Re: A proof for !(p^q) |- !p v !q

From: "Paul Holbach" <paulholbachDELETETHENAME@xxxxxxxxxx>
Date: 7 Sep 2005 09:16:32 -0700

> George Dance wrote:
> > Newbie wrote:

> > Anyone knows how to proof it by natural deduction ?

> I assume you're trying to prove this for a course, and are only allowed
> to use certain rules (otherwise you could prove it directly using
> DeMorgan's rules).

~(p & q) <-> (~p v ~q)

This is de Morgan 2!

#PH

.

References:
- A proof for !(p^q) |- !p v !q
  - From: Newbie
- Re: A proof for !(p^q) |- !p v !q
  - From: George Dance

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