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From: Newberry on 14 Feb 2010 00:00
I had claimed that if for all a in the range of x

(y)Aay (1)

is vacuously true then

(x)(y)Aay (2)

is vacuously true. I have got an objection that this is scope fallacy
and that the inference is incorrect. I do not know, but I no longer
claim that (2) is vacuously true. I do claim that if for all a in the
range of x

(y)Aay (1)

is neither true nor false then

(x)(y)Aay (2)

is neither true nor false. A new version of my paper titled 'When Are
Relations Neither True Nor False?' is posted here
http://www.scribd.com/doc/26833131/RelationsAndPresuppositions-2010-02-13.

Note that given

~(Ex)(Ey)[(x + y < 6) & (y = 8)]

in Strawson's logic of presuppositions the formula is neither true nor
false for any choice of y. Let y be 8:

~(Ex)[(x + 8 < 6) & (8 = 8)]
i.e.
(x)[(x + 8 < 6) -> ~(8 = 8)]

the subject class is empty and hence the sentence is neither true nor
false.

Let y be anything but 8, say 9:

~(Ex)[(x + 9 < 6) & (9 = 8)]
i.e.
(x)[(9 = 8) -> ~(x + 9 < 6)]

the subject class is empty and hence the sentence is neither true nor
false.

Comments appreciated.
From: Jesse F. Hughes on 14 Feb 2010 06:11
Newberry <newberryxy(a)gmail.com> writes:

> I had claimed that if for all a in the range of x
>
> (y)Aay (1)
>
> is vacuously true then
>
> (x)(y)Aay (2)

That's supposed to be (x)(y)Axy, I suppose?

--
Jesse F. Hughes
"[I]t's the damndest thing. There's something wrong with every last
one of you, and I *never* thought that was a possibility. But now I
feel it's the only reasonable conclusion." --JSH sees some sorta light
From: Jesse F. Hughes on 14 Feb 2010 06:18
Newberry <newberryxy(a)gmail.com> writes:

> Note that given
>
> ~(Ex)(Ey)[(x + y < 6) & (y = 8)]
>
> in Strawson's logic of presuppositions the formula is neither true nor
> false for any choice of y. Let y be 8:
>
> ~(Ex)[(x + 8 < 6) & (8 = 8)]
> i.e.
> (x)[(x + 8 < 6) -> ~(8 = 8)]
>
> the subject class is empty and hence the sentence is neither true nor
> false.
>
> Let y be anything but 8, say 9:
>
> ~(Ex)[(x + 9 < 6) & (9 = 8)]
> i.e.
> (x)[(9 = 8) -> ~(x + 9 < 6)]
>
> the subject class is empty and hence the sentence is neither true nor
> false.

Plug in 0 for y and you get

(x)[(x + 0 < 6) -> ~(0 = 8)]

which is surely true. Unfortunately, this is equivalent to

(x)[(0 = 8) -> ~(x + 0 < 6)],

which has an empty subject class[1], and so (according to you) is
neither true nor false. Oops!


Footnotes:
[1] I think that when you write "(x)(Px -> Qx)" has an empty subject
class, you mean simply that (x)~Px. Correct me if I'm mistaken.

--
Jesse F. Hughes
"Being wrong is easy, knowing when you're right can be hard, but
actually being right and knowing it, is the hardest thing of all."
-- James S. Harris
From: calvin on 14 Feb 2010 09:14
The 'continuum hypothesis' is neither true nor false,
for example.
From: Aatu Koskensilta on 14 Feb 2010 09:34
calvin <crice5(a)windstream.net> writes:

> The 'continuum hypothesis' is neither true nor false,
> for example.

This piece of philosophical reflection -- which stands in need of some
argument -- has no apparent relevance to Newberry's original post.

--
Aatu Koskensilta (aatu.koskensilta(a)uta.fi)

"Wovon man nicht sprechan kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
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Prev: geometry precisely defining ellipsis and how infinity is in the midsection #427 Correcting Math
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