Re: Small Set theory:Revised.


hagman wrote:
zuhair schrieb:
From all of this dicussion, this Russell paradox is becoming clearer
to me. I think that this paradox is a deeper one that I originally
thought, now I am thinking that we can solve this paradox in one of
three aproaches.

1) Limit P. so we should have a kind of a type theory that allowes some
kinds of P and do not allow other kinds.

Maybe. Type theory is one approach against the Russel paradox.
But that's a completely different story than condensing ZFC into 4
axioms...


2) Apply a non binary logical concepts, I mean we lessen the
application of the law of excluded middle as regards this case.

Maybe. Again converting to become intuitionist is a valid but totally
different story...


3) "Existance" predicated. i.e. we treate "existance" as a predicate.

for example.

AxAy (((P[y]->yex)&(~P[y]->(~yex&~y=x))) <-> x Exist.)

What is the quantor Ax quantisizing about if not /existing/ sets x??
And if Ax runs over non-existing sets, does that mean that you want
((P[y]->yex)&(~P[y]->(~yex&~y=x)))
iff x exists whereas y may or may not exist?

If you start from ...
AxAy (((P[y]->yex)&(~P[y]->(~yex&~y=x))) <-> x Exist.)

Now if P[y]<->y=0

Ay (((y=0->0ex)&(~y=0->(~yex&~y=x))) <-> x Exist.)
or simply
Ay ((0ex&(~y=0->(~yex&~y=x))) <-> x Exist.)

if x = 0
... and specialize for x=0 ...

Ay (((y=0->0e0)&(~y=0->(~ye0&~y=0))) <-> x Exist.)
rather
Ay ((0e0&(~y=0->(~ye0&~y=0))) <-> 0 Exist.)

Ay(false & true <-> x Exist)

Ay(false<-> x Exist)

Ay(true<->x do not Exist).


... and obtain this ^^ (which is equivalent to "x does not exist")

No. ^^ is not equivalent to x does not exist.

It seem that you are not differentiating between ~x and x do not exist.

what is equivalent to x do not exist here is if you say that x=y were

P[y]<->y=0.
you have shown that the empty set does not exist.

No. I didn't

Let P[y]<->y is nothing.

from AxAy (((P[y]->yex)&(~P[y]->(~yex&~y=x))) <-> x Exist.)

AxAy(((y is nothing ->yex)& ( y is a thing -> ( ~yex & ~y=x))) <-> x
Exist.

if x=/=y.

we have x= {}.

so your reply is erronous.
Congratulation.

For the first time you made bad replies, while all your replies in this
thread were in reality fabulous and decisive, these later replies are
very weak.

Zuhair

.

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