Re: ZFC in 4 Axioms.


MoeBlee wrote:
zuhair wrote:

Definition:

Ax(x is P_embeded <-> (Ay: y e x <-> P(y)) /\ P(x))

What does 'P' range over? Formulas? If so, then I'm concerned whether
we can justify your definition even as a definitional schema. But let's
see what we find.

But first, please use parentheses rather than the ':' colon symbol. It
will be much clearer.

Now, I don't know whether you want:

x is P_embedded
stands for
Ay(yex <-> (P(y) & P(x)))

or

x is P_embedded
stands for
Ay(yex <-> P(y)) & P(x)

or

x is P_embeddd
stands for
Ay yex <-> (P(y) & P(x))

or

x is P_embdded
stands for
(Ay yex <-> P(y)) & P(x)

MoeBlee

let me present it in an informal manner and you choose.

x is embeded with respect to formula P means that every member of x
satisfy P and x itself satisfy P, and every z : z!ex /\ z!=x then
P(z)is false.

Let me try write it in symboles:

x is P_embeded <-> ( [ Ay(yex->P(y)) ] /\ [P(x)] /\ [ Az((z!ex /\ z!=x)
-> ~P(z)) ] )

I think this is the best way to put it in symboles you people here
odour and never work without them.

Zuhair

.

Relevant Pages

  • Re: ZFC in 4 Axioms.
    ... we can justify your definition even as a definitional schema. ... satisfy P and x itself satisfy P, and every z: ... I think this is the best way to put it in symboles you people here ... Ax.4: Infinity:As in ZFC. ...
    (sci.math)
  • Re: ZFC in 4 Axioms.
    ... zuhair wrote: ... we can justify your definition even as a definitional schema. ... will be much clearer. ...
    (sci.math)