A hidden assumption

[PixyMisa]

Your misunderstanding is profound because you are trained to ignore consciousness\object relations, which is the basis of any reasoning.

My misunderstanding? You can't even maintain consistency within a single post, much less from one post to the next, and you speak of my misunderstanding?

Oh, and: Meaningless.

 

[doronshadmi]

Bzzt! Try again:

(My emphasis.)

So, what's it to be, 2≠2 or 2∉{2}?

Or was your earlier statement just another bit of nonsense that you immediately abandon when called on it?

There is X.

By using only X, please prove that it is a member of {X}.

Be aware that without "{" and "}", the term "a member of ..." does not hold.

 

[doronshadmi]

My misunderstanding? You can't even maintain consistency within a single post, much less from one post to the next, and you speak of my misunderstanding?

Oh, and: Meaningless.

How do you know that (after all no consciousness is involved by your school of reasoning)?

The amazing thing is that you are unable to understand the question.

Here is the answer: you are using your consciousness.

 

[PixyMisa]

There is X.

Okay.

By using only X, please prove that that it is a member of {X}.

That doesn't mean anything.

Be aware that without "{" and "}", the term "a member of ..." does not hold.

Sure it does. { and } are merely notation. I can denote a set any way I like, and in any case, the definition of set membership is independent of notation.

 

[PixyMisa]

How do you know that (after all no consciousness is involved by your school of reasoning)?

How do I know what? That your posts are inconsistent? Simple mechanical application of the rules of propositional logic can show that. Though your parenthetical statement is baseless and false.

The amazing thing is that you are unable to understand the question.

What question?

Here is the answer: you are using your consciousness.

I am using my consciousness, but you have failed to show that this is either necessary or relevant.

And your mathematics is still garbage.

 

[Apathia]

I don't think Goedel's theorem applies to Doron's work, because it addresses sufficiently powerful consistent formal systems. Doron's work is incomplete for an entirely different reason

You're right. I was way to generous.

 

[doronshadmi]

Okay.


That doesn't mean anything.


Sure it does. { and } are merely notation. I can denote a set any way I like, and in any case, the definition of set membership is independent of notation.

"{" and "}" is some representation of the notion of membership. You can use any representation that you like, but it does not change the fact that also the definition needs notations in order to be discussed.

At the moment that you understand this necessity (which holds both in "{"and "}" and the definition of membership), you can leave it behind and look only on the notion that it represents.

There is nothing in 2 that leads to the conclusion of {2}.

 

[doronshadmi]

How do I know what? That your posts are inconsistent? Simple mechanical application of the rules of propositional logic can show that. Though your parenthetical statement is baseless and false.

The simple mechanical application of the rules of propositional logic is nothing but some particular reflection of your consciousness.

These reflections are the result of the universal principle of the researchable, which is: Symmetry\Asymmetry complementation. Without this complementation nothing (abstract or not) is researchable.

Since you are not aware of the universal principle of the researchable (because of your trained mind that ignores itself as a factor of the mathematical research) you get your own reflection as something that is independent of you.

I am using my consciousness, but you have failed to show that this is either necessary or relevant.

This is simply unbelievable.

I have failed? I have failed?

My frailer is not relevant to your frailer to get the simple fact that you are using your consciousness also during your mathematical work.

Please show even a single thing of your reasoning which is independent of your consciousness.

 

[Don At Work]

does this help at all?
Standard definitions and symbols:
http://www.purplemath.com/modules/venndiag2.htm

A lot of this math is beyond my training. There are standard symbols used in math (and set theory), and they do not depend on your native language. If you could use those symbols, perhaps some of these math professors could then see clearly what you are trying to say.

 

[nathan]

Please show even a single thing of your reasoning which is independent of your consciousness.

2∈{2}

 

[doronshadmi]

does this help at all?
Standard definitions and symbols:
http://www.purplemath.com/modules/venndiag2.htm

A lot of this math is beyond my training. There are standard symbols used in math (and set theory), and they do not depend on your native language. If you could use those symbols, perhaps some of these math professors could then see clearly what you are trying to say.

Dear Don At Work,

First, thank you.

What I say is very simple and it does not depend on any particular representation of it.

In order to get it, please ask yourself a simple question:

How two opposites (abstract or not) can interact with each other without eliminating each other?

Please be aware that according to Classical Logic X AND X's negation is a contradiction, or in other words if X is definable, then X's negation cannot be found (and vice versa), otherwise we get a contradiction (no one of them can be found).

Since this is not the case, then Classical Logic is not sufficient enough to provide the answer.

You may claim that X OR X's negation enables to provide the answer.

In that case I agree with you, but first we have to understand what is OR.


Let X be True = T

Let X's negation be False = F


The truth table of OR is:

FF --> F
FT --> T
TF --> T
TT --> T


It can be reduced (without a loss of generality) to TT,TF (or TF).

What we get is invariant state (represented as TT) and variant state (represented as TF or FT).

So the principle that enables T and F to share the same universe without eliminating each other is based on invariant\variant complementation.

You may claim: but OR, AND, NOR, NAND, etc. … are all connectives of Classical Logic, so you have used Classical Logic in order to provide the answer.

My answer: Classical Logic cannot work if X and X's negation contradicting each other.

Any connective of Classical Logic can be used because of the invariant\variant complementation, and not vice versa.

Try to avoid the invariant\variant complementation and Classical Logic does not hold.

 

[doronshadmi]

2∈{2}

A reflection of your consciousness nathan.

You read my post and replied.

Even if a computer is used to do it, it is nothing but your agent's consciousness.

 

[Bodhi Dharma Zen]

You cannot get that without using your consciousness, and a consciousness gets the researchable if and only if both the memory without thoughts (the symmetrical side of the consciousness) and a thought without memory (the asymmetrical side of the consciousness) complement each other into a one_of many_faces realm, which is open by nature (it has an evolution).

I am not talking about any subjective experience here but about Symmetry\Asymmetry complementation, which is the universal principle of the researchable, whether it is abstract or not.

Without this fundamental understanding, you will not be able to understand that any researchable result (abstract or not) is actually the reflection of Symmetry\Asymmetry complementation, which stands in the basis of the researcher and the researched thing as a one organic realm.

Lets see... "memory without thoughts" How can you reach it? "thoughts without memory" Again, how? Who invented such labels? Why symmetrical or asymmetrical? How come consciousness is related to the "researchable"?

What is this implementation of "symmetry/asymmetry"? What do you mean as "one organic realm"?

Thanks.

 

[nathan]

A reflection of your consciousness nathan.

You asked for a statement that was independent of Pixy's consciousness. I provided one. Also, 2∈{2} continues to be true even when I'm unconscious.

If however you are taking *any* response as showing consciousness is necessary for true statements to be true, then you're setting up an unfalsifiable position.

 

[nathan]

How two opposites (abstract or not) can interact with each other without eliminating each other?

It depends what you mean by 'opposite', 'interact', and 'eliminate.

eg)
*) me and mrs me are opposites (man/woman), and we interact very nicely thank you
*) wet tea and dry carpet (wet/dry opposites) interact in such a way as the tea remains wet, but the carpet does not remain dry
*) proton and anti-proton are going to interact very forcibly in LHC. In some cases they'll survive and in some cases they wont.
*) +N and -N can interact in a number of ways, one way they sum to zero, another way they multiply to -N^2, another way they divide to -1, another way they subtract to + or - 2N ...

You need to be more precise with your language.

 

[jsfisher]

I need to archive these:

Doronshadmi says:

1. A set is equal to the union of its members.
2. 2 is not part of { 2,3,4 } much like a severed finger is not part of your body.
3. 2 may not be a member of { 2 }.

 

[jsfisher]

What I say is very simple and it does not depend on any particular representation of it.

Yes, what you say is usually independent of its representation (i.e. your use of language). It is gibberish.

In order to get it, please ask yourself a simple question:

How two opposites (abstract or not) can interact with each other without eliminating each other?

Some marriages are like this; apparently, not nathan's. As a mathematical principal, though, this isn't true. Of course, you'd need to explain what you actually mean by 'opposites' and 'interact'. Even though there is no hope of that ever happening, you'd still be incorrect in your statement, but at least then it might possibly be clearer to you as to why.

Please be aware that according to Classical Logic X AND X's negation is a contradiction, or in other words if X is definable, then X's negation cannot be found (and vice versa), otherwise we get a contradiction (no one of them can be found).

Since this is not the case, then Classical Logic is not sufficient enough to provide the answer.

Complete and utter nonsense. Define X as "Cows are purple". Its negation is "Cows are not purple". Look, I found X's negation.

...
The truth table of OR is:

FF --> F
FT --> T
TF --> T
TT --> T

Ok, so far so good.

It can be reduced (without a loss of generality) to TT,TF (or TF).

No. The truth table for the OR operator cannot be reduced. Were you to explicitly state the OR is a commutative operator, then you could get by with only three lines, the fourth being implied, but that isn't the same as reducing it. Moreover, false ∨ false cannot just be ignored.

...
My answer: Classical Logic cannot work if X and X's negation contradicting each other.

Good thing they don't then.

 

[PixyMisa]

How two opposites (abstract or not) can interact with each other without eliminating each other?

They don't. You have fundamentally misunderstood the nature of formal systems.

Please be aware that according to Classical Logic X AND X's negation is a contradiction, or in other words if X is definable, then X's negation cannot be found (and vice versa), otherwise we get a contradiction (no one of them can be found).

X cannot be true and false simultaneously. That doesn't mean that X can't be true when Y is false, or that X can't be true at one time and false at a different time, or under different conditions.

Since this is not the case, then Classical Logic is not sufficient enough to provide the answer.

Baloney.

You may claim that X OR X's negation enables to provide the answer.

That doesn't mean anything.

In that case I agree with you, but first we have to understand what is OR.

No, it's only you that has a problem here.

Let X be True = T

Let X's negation be False = F

Whatever.

The truth table of OR is:

FF --> F
FT --> T
TF --> T
TT --> T

Yes.

It can be reduced (without a loss of generality) to TT,TF (or TF).

NO! This is complete nonsense. It's not merely a question of loss of generality, it's a question of loss of any meaning whatsoever.

What we get is invariant state (represented as TT) and variant state (represented as TF or FT).

No.

So the principle that enables T and F to share the same universe without eliminating each other is based on invariant\variant complementation.

No, this is idiocy. YOU HAVE TWO VARIABLES HERE.

You may claim: but OR, AND, NOR, NAND, etc. … are all connectives of Classical Logic, so you have used Classical Logic in order to provide the answer.

You haven't used any sort of logic.

My answer: Classical Logic cannot work if X and X's negation contradicting each other.

Wrong. Propositional logic says that X is X, and that X isn't NOT X. It's not a question of logic not working, it's that logic tells you that you have generated a contradiction. A contradictory result is a valid and useful one, because it tells you that you have something wrong (e.g. proof by contradiction).

Any connective of Classical Logic can be used because of the invariant\variant complementation, and not vice versa.

Meaningless.

Try to avoid the invariant\variant complementation and Classical Logic does not hold.

Meaningless.

 

[ArmillarySphere]

Correct me if I'm wrong, but isn't a set a member of its own complement?

So if a set is the union of its members, A's complement contains all the members of A. Which contradicts the definition of complement.

Therefore, a set is not the union of its members.
Q.E.D.

Edited to add:
I just realised that "the set of all sets" actually does contain itself, so its complement doesn't.
Guess that means my proof only holds true for all the sets which don't contain themselves.

The proof for the other case is left as an exercise for the student...

 

[doronshadmi]

Correct me if I'm wrong, but isn't a set a member of its own complement?

So if a set is the union of its members, A's complement contains all the members of A. Which contradicts the definition of complement.

Therefore, a set is not the union of its members.
Q.E.D.

No, a set is a particular case of Symmetry\Asymmetry complementation, which is based only on distinct objects.

Symmetry itself cannot be a member because it is domain less (given any domain, Symmetry is the same both in(belongs) AND out (does not belong) of it.)

This sameness is not researchable exactly as the question "what an atom is made of?" is meaningless.

The same holds also in the case of total isolation, where some object cannot be examined according to some domain (because by total isolation there cannot be any connection between the object and the domain).

In other words, a set (which is a researchable thing) cannot be but the result of Symmetry\Asymmetry complementation.

Edited to add:
I just realised that "the set of all sets" actually does contain itself, so its complement doesn't.
Guess that means my proof only holds true for all the sets which don't contain themselves.

1. Symmetry itself (total unity) or Asymmetry (total isolation) itself are not members of any domain, or in other words, Membership does not hold at the atomic state, and nothing is researchable at the atomic state.

2. A set is the result of Symmetry\Asymmetry complementation, and no complementation has the completeness of the atomic state.

3. By using this fundamental notion as an axiom, it is clearly understood that the set of all sets does not hold, because the term all is meaningless (no collection has a cardinality which is equivalent to the completeness of the atomic state (which is not researchable, but can be used as the best point of view of the researchable)).

4. By using the atomic notion, a set is an incomplete object and Cantor's transfinite Cardinality does not hold.

5. The atomic notion is inevitable if we understand the researchable and its negation.