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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 9:58 AM
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On Feb 4, 9:32 am, billh04 <bill...@gmail.com> wrote:
> On Feb 4, 6:26 am, Charlie-Boo <shymath...@gmail.com> wrote:
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> > On Feb 3, 11:53 pm, camgi...@hush.com wrote:> On Feb 4, 2:19 pm, Charlie-Boo <shymath...@gmail.com> wrote:
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> > > > > RELATION
> > > > > p(a, b, e)
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> > > > If wffs are built on relations then { x | x ~e x } is not a wff
> > > > because ~e is not a relation.
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> > > if e(x,y) is a predicate
> > > then not(e(x,y)) is a predicate
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> > And more importantly not(e(x,x)) is a predicate (diagonalization.)
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> > Yes, that is Naïve Set Theory, which is correct. But the IF fails.
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> > "e(x,y) is a predicate" is not correct due to diagonalization. There
> > is no Russell Paradox, only Russell's Diagonalization.
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> > If e(x,y) were a predicate then not(e(x,x)) would be a predicate but
> > because of diagonalization it is not.
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> 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. It depends on how you define wff,
including substitution for (aka interpreting) symbols for these
primitives. How do you define it?
C-B
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