Re: An uncountable countable set
- From: Tony Orlow <tony@xxxxxxxxxxxxx>
- Date: Tue, 05 Sep 2006 12:29:51 -0400
MoeBlee wrote:
Tony Orlow wrote:MoeBlee wrote:Tony Orlow wrote:When you say "a=b", you say "A a A b P(a)=P(b)". The two are equivlentNo, it all rests on the notions of identity and equality. As LeibnizHa! The fallacy of reversing implication right there! An example of
pointed out, when the properties of two objects are all exactly the
same, then they are the same object. So, when we say two numbers are
equal, that means all properties of the two are equal.
just about the most basic fallacy.
statements, and therefore imply each other.
I explained to you a long time ago that in general, in first order
logic we cannot state the identity of indiscernibles, even as a schema,
let alone have it implied from something else in first order logic.
However, as exception to the generalization just mentioned, in a
language with only finitely many non-logical primitive symbols, we can
state the identity of indiscernibles as a schema. And in very general
terms not tied to any specific kind of system, we may say that the
indiscernibility of identicals implies the identity of indiscernibles
only in the sense that the identity of indiscernibles is a logical
priniciple (or at least taken by many people to be a logical
principle), thus a given.
But what was incorrect in your original statement was the word 'so' in
the sense that you were RELYING on one principle to infer the other. In
that sense, you committed the fallacy of inferring B -> A from A -> B.
Looking at what I wrote now (I've beeen tied up for a bit) it's not correct. I should have said a=b = A P P(a)=P(b). Since one cannot quantify over sets, or the properties that define them, in first order logic, this is not a first order statement, but second order. In any case, I agree with Leibniz that the unique identity of an object is defined by its unique set of properties, and that equivalence between two objects is the SAME as equivalence between the entire set of properties of each. So, it's not that a->b -> b->a, but that a=b <-> b=a. The object IS the set of properties which defines it, and the inability to discern two objects by their properties makes them equal, at least until some property is discovered which can discriminate between the two.
No, the indiscernibility of identicals does NOT imply the identity ofYou prove two quantities equal by showing there is no difference, do you
indiscernibles. You need both implications; you can't derive one from
the other. And, in first order logic, one direction can be posited only
in the semantics not in the axioms.
not?
It depends on the specific theory. In a first order theory with
infinitely many primitive predicate symbols, we have no theorem schema
for doing what you suggest. But set theory has only two primitive
predicates (one if you take equality as defined) so we can state such a
theorem schema. However, we don't need to do that since the axiom of
extensionality allows us to prove x=y merely by proving Az(zex <->
zey).
And what is z besides one of the set of properties which defines the sets x and y? The distinction between elements and properties is rather tenuous. Is it not a property of y that z e y?
Anyway, I don't disagree with the principle of the identity of
indiscernibles; my point is that we don't infer it (or if we do, then a
demonstration is required) from the principle of the indiscernibility
of identicals (except in the trivial sense that we can infer a logical
principle from anything at all).
No, but we can take as axiomatic that a=b = A P P(a)=P(b), and that P=Q = A a P(a)=Q(a). Thus properties are defined by the objects to which they pertain, and objects are defined by the properties which pertain to them. Despite the fact that this statement is not first-order, I see no problem arising from it.
MoeBlee
:)
TOny
.
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