posted by matthayzon89
I made a lil mistake, but i just fixed it, in my post above....

In order to negate, like you said, you CAN put brackets around the entire proposition and then apply the laws of negation.

Are you suppose to negate [ (~p v ~q) <-> p A q ] or [~(~p v ~q) <-> p A q] ????

If the second option is what you are suppose to NEGATE then my post above should work. I applied the laws of negation to that, like you said. Be careful when negating And and OR signs.

For example:
(p v q) <---- negation of= ~p AND ~q
(p A q) <--- negation of= ~p OR ~ q
Sorry for any misunderstanding. I am suppose to negate the WHOLE PROPOSITION. so: $\sim \big{(} \sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q) \big{)}$ with the outta negation sign meaning to negate the whole proposition..

12. Originally Posted by jvignacio
Sorry for any misunderstanding. I am suppose to negate the WHOLE PROPOSITION. so: $\sim \big{(} \sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q) \big{)}$ with the outta negation sign meaning to negate the whole proposition..
GOOD so thats exactly what I did... I didn't misunderstand you afterall

Re-read my main post explaining the negation. That should be your answer. If you don't understand, state what you don't understand about it and ill try to clarify...

13. Originally Posted by matthayzon89
GOOD so thats exactly what I did... I didn't misunderstand you afterall

Re-read my main post explaining the negation. That should be your answer. If you don't understand, state what you don't understand about it and ill try to clarify...
See my answer was exactly like yours but I ended up with a negation sign infront of $\sim ( \sim p \vee \sim q) \leftrightarrow ( \sim p \vee \sim q)$ and yours was $( \sim p \vee \sim q) \leftrightarrow ( \sim p \vee \sim q)$ OHHHHHHHH RIGHT.... i think i understand you.. You simplied the given stuff first, then negated it, then you negated the whole thing...

14. Hello, jvignacioQ!

Negate: . $\sim ( \sim\!p \:\vee \sim\! q) \Longleftrightarrow (p \wedge q)$

We have: . $\sim\bigg[\sim(\sim\!p \:\vee \sim\!q) \:\Longleftrightarrow\p \wedge q)\bigg] " alt="\sim\bigg[\sim(\sim\!p \:\vee \sim\!q) \:\Longleftrightarrow\p \wedge q)\bigg] " />

. . . . . . . . $\equiv\;\;(\sim\!p \:\vee \sim\!q) \;\rlap{\quad /}\Longleftrightarrow\; \sim(p \wedge q)$

15. Originally Posted by jvignacio
Hey guys, the question is to negate:

$\sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q)$
Maybe I am missing something, but I get a trivial result:

$\sim\Big[\sim ( \sim p \vee \sim q) \leftrightarrow (p \wedge q)\Big]$
is equivalent to
$\sim\Big[(p\wedge q)\leftrightarrow (p \wedge q)\Big]$
and hence the negation of a tautology
$\sim T$
and thus always false
$F$