Date: Feb 5, 2013 8:51 AM
Author: Charlie-Boo
Subject: Re: This is False. 0/0 {x | x ~e x} e {x | x ~e x} A single Principle<br> to Resolve Several Paradoxes

On Feb 4, 2:33 pm, William Hale <> wrote:
> In article
> <>,
>  Charlie-Boo <> wrote:

> > On Feb 4, 9:32 am, billh04 <> wrote:
> > > On Feb 4, 6:26 am, Charlie-Boo <> wrote:
> > > > On Feb 3, 11:53 pm, wrote:> On Feb 4, 2:19 pm,
> > > > Charlie-Boo <> 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.

Except that it?s not a set i.e. relation. (Set = Relation =
Expressible = Has negation, as discussed.)

> 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.

Eggs. Eggs actly. And reduce to x ~e x (work backwards from x ~e x)
to show what else is not a set, just as we reduce to the Halting
Problem. But actually reduction to the set of programs that do not
halt yes on themselves is more primitive and thus more general. It
applies to systems that may be inconsistent while consistency is vital
to the Halting Problem. No program can halt yes and halt no, so the
Halting Problem is unsolvable.

Set Theory is just the Theory of Computation structurally. The axiom
is that the set of programs that do not halt yes on themselves is not
recursively enumerable. Therefore the set of programs that do not
halt yes is not r.e.

The Russell Expression is the wff that expresses the set of programs
that do not halt yes on themselves. It is the axiom because it is
proven to not be a set by just diagonalization ? using a more
detailed, lower level system of proof. But The Set Theory axioms must
simply state that x ~e x is not a set, and the rules of inference do
the rest.

That, BTW, would be the same rules used to prove what is a set.
Marvin Minsky and Hartley Rogers were very wrong to say that the
Theory of Computation has no practical use. It overlaps Program
Synthesis, Program Analysis, Program Debugging etc.

> > 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.

What does suspicion have to do with mathematics? (They are

No, I am not asking ?Do you (or others) know how to define a wff??
After all, that is but a trivial task, wouldn?t you say? Actually,
the Frege/Russell problem is that, for whatever reason, Frege didn?t
formally define wff. So the real lesson (ironically) is that Frege
wasn?t formal enough.

The first answer offered by someone was to allow only relations in
wffs. You define a wff to allow most any symbols and allow any
substitution. Of course, at some point a system may have to
substitute, so we need a determination there as to what can be used.
If ?anything?, then you may have a problem just from saying that.
It?s certainly not formal.

So we have 2 alternatives to ZF (which may not be incompatible by
introducing a second kind of expression or by playing close attention
to whether an expression contains a non-relation or not.)

The rest is just details. Who wants to go first to complete the
consistent formalization of Naïve Set Theory ? a massively simpler
solution to the problem brought to light by Russell?s Paradox?