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Re: This is False. 0/0 {x | x ~e x} e {x | x ~e x} A single Principle to Resolve Several Paradoxes
Posted:
Feb 4, 2013 2:43 PM
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"William Hale" wrote in message
news:billh04-154C6A.13334504022013@news.eternal-september.org...
In article
<c4a53c56-e96f-4524-b9dd-b7ae8c75160a@r14g2000yqe.googlegroups.com>,
Charlie-Boo <shymathguy@gmail.com> wrote:
> On Feb 4, 9:32 am, billh04 <bill...@gmail.com> wrote:
> > On Feb 4, 6:26 am, Charlie-Boo <shymath...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > On Feb 3, 11:53 pm, camgi...@hush.com wrote:> On Feb 4, 2:19 pm,
> > > Charlie-Boo <shymath...@gmail.com> wrote:
> >
> > > > > > RELATION
> > > > > > p(a, b, e)
> >
> > > > > If wffs are built on relations then { x | x ~e x } is not a wff
> > > > > because ~e is not a relation.
> >
> > > > if e(x,y) is a predicate
> > > > then not(e(x,y)) is a predicate
> >
> > > And more importantly not(e(x,x)) is a predicate (diagonalization.)
> >
> > > Yes, that is Naïve Set Theory, which is correct. But the IF fails.
> >
> > > "e(x,y) is a predicate" is not correct due to diagonalization. There
> > > is no Russell Paradox, only Russell's Diagonalization.
> >
> > > If e(x,y) were a predicate then not(e(x,x)) would be a predicate but
> > > because of diagonalization it is not.
> >
> > But, in ZFC, the statement "Ax.not x e x" is true and the statement
> > "Ex. x e x" is false, among many other such statement. Certainly, e(x,
> > y) and e(x, x) must be a predicate in ZFC. How can it not be?
>
> In this case, because primitives of logical expressions must be
> relations and ~e is not a relation.
I (1) don't make the assumption that primitives of logical expressions must
be relations. I (2) assume you mean the relation "~e" to be the set of
ordered pairs (x, y) such that x ~e y.
Since I (3) don't take logical expressions to be sets, I (4) certainly don't
take logical expressions to be relations. I (5) would prefer to say that a
logical expression may sometimes determine a set. But sometimes a
logical expression won't determine a set (e.g., the logical expression
"x ~e x" wont' determine a set.)
Thus, I (6) say that "x ~e x" is a wff, but "x ~e x" cannot be used to
define a relation that corresponds to it.
> It depends on how you define wff,
> including substitution for (aka interpreting) symbols for these
> primitives.
I (7) don't include the substitution for symbols in my consideration of
whether a statement or expression is a wff. For me (8), whether something is
a wff is a syntactical question.
> How do you define it?
Do you want to know or do you want to see if I (9) know? If you want to see
if I (10) know, I (11) will concede that I (12) am not an expert on wffs. I
(13) suspect
that you want to show that it cannot be explained even though people
claim that a wff can be defined.
==============================================================
Thirteen references to yourself.
You really are an opinionated fuckwit, aren't you?
*plonk*
-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway.
When the fools chicken farmer Wilson and Van de faggot present an argument I
cannot laugh at I'll retire from usenet.
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