# The Negation of an Implication Statement?

Hello,

So someone just asked me for assistance on a proof, and while I'm fairly certain you can't do what he did, I am not completely sure on the reasons.

To state it as formal logic,
If you have proposition A:
$$P \rightarrow Q$$
And let's call proposition B
$$\neg (P \rightarrow Q)$$
If you were to show B was false, then I think that does not imply A is true.

Am I right? And what logic is really going on above?

EDIT:
I tried looking at the implication as,
$$P \rightarrow Q \equiv \neg P \vee Q$$
which means that
$$\neg (P \rightarrow Q) \equiv P \wedge \neg Q$$
which no longer seems to be really an implication statement.

Last edited:

Related Set Theory, Logic, Probability, Statistics News on Phys.org
it is always true that

$$a\lor \neg a$$

¬¬a is equivalent to a

disregardthat
Hello,

$$\neg (P \rightarrow Q) \equiv P \wedge \neg Q$$
which no longer seems to be really an implication statement.
Why would it be, it's the negation of an implication statement.

I'm a little unclear on exactly what your question is and what system of logic you are talking about.

If you are talking about intuitionistic/constructive logic then a proof that B is false (i.e. a proof of the negation of B, which I believe in intuitionistic terms would be "a proof that B cannot be proven") would not be a proof of A, since intuitionistic logic does not have a double negation elimination rule. Intuitionistic logic also has a different definition of negation than classical logic (part of the reason there is no double negation rule).

However, in classical logic, a proof of the negation of B would be the double negation of A, which is equivalent to A via a rule of double negation elimination.