'When Are Relations Neither True Nor False?
in [Math]
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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 |