I have no clue what you mean by this. Grammatically you are saying that " " is the same as " " but surely you don't mean that. What is true, is that " " is never true.

We can't redefinexϵy →yϵ’x Axiom, 1ϵ{1,2} →{1,2}ϵ’1 (‘ means not)

xϵy & yϵz →’xϵz Axiom, 1ϵ{1} & {1}ϵ{{1},2} but 1ϵ'{{1},2}

IDENTITY: x=x

EQUALITY: x=y, “x” and “y” represent the same thing.

Examples:

Active =A = {on, green,pw,hum,hot}

Def: x=A iff xϵA

xϵA & yϵA → x=y

A={1/3,2/6,3/9,….}

Def: x=A iff xϵA

xϵA & yϵA → x=y

Equality of Sets: A=B: A and B represent the same thing, by Def above.

Standard Definition: every member of A is a member of B and ----.

BUT “is” is not defined, is it identity or equality?everyEnglish word for mathematics. How are you distinguishing between "identity" and "equality". Mathematically, two objects are identical if and only if they are equal. Where have you ever seen a different definition for either "identical" or "equal"?

(We do, in some cases, give different definitions for "equivalent" in different fields of mathematics. The general category of "equivalence relations" includes "equal" (or "identity") but also other things.)

Yes, of course.Suppose A ={x,y,z}. if x=w, is w a member of A?

By the 'above Def.", yes, but "blue= 100 lb" is a very strange definition! Assuming that "blue" and "100 lb", etc. are simplyA={blue, green,yellow}

blue =100lb, green =200lb, yellow =300lb

B={100lb, 200lb, 300lb}

Does A=B. By above Def, yes. By Standard Def, depends on what you mean by “is.”labelsfor the same thing and have nothing to do with our usual definitions of those words, then, yes, A= B. Now, what do you mean by the "standard definition"? The "standard definition" ofwhat? What you giveisthe "standard definition" of "equals" (and "are indentical) for sets. If you mean "the standard definition(s)" of "blue", "200 lb", etc. then , but then "blue= 200 lb" is no longer true.

From Suppes AXIOMATIC SET THEORY

Primitive Atomic Formulas (paf): (vϵw), (v=w)

Primitive Formulas (pf), recursive definitions

a) every paf is a pf.

b) if P is a pf, then –P is a pf

c) (P&Q), (PVQ), (P→Q), and (P<̶̶ ̶̶>Q) are pf.

d) (for all v)P, (v exists)P are pf.

Any primitive formula of the form---------------is an axiom. (too difficult to type out, upside down A’s and backward E’s, but irrelevant to the point).

THE POINT: The trouble with the whole scheme is that you are defining truth functions in terms of variables (vϵw), (v=w) which are not defined as truth variables, and that leads to the trouble with

Russells Paradox, derived from the above axiom scheme by using (xϵx) to get:

(xϵx) iff (xϵ’x).

But unless you show (xϵx) is a truth variable with value T (you can’t), Russels paradox is simply a meaningless symbolic axiom and not a Paradox.[/QUOTE]