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Topic: This is False. 0/0 {x | x ~e x} e {x | x ~e x} A single Principle

Replies: 53   Last Post: Feb 13, 2013 3:53 PM

 Messages: [ Previous | Next ]
 William Hale Posts: 49 Registered: 5/2/12
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 2:33 PM

In article
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 don't make the assumption that primitives of logical expressions must
be relations. I assume you mean the relation "~e" to be the set of
ordered pairs (x, y) such that x ~e y.

Since I don't take logical expressions to be sets, I certainly don't
take logical expressions to be relations. I 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 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 don't include the substitution for symbols in my consideration of
whether a statement or expression is a wff. For me, 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 know? If you want to see
if I know, I will concede that I am not an expert on wffs. I suspect
that you want to show that it cannot be explained even though people
claim that a wff can be defined.

Date Subject Author
2/1/13 Graham Cooper
2/3/13 Charlie-Boo
2/3/13 Graham Cooper
2/3/13 Charlie-Boo
2/3/13 Graham Cooper
2/3/13 Graham Cooper
2/3/13 Charlie-Boo
2/3/13 Graham Cooper
2/3/13 Charlie-Boo
2/3/13 camgirls@hush.com
2/4/13 Charlie-Boo
2/4/13 billh04
2/4/13 Charlie-Boo
2/4/13 William Hale
2/4/13 Lord Androcles, Zeroth Earl of Medway
2/9/13 Graham Cooper
2/5/13 Charlie-Boo
2/4/13 Graham Cooper
2/5/13 Charlie-Boo
2/5/13 Graham Cooper
2/5/13 Brian Q. Hutchings
2/6/13 Graham Cooper
2/6/13 Charlie-Boo
2/4/13 fom
2/4/13 Charlie-Boo
2/4/13 fom
2/5/13 Charlie-Boo
2/7/13 fom
2/9/13 Charlie-Boo
2/9/13 Graham Cooper
2/11/13 Charlie-Boo
2/10/13 fom
2/10/13 Graham Cooper
2/10/13 fom
2/10/13 Graham Cooper
2/11/13 Charlie-Boo
2/11/13 Charlie-Boo
2/11/13 Charlie-Boo
2/11/13 Graham Cooper
2/13/13 Charlie-Boo
2/11/13 Charlie-Boo
2/11/13 fom
2/5/13 Charlie-Boo
2/5/13 fom
2/6/13 fom
2/11/13 Charlie-Boo
2/11/13 fom
2/13/13 Charlie-Boo
2/13/13 fom
2/4/13 Graham Cooper
2/4/13 Charlie-Boo
2/5/13 Charlie-Boo