Re: Small Set Theory,Updated.


MoeBlee wrote:
zuhair wrote:
were P is a Predicate in one variable, that doesn't use x,
nor it is equivalent to any Predicate Q in one variable that use x.

It would be a quagmire just trying to sort that out with you.

MoeBlee

You can understand this if you contemplate that x is a set, it is not a
variable.
x is a set that is defined intentionally by the predicate P.

what I am saying is that if you want to defined a set x by using the
predicate P. then this predicate shouldn't use x nor it should be
bicondioned with a predicate that uses x, because this would be a
circular (cyclical) definition.

As an example to see what I mean: take for example the empty set { } ,
i.e the set which has no member in it. now using the formula x is
P_defined were P is any Predicate that does't use x.we can define { }
by two predicates .

i) P[y]<-> ~y=y.
ii) P[y]<->(Az(~zey)).

so we have { } is (~y=y)_defined and we have { } is (Az(~zey))_defined.

You see neither ~y=y , nor Az(~zey) uses { }. ( more acurratly speaking
the symbole { } ).

and we will reach to the problem of non uniquenss outlined by hagman.

BUT!!! if we add another condition on P, that is P cannot be equivalent
(biconditioned)

to a Predicate that uses the symbole x.

Then here you see that the Predicate z(~zey) though it doesn't use the
symbole x in it, yet

it is equivalent ( i.e biconditioned) to another Predcate Q that uses
the symbole x in it, and that
Predicate is Q[y]<-> y= { }.

see that y={ } <-> Az(~zey)

and since P[y]<->Az(~zey) and Q[y]<->y={ } , then it follows that
P[y]<->Q[y].

so Az(~zey) is equivalent to a Predicate Q that uses the symbole { }.
and therefore neither

P nor Q can define { } , i.e "{} is Az(~zey)_defined" IS not
acceptable Nor

" {} is (y={})_defined"is acceptable.

I hope now you got what I mean. It is fairly simple and clear. It
mounts to a very simple
fairly intuitive premiss, Circular definitions are not acceptable.

Zuhair

.

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