Deeper than primes

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doronshadmi said:
See what? that a mathematician thinks that he/she avoids circular reasoning just because he/she ignores it?
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If I recall correctly, the last time you hyper-ventilated about the empty set, it wasn't over this latest lunacy, circular reasoning. It was because of the word, 'the', instead of 'an', as if the name used to refer to an axiom somehow negated it.
 
jsfisher said:
By the way:

Emptiness is that has no predecessor. [sic]​

In other words:

Emptiness is a thing that has no thing less than it.​

Must not emptiness be one of the things that isn't less than emptiness? Is this circular reasoning? OMG!!!!
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Emptiness is no-thing, and no-thing does not have predecessor.

You are still get things at the level of the empty set, which is an existing empty thing, which its predecessor is Emptiness.
 
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HatRack said:
Here is the argument in an even more basic form, with (1) the definition, (2) the axiom, and (3) the conclusion:

(1) {} is the set such that if the set X exists then X is not its member.
(2) {} exists.
(3) Therefore, {} is not a member of {}.
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According (1) X is any set, including {}, and {} exists also by (1), otherwise set X can't be "not its member". So right at the level of the definition {} is used to define its own property by not being one of the members, which is a circular reasoning because {} is defined by using {} as a part of its own definition.

Since this is the case then (2) and (3) do not hold.
 
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jsfisher please reply to:

HatRack said:
doronshadmi said:
The axiom of the empty set uses the definition of being a set with no members.
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Correct.
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HatRack said:
"X exists" certainly does provide the "terms of X's existence" because I defined X, the empty set, before I asserted its existence.
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doronshadmi said:
HatRack said:
Here is the argument in an even more basic form, with (1) the definition, (2) the axiom, and (3) the conclusion:

(1) {} is the set such that if the set X exists then X is not its member.
(2) {} exists.
(3) Therefore, {} is not a member of {}.
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According (1) X is any set, including {}, and {} exists also by (1), otherwise set X can't be "not its member". So right at the level of the definition {} is used to define its own property by not being one of the members, which is a circular reasoning because {} is defined by using {} as a part of its own definition.

Since this is the case then (2) and (3) do not hold.
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doronshadmi said:
What strawman?

The axiom of the empty set is based on the definition of the empty set.
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No, it isn't.

Perhaps you should actually consider the real axiom:
[latex]$$$\exists x\, \forall y\, \lnot (y \in x)$$$[/latex]​
The first part of the axiom stipulates the existence of something, identified only by the letter x, and the remainder of the formula establishes a property of x.

I show that this definition (which is step 1) is based on circular reasoning.
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No, you didn't.

As zooterkin has already observed, you haven't a clue what circular reasoning actually is, but if you did you might also realize it is not possible within first-order predicate calculus. However, since you haven't, I'm sure you won't.
 
jsfisher said:
As zooterkin has already observed, you haven't a clue what circular reasoning actually is, but if you did you might also realize it is not possible within first-order predicate calculus. However, since you haven't, I'm sure you won't.
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Cut the nonesense.

http://en.wikipedia.org/wiki/Circular_reasoning
Circular reasoning is a formal logical fallacy in which the proposition to be proved is assumed implicitly or explicitly in one of the premises.
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jsfisher said:
The first part of the axiom stipulates the existence of something, identified only by the letter x, and the remainder of the formula establishes a property of x.
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It establishes a property of x by using x as one of the elements that establish a property of x, by not being a member of x, which is a circular reasoning.
 
doronshadmi said:
It establishes a property of x by using x as one of the elements that establish a property of x, by not being a member of x, which is a circular reasoning.
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More confirmation you haven't a clue what circular reasoning means. Be that as it may, let's stick to the actual axiom, please:

[latex]$$$\exists x\, \forall y\, \lnot (y \in x)$$$[/latex]​

What premise are you claiming is implicitly or explicitly used in the proposition? By the way, few axioms have premises. If they did, we would probably call them theorems instead of axioms.


Oh, and how is that definition of "at the domain of" coming along?
 
doronshadmi said:
Cut the nonesense.
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By all means, please, you first.

doronshadmi said:
http://en.wikipedia.org/wiki/Circular_reasoning

Circular reasoning is a formal logical fallacy in which the proposition to be proved is assumed implicitly or explicitly in one of the premises.
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That is exactly why people have been telling you “you haven't a clue what circular reasoning actually is”. That the empty set is empty is not a “proposition to be proved” it is explicitly what defines the empty set as being, well, empty.

doronshadmi said:
It establishes a property of x by using x as one of the elements that establish a property of x, by not being a member of x, which is a circular reasoning.
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So tell us Doron does your “empty set” have itself as a member? If so then how is it empty? If not then you are just using “a circular reasoning” that you claim is invalid. You have simply painted yourself into a corner which is evidently why you think you need the contradiction of your “non-locality” to extricate you. Too bad though, as that very contradiction which you claim “non-locally” is not a contradiction makes your “non-locality” “at AND not at the domain of” your locality. So at “the domain of" your locality it still remains a contradiction while “not at the domain of” your locality it simply has no relation to the corner you have painted yourself into and thus can’t extricate you. After all this time you still fail to relize that your assertions and notions just fail when applied to themselves.

Doron if one “establishes a property of x by” not “using x as one of the elements that establish a property of x” then one has simply not established a property of x. x is always an element of the properties of x, specifically being the element that has said properties of x.


This exchange over the past couple of days seems centered around your usual misinterpretations. As noted above That the empty set is empty is not a “proposition to be proved” as you seem to want to consider it.

Also in this assertion before…


doronshadmi said:
It asserts about sets that are not its members. If one of the sets that are not the members of the empty set is the empty set, then a circular reasoning is used, because we cannot determine something about the empty set by using the empty set.
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…you seem to think we need to “determine something about the empty set” before one defines the empty set.

Please tell use Doron how one can “determine something about the empty set by” not “using the empty set” or without at least having a definition of the empty set?
 
doronshadmi said:
According (1) X is any set, including {}, and {} exists also by (1), otherwise set X can't be "not its member".
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Incorrect. I did not declare {} to be an existent set in step 1, that happened in step 2. At the point of step 1, we do not know any set to be an existent set. Therefore, you cannot conclude that {} is not a member of {} at this point. That cannot be done until after step 2, when the existence of {} is asserted.

Your incessant claims of this argument being incorrect boils down to your unrelenting misunderstanding of definitions and existence. It's definitely possible to state this axiom without using a definition at all, as I have done in previous posts informally and jsfisher has done formally. And then, the definition can be given after. Although I personally find it to be in better style to give definitions first, I apparently can't do that because you are having a really hard time grasping the difference between a definition and a statement of existence.

The Axiom of the Empty Set
There exists a set X such that if Y is a set then Y is not a member of X.

Let X in the previous step be known as the empty set, denoted by {}.

For the last time: An axiom is a statement that you accept as true without question, there is no need to prove them. Given that you can only use circular reasoning in a proof, there can be no circular reasoning here.
 
Historically, when Doron finds himself painted in a corner, he harps on irrelevant shades of meaning in colloquial expression, exploits various forms of equivocation, revels in word shifting and new-found gibberish, moves the goal posts, and attempts to change the subject. All while avoiding any formalisms that would expose his thirty-year nonsense for the nonsense it is.

We seem to be again in that phase now.
 
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HatRack said:
Incorrect. I did not declare {} to be an existent set in step 1,
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You don't have to declare anything in order to determine the existence of {}.

The existence of {} is determined right at the definition, even if one not explicitly declares it. All we need is elementary reasoning as follows:

Definition 1: {} is the set such that if the set X exists then X is not its member.

According to Definition 1, X must be any existing set, including {}, otherwise set X can't be "not a member" of {}.

So right at the level of definition 1 {} is used to define its own property by not being one of the members, which is a circular reasoning, because {} is defined by using {} as a part of its own definition.

So, axiom 2 is already based on an existing object that is defined by circular reasoning, and therefore axiom 2 and conclusion 3 do not hold.

HatRack said:
It's definitely possible to state this axiom without using a definition at all, as I have done in previous posts informally and jsfisher has done formally.
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HatRack said:
The Axiom of the Empty Set
There exists a set X such that if Y is a set then Y is not a member of X.

Let X in the previous step be known as the empty set, denoted by {}.
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HatRack, if Y is a set, then Y is also X because X is also a set.

In that case X establishes a property of X by using X as one of the elements that establish a property of X, by not being a member of X, which is a circular reasoning.

Furthermore, you did not avoid the definition of {}, because this definition is an inseparable part of the axiom, as follows:

By using your of words The Axiom of the Empty Set is:

"There exists a set X such that if Y is a set then Y is not a member of X" where the blue part is actually Definition 1.

There is no previews step here, because the axiom is not less than all of its parts at once, exactly as expressed by:

[latex]$$$\exists x\, \forall y\, \lnot (y \in x)$$$[/latex]​
 
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doronshadmi said:
You don't have to declare anything in order to determine the existence of {}.

The existence of {} is determined right at the definition, even if one not explicitly declares it.
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Wow. Just, wow. So, if I define IPU as horse that is invisible, pink and has exactly one horn on its head then I also implicitly assert that it exists? In what kind of twisted universe makes that sentence any sense at all?
 
doronshadmi said:
You don't have to declare anything in order to determine the existence of {}.

The existence of {} is determined right at the definition, even if one not explicitly declares it. All we need is elementary reasoning as follows:
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No. Definitions do not determine existence. A definition in mathematics is a word substitution in informal language, and a symbolic substitution in formal language, nothing more. The actual axiom can be stated without making use of definitions at all:

The Axiom of the Empty Set: There exists a set X such that if Y is a set then Y is not a member of X.
 
HatRack said:
No. Definitions do not determine existence. A definition in mathematics is a word substitution in informal language, and a symbolic substitution in formal language, nothing more. The actual axiom can be stated without making use of definitions at all:

The Axiom of the Empty Set: There exists a set X such that if Y is a set then Y is not a member of X.
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You simply ignore http://www.internationalskeptics.com/forums/showpost.php?p=6602376&postcount=12742 (it was updated in order to be clearer).
 
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The Man said:
So tell us Doron does your “empty set” have itself as a member?
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If {} is used to determine the empty state of {} (by {} as not being a member of {}), then a circular reasoning is used, because one can't define X by using X as a part of the definition of X.
 
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doronshadmi said:
If {} is used to determine the empty state of {} (by {} as not being a member of {}), then a circular reasoning is used, because one can't define X by using X as a part of the definition of X.
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Answer the question, Doron. Does your empty set have itself as a member? There is no 'if'.
 
zooterkin said:
Answer the question, Doron. Does your empty set have itself as a member? There is no 'if'.
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The answer no, but it used as one of the elements that determine the property of being empty existing thing, by not being a member of the empty set.

Any use (even negative use) of X as a part of the definition of X, is a circular reasoning.
 
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doronshadmi said:
You don't have to declare anything in order to determine the existence of {}.
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Yes you do. Sets don't just exist in a logical framework unless you explicitly assert that they do.

doronshadmi said:
The existence of {} is determined right at the definition, even if one not explicitly declares it.
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Definitions do not assert anything, they are word substitutions.

doronshadmi said:
Definition 1: {} is the set such that if the set X exists then X is not its member.

According to Definition 1 X must be any existing set, including {}, otherwise set X can't be "not a member" of {}.
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We don't know that the empty set exists at this point, so this is of no significance.

doronshadmi said:
So right at the level of definition 1 {} is used to define its own property by not being one of the members, which is a circular reasoning, because {} is defined by using {} as a part of its own definition.

So, axiom 2 is already based on an existing object that is defined by circular reasoning, and therefore axiom 2 and conclusion 3 do not hold.
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Still of no significance because definition 1 does not assert the existence of the empty set.


doronshadmi said:
HatRack, if Y is a set, then Y is also X because X is also a set.

In that case X establishes a property of X by using X as one of the elements that establish a property of X, by not being a member of X, which is a circular reasoning.
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No. All we are doing here is declaring that X is not a member of itself. That is not circular reasoning. Every time we define almost any set in mathematics, we are declaring that it is not a member of itself. For example:

Let B = {1,2,3}. Here, we declared that 1, 2, and 3 are elements of B. But, we also implicitly declared that B is not an element of B. This is common practice, and it is completely unavoidable when working with any set that is not an element of itself to implicitly declare that it is not an element of itself. Thus, your entire argument against the existence of the empty set fails.

So, if you turn your own ass backwards logic onto your own "theory", it turns out that your theory of "Organic Math" can't exist after all if it leads to any sets that do not contain themselves. Because, according to you, it is a flaw to state that a set does not contain itself in its definition, which I just demonstrated is implicitly implied in just about every case imaginable.

doronshadmi said:
Furthermore, you did not avoid the definition of {}, because this definition is an inseparable part of the axiom, as follows:

By using your of words The Axiom of the Empty Set is:

"There exists a set X such that if Y is a set then Y is not a member of X" where the blue part is actually Definition 1.
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Logically, it doesn't matter whether I state the definition first or combine it into the axiom. I figured I'd do it both ways due to your unrelenting confusion as to the difference between a definition and a statement of existence.

doronshadmi said:
There is no previews step here, because the axiom is not less than all of its parts at once, exactly as expressed by:

[latex]$$$\exists x\, \forall y\, \lnot (y \in x)$$$[/latex]​
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Nonsensical gibberish that I don't feel like deciphering, and it doesn't matter anyway because I already explained why your argument is flawed.
 
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doronshadmi said:
The answer no, but it used as one of the elements that determine the property of being empty existing thing, by not being a member of the empty set.
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Here we go again with your obsession with enumerating. We do not have to list every possible set that isn't in the empty set. Since we do not have to do that, the fact that the empty set is one of those sets is not relevant. Even if we did, it still would not be a circular definition.
 
HatRack said:
Let B = {1,2,3}. Here, we declared that 1, 2, and 3 are elements of B. But, we also implicitly declared that B is not an element of B.
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It does not change the fact that you can't use B as a part of the definition that defines B.

HatRack said:
This is common practice, and it is completely unavoidable when working with any set that is not an element of itself to implicitly declare that it is not an element of itself. Thus, your entire argument against the existence of the empty set fails.
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It has nothing to do with elements that are members or not members of themselves, it is about not using the concept of X as a part of the definition that enables X.

Its is obvious that {} must be empty, but in order to define that {} is empty, {} must be one of the elements that are not members of {}, and in this case we are using {} ("{} is not a member of {}") in order to define {}, which is a circular reasoning.

B = {1,2,3}

C = {1,2,3,{1,2,3}}

E = {}

Furthermore, are you claiming that C is B as a member of itself?

In that case {{}} is E as a member of itself, which is false, because {} is empty and {{}} is non-empty.

Also C is not B, so your
HatRack said:
But, we also implicitly declared that B is not an element of B.
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is an utter nonsense.

HatRack said:
Nonsensical gibberish that I don't feel like deciphering, and it doesn't matter anyway because I already explained why your argument is flawed.
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As explicitly shown, you are wrong.
 
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doronshadmi said:
It does not change the fact that you can't use B as a part of the definition that defines B.


It has nothing to do with elements that are members or not members of themselves, it is about not using the concept of X as a part of the definition that enables X.

Its is obvious that {} must be empty, but in order to define that {} is empty, {} must be one of the elements that are not members of {}, and in this case we are using {} ("{} is not a member of {}") in order to define {}, which is a circular reasoning.

B = {1,2,3}

C = {1,2,3,{1,2,3}}

E = {}

Furthermore, are you claiming that C is B as a member of itself?

In that case {{}} is E as a member of itself, which is false, because {} is empty and {{}} is non-empty.
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You've really backed yourself into a corner this time Doron.

B = {1,2,3} is the set such that if X is any object but 1, 2, or 3 then X is not an element of B.

Compare this to the definition of the empty set:

A = {} is the set such that if X is any object then X is not an element of A.

Completely analogous, yet you claim that B is a valid set and A is not. But, simply let X = B in the definition of B and you have the exact same situation that you claim is "circular reasoning" in the definition of A.
 
HatRack said:
B = {1,2,3} is the set such that if X is any object but 1, 2, or 3 then X is not an element of B.
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In this case you explicitly excluded the elements that define B (where no one of them alone is B) from the elements that are not members of B.
HatRack said:
A = {} is the set such that if X is any object then X is not an element of A.
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In this case no element is excluded from the elements that are not members of {}, and one of them is {}, so "{} is not a member of {}" is still a part of the definition of {}, which is a circular reasoning.
 
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doronshadmi said:
In this case you explicitly excluded the elements that define B (where no one of them alone is B) from the elements that are not members of B.

In this case no element is excluded from the elements that are not members of {}, and one of them is {}, so ("{} is not a member of {}") is still a part of the definition of {}, which is a circular reasoning.
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Doesn't matter what's excluded and what isn't. Your sole claim of the empty set being invalid is that "{} is not a member of {}" is part of its definition. You simultaneously claim that {1,2,3} is valid, yet "{1,2,3} is not a member of {1,2,3}" is part of the definition of {1,2,3} as I have demonstrated.
 
HatRack said:
Doesn't matter what's excluded and what isn't. Your sole claim of the empty set being invalid is that "{} is not a member of {}" is part of its definition. You simultaneously claim that {1,2,3} is valid, yet "{1,2,3} is not a member of {1,2,3}" is part of the definition of {1,2,3} as I have demonstrated.
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If the definition of {1,2,3} is based on the use of {1,2,3} as "{1,2,3} is not a member of {1,2,3}" then also in this case a circular reasoning is used, because you can't define X by using X as one of the things that define X.

As you see there is a fundamental failure in the attempt to define X in terms of X, and this is exactly the reason that you do not understand the concept of Collection because you are closed under this concept.
 
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doronshadmi said:
If the definition of {1,2,3} is based on the use of {1,2,3} as "{1,2,3} is not a member of {1,2,3}" then also in this case a circular reasoning is used, because you can't define X by using X as one of the things that define X.
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Well then you are in a world of trouble Doron. The following definition of {1,2,3}:

Let {1,2,3} be the set such that if X is 1, 2, or 3 then X is an element of {1,2,3}.

Is logically equivalent to this definition:

Let {1,2,3} be the set such that if X is not 1, 2, and 3 then X is not an element of {1,2,3}.

ETA: The first definition reads P if and only if Q and the second definition reads ~P if and only if ~Q, which is why they are logically equivalent.

So Doron, we have achieved a milestone in this thread. You are now officially claiming that every set which does not contain itself is invalid. Organic Mathematics is looking really useful now. Congratulations.

:bigclap
 
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