The Negation of an Implication Statement?

  • #1
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:
[tex]P \rightarrow Q[/tex]
And let's call proposition B
[tex]\neg (P \rightarrow Q)[/tex]
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?

Thanks for any help you can provide.

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

Answers and Replies

  • #2
it is always true that

[tex]a\lor \neg a[/tex]


¬¬a is equivalent to a
 
  • #3
disregardthat
Science Advisor
1,854
33
Hello,

[tex]\neg (P \rightarrow Q) \equiv P \wedge \neg Q[/tex]
which no longer seems to be really an implication statement.
Why would it be, it's the negation of an implication statement.
 
  • #4
2
0
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.
 

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