Question

Claim: For any two propositions P, Q, ¬P ∧ ¬Q is equivalent to ¬[P ∧ Q].

Answer 1

Proof: Suppose that ¬P ∧ ¬Q is true. Then both ¬P and ¬Q are true. So P and Q are both false. Thus P ∧Q is false. Hence ¬[P ∧Q] is true.
This argument clearly works the other way. So we have implication in both directions, which proves the claim.

Is the proof correct or am I missing something?

Answer 2

You're missing something.

Answer 3

Don't assume that the argument "clearly" works backwards when it doesn't. lol

Answer 4

I was just checking the backwards ...

Answer 5

¬[P ∧Q] is true
P ∧Q is false
That means either P or Q is false, but one of them could be true.

Answer 6

please elaborate ...

Answer 7

P ∧Q is false
P false, Q false
P true, Q false
P false, Q true

Answer 8

Well, the best way would just be to start from the beginning and suppose that the statement is false to start with, since if you're claiming "For any two propositions" then it must also work when you suppose that ¬P ∧ ¬Q is false.

Answer 9

It works when you suppose it's true and go that direction, but going backwards or supposing it's false will lead to a different conclusion. In fact, this one's so popular it's got a special name, DeMorgan's law. http://en.wikipedia.org/wiki/De_Morgan's_laws

Answer 10

¬[P ∧Q] equivalent to ¬P v¬Q which is not equivalent to ¬P∧¬Q . Correct ?

Answer 11

Exactly. Just trying to equate the last two statements is obviously wrong, since you can replace ~P and ~Q with just P and Q and you will basically be saying the and and or operators are the same!

Answer 12

Should I keep the first part of the proof or should I change it as you suggested ?

Answer 13

you cant prove this

Answer 14

trying to prove it False ?

Answer 15

Now I'm really confused ... lost in logic :{