Witt

Witt :
How can you deny the existence of that class which contains all existent things?

John:
Russell's Antinomy is what occurs when you insist that a representational system is reality:

No, it is not.
It is false to say: if we admit the existence of the universal set then the the system is inconsistent.

Russell's antinomy occurs if, y e {x:Fx} <-> Fy is a theorem, which is incorrect.

1. EyAx((x e y) <-> Fx), is invalid.
2. (y e {x:Fx}) <-> Fy, is invalid.
3. Ax(Fx <-> Gx) -> {x:Fx}={x:Gx}, is invalid.


quote:
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Russell's paradox arises as a result of naive set theory's so-called unrestricted comprehension (or abstraction) axiom. Originally introduced by Georg Cantor, the axiom states that any predicate expression, P(x), which contains x as a free variable, will determine a set whose members are exactly those objects which satisfy P(x). The axiom gives form to the intuition that any coherent condition may be used to determine a set (or class). Most attempts at resolving Russell's paradox have therefore concentrated on various ways of restricting or abandoning this axiom.
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John:
With the "set of all sets" constrained to being a mental concept, reality continues unchanged. The irony is that Cantor's axiom of abstraction is a concept for abstration but Russell tried to use it to categorize "things-in-themselves" when the analysis only applies to "things-as-they-appear-to-us".

Your apparent resolution of the Russell's antinomy will not do.

The answer to Russell's question: Is the class of those classes that are not members of themselve a member of itself or not...is No it cannot be a member of itself.

The solution to Russell's Paradox is: the Russell class does not exist. Nor does the Barber who shaves all and only those who do not shave themselves, exist.

Witt

Witt

Hi Witt,


Witt :
John: IMO this is where all truths stem from. Of course, you are free to invent any system of divining the truth that you wish.

"divining" what is that?

Telling, deciding, determining.

Witt :
I don't agree that A=A is required to prove any theorem of: propositional logic, boolean logic, syllogistic logic, etc..

John: How do you prove, for example P or P = P without the Law of Identity?

(p v p) = p, is not provable within first order propositional logic.

(p v p) <-> p, is provalble and does not require any theorm from identity theory to prove it.

Witt :
Axiom 1. (All x)(x=x), is the law of identity.

Surely A=A means, the object named by A is exactly identical with the object named by A.
There is no 'similar enough' to be identical here.
They are identical because there is no difference.
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John: Agreed, an object can have two names - but it is not the names we are comparing.

Witt:
I don't agree.
Objects have unique names.
An object with two names is not unique.
Although, it is true that objects may have many different 'descriptions'.

John:
Sorry, I don't understand. Naming an object doesn't change the object (it merely asserts that the definition associated with the name can be applied to the object). Also, regarding the predicate logic example of the LOI you give (For all x)(x=x), this implies that there may be more than one x - in which case how can they be identical.

??

(all x)(x=x) means a=a & b=b & c=c & .. for all values of the variable x.

Hopefully there is more than one value of x.

John: Yes, this is a tautology or truism, a self-defining statement.

Witt: What does self-defining mean?

John: Tautological.

OK.

Witt :
If we cannot show that a given proposition is true then we cannot know that it is true.
To know is to be able to show.

John:
I disagree on this one, I know I exist from my own subjective experience but I cannot show this to you.

Knowledge is not attained by subjective experience, at all.

Yes, you can show that you exist by demonstrating that you do have a particular property.
If it is true that you do have some property then you do exist. (descartes dictum )


Witt :
x=y, is defined, x has a property iff y has the same property..for all properties.

D1. x=y, defined, (All F)(Fx <-> Fy)

a=a, means, a has a property iff a has the same property..for all properties.
Which is tautologous. (All F)(Fa <-> Fa), is a theorem.
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John: Agreed, but what is F?

F is a predicate.
eg. Fx might be: x is wise, x=x, x>y, x exists, and so on.

Witt

John Page

But this is not my claim.

I admit the existence of the concept of the universal set. It is a way of representing/denoting everything that exists. However, it is not everything that exists.

Russell's Antinomy indicates that if (the set of all sets) is (a member of itself) a contradiction to the laws of predicate logic arises. To state that (the set of all sets) is (not a member of itself) is to undermine the concept of set membership. It is the observation of others that taking the second of these positions is to say that predicate logic is inconsistent. Others still are willing to treat the set of all sets case as an exception - but on what grounds?

My response is simply that something can be itself but cannot at the same time be a member of itself. These things exist at a different level of abstraction.

But what is the reason for this exceptional treatment? What does this tell us about the nature of truth? What are the real conditions required for set membership? (Isn't this just matching a template and recognizing that the template is an archetype and not the thing-in-itself?)
Sorry but I don't get the jump to line 3 - unless F and G are identical then (Fx <-> Gx) is bound to fail. However, I'm not disagreeing with you because this is pretty much my point!
The Russell class? Please tell me what this is.
If you assume that the barber is a woman the paradox dissolves.
The paradox arises because we mentally assume that the barber must be in one of two categories that have been set up to exclude the barber. Given the rules set up to disqualify from the barber from both definitions then, if the barber is ever shaved, one has to change the definitions or invent a barber (such as a female barber) that is not disqualified.

Cheers, John

John Page

but . Then x is just a name that can refer to one or more objects. A symbolically equivalent name says nothing about the identity of the object(s) under scrutiny. The very act of substitution (into the variable x) seems to violate the Law of Identity - all one is doing here is an abstract manipulating properties (or collections of properties) of things. I'm still having trouble with your statement above that an object with two names is not unique.
Then where does mine come from?
...and a and x are also predicates, consistent with (a <-> a), so (Fa <-> Fa) just adds more information about a (which has the property "a'ness").

I'm very interested that we see the LOI differently. Your expression "Axiom 1. (All x)(x=x), is the law of identity." seems to say nothing about uniqueness, but if things aren't unique (in spacetime) how can the mind conflate the universe?

Cheers, John <-> John

GrandDesigner

The only universal truth one can know is that they exist. Some people may even find that hard to prove. Whats infinitely harder to prove is your existence to someone else.

I know I exist. My faith is that others exist, too.

Grand Ol Designer

sophie

If the universal truth is that one exists, then would it not be a universal truth concerning all the things which regularily support that existence?

Witt

John:
Russell's Antinomy is what occurs when you insist that a representational system is reality:

Witt:
No, it is not.
It is false to say: if we admit the existence of the universal set then the the system is inconsistent.

John: But this is not my claim.

It's a common objection, see Zermelo.

John: I admit the existence of the concept of the universal set. It is a way of representing/denoting everything that exists. However, it is not everything that exists.

Agreed, properties and sets are different.

The class of those things that exist is the universal set.

John:
Russell's Antinomy indicates that if (the set of all sets) is (a member of itself) a contradiction to the laws of predicate logic arises.

What contradiction? V e V is theorem for Quine.

John: My response is simply that something can be itself but cannot at the same time be a member of itself. These things exist at a different level of abstraction.


Quine proves that (V=V) and (V e V) are theorems.
See: Quine, Mathematical Logic, page 163.


Witt : The answer to Russell's question: Is the class of those classes that are not members of themselve a member of itself or not...is No it cannot be a member of itself.

John: But what is the reason for this exceptional treatment?

The issue is existence. Some described: objets, sets, sets of sets, etc. do not exist. Those descriptions require exceptional logical treatment.

John: What does this tell us about the nature of truth? What are the real conditions required for set membership?

(y e {x:Fx}) <-> (Exists{x:Fx} & Fy), is a theorem for me.


Witt : Russell's antinomy occurs if, y e {x:Fx} <-> Fy is a theorem, which is incorrect.

1. EyAx((x e y) <-> Fx), is invalid.
2. (y e {x:Fx}) <-> Fy, is invalid.
3. Ax(Fx <-> Gx) -> {x:Fx}={x:Gx}, is invalid.
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John: Sorry but I don't get the jump to line 3 - unless F and G are identical then (Fx <-> Gx) is bound to fail. However, I'm not disagreeing with you because this is pretty much my point!

1. is the naive axiom of comprehension, Cantor.
2. is axiom 2.1 of Quine's Set Theory and its Logic.
3. is the axiom (V) of Frege's system.

Witt: The solution to Russell's Paradox is: the Russell class does not exist.

John: The Russell class? Please tell me what this is.

The class of those classes that are not members of themselves.

R={x:~(x e x)} Df.

{x:~(x e x)} has no members and is not a member of any class.
It is not self-identical.
There is no described object that is equal to it.

Witt:
Nor does the Barber who shaves all and only those who do not shave themselves, exist.

John: If you assume that the barber is a woman the paradox dissolves.

The puzzle talks about 'those' that shave, male or female.

quote:
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Consider all of the men in a small town as members of a set. Now imagine that a barber puts up a sign in his shop that reads I shave all those men, and only those men, who do not shave themselves.
Obviously, we can further divide the set of men in this town into two further sets, those who shave themselves, and those who are shaved by the barber. To which set does the barber himself belong?

The barber cannot shave himself, because he has said he shaves only those men who do not shave themselves. Further, he cannot not shave himself, because he shaves all men who do not shave themselves!
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The paradox arises because we mentally assume that the barber must be in one of two categories that have been set up to exclude the barber. Given the rules set up to disqualify from the barber from both definitions then, if the barber is ever shaved, one has to change the definitions or invent a barber (such as a female barber) that is not disqualified.

Agreed, the conditions that this presumed barber must satisfy cannot be satisfied. That is to say the described barber cannot exist.

Witt

John Page

Witt:

Thanks for your response - much for me to consider here. I've had a v busy day - just got home. Will catch up later but here's a brief response on some points.

Cheers, John
Are you saying All x = {x e x}?
Thanks - I need to look at this. (I've a sneaking suspiscion that Quine's V is a logical object only and cannot be regarded as a universal)
With you mostly on this, but what do you mean by existence in this context? Is this materially exist? If so one might argue that in this context nothing exists since the mind merely apprehends and analyzes sense data and ascribes the properties that are manipulated using set theory etc.

My contention is that logic systems must accomodate the abstraction that their subject matter necessarily has undergone. For example, 3 is not an absolute - it arises through the abstraction of the property 3'ness from a set of sense data being analyzed.
I need to think about this one.
I need to think about this - wasn't he saying there must be a function of x that is biconditional with x being a member of y where there is a "one to all" relationship between y and x? How do you consider the axiom invalid - I thought soemthing like this was necessary to explain how logical variables related to each other.
OK, but it is a class! As expressed though, it doesn't make sense to me I would say more {x:~(y e y)} because of the different levels of abstraction which brings us back to Frege's axiom V and AC (I think).
Agreed, I was refering to the Russell version quoted! Would it count if the barber was shaving in his spare time?