[FOM] The denial of '~p'

[Richard Heck]

On 08/28/2010 02:12 PM, Alex Blum wrote:
> Richard Heck objects to my:
> One anomaly in thinking of 'p' as the denial of '~p' is that while it is
> immediately clear that it must be the case that one of the two is true, it
> is not immediately clear that this disallows them from being jointly true.
> However, it is immediately clear that this disallows, the denial of each,
> i.e.,'~p' and '~~p', from being jointly true.
>
> He writes:
>
> | I'm confused. How can it be any less clear that `p' and `~p' cannot both
> | be true than it is that `~p' and `~~p' cannot both be true? Surely what
> | makes it clear that the latter cannot both be true (in so far, with a
> | nod to dialetheists, it is) is the fact that one is the negation
> | (denial, if you wish) of the other. But that is true in the other case,
> | too: whether `p' is the denial of `~p' or not, surely `~p' is the denial
> | of `p', and so they cannot both be true.
> |
>
> But in one case we are assuming that a statement and its denial cannot both
> be true and in the other we are not.
>
>    
My question was why you think that "it is immediately clear" that `~p' 
and `~~p' cannot both be true. It's not that I was disagreeing (though 
some would). I just did not see any reason to think that that would not 
apply to `p' and `~p'. The only reason I could see was that you think, 
quite generally, that a statement and its negation (denial, if you wish) 
cannot both be true. If so, however, then I was pointing out that the 
very same reasoning would show that `p' and `~p' cannot both be true.

I appreciate that you may have had some other reasoning in mind, but 
this comment does not clarify it. So let's move slowly. Let's write "Dp" 
for the denial of p, whatever that is.

Question 1: Is it supposed to be clear, in general, that Dp and p cannot 
both be true, whatever p may be (i.e., even if it is itself a complex 
formula)?
Question 2: Is ~p the same as Dp, in your view (so, e.g., that D(~p) 
would be: ~~p) If not, what is their relation? Does ~p entail Dp, in 
some sense?

If the answer to (1) is "yes", and we at least have the entailment in 
(2), then my argument goes through.

The point is that the intuition that two statements cannot both be true 
requires only that ONE be the denial of the other; you do not need to 
know that each is the denial of the other. So the fact (if it is one) 
that `p' is not the denial `~p' does not undermine the intuition that 
they cannot both be true, since `~p' looks for all the world like it is 
(or in some sense entails) the denial of `p'.

> | I would have thought, moreover, that if anything isn't clear here, it is
> | that one of `p' and `~p' must be true.
>
> That's not me. I wrote: "it must be the case that one of the two is true".
> Surely there is a difference.
>
>    
If what you have in mind is the a difference between "One of `p' and 
`~p' must be true" and "It must be that one of `p' and `~p' is true", 
the former can have the latter reading---that is, "must" can scope over 
"one", as in "one of us must pay the bills"---and it was the latter 
reading that I intended. As others have said, there are plenty of folks 
who think it is NOT the case that it must be the case that one of `p' 
and `~p' is true. I'm not necessarily among them, though I might be.

Richard