The Man
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Like we haven't heard that before, though there is always just the slightest possibility that this time you aren’t just lying.
Deeper than primes
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Like we haven't heard that before, though there is always just the slightest possibility that this time you aren’t just lying.
Lying (or giving someone "asymmetric information") doesn't seem to be just Doron's way of smoothing out the wrinkles on the face of his arguments. When I clicked yesterday on the main news, I saw president Obama in the middle of some speech of his. So I listen and then, to my mild surprise, he said, “Be still, and know that I am God. That statement got my full attention, coz the atheists maintain that there is no God. But such an assertion is false, coz the prez finished his speech without having disappeared. To put it simply, the claim that there is no God is simply a lie.
I know that you, as a realistic person, have hard time to believe that the president would publicly declare himself God, but I swear it's true. Unfortunately, my thread with the video has been relocated to a special place called "Abandon All Hope" and then, probably on the request of the White House, entirely deleted. But there is a chance that you may find the speech somewhere on the web. I wouldn't give you asymmetric info, would I?
doronshadmi
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Happy Birthday The Man.
The Man
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Thank you very much Doron.
The Man
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While I may have some specific distain for some particular instance of lying I generally don’t have any particular distain for lying in and of itself. As knowingly making a false statement with the intent of deception, a lie, by that definition, confirms the liar has at least some accurate knowledge. They merely and deliberately choose not to express that accurate knowledge correctly. On the other hand an honest person who is simply wrong believes themselves to be correct even though the statement they make is in fact false. So while the liar has some true and perhaps useful knowledge the honestly wrong just doesn’t have a clue, even to their cluelessness. Doron has combined the two (liar and simply wrong) into his own symmetrical superposition of their identities without symmetry, superposition or identity.
doronshadmi
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President Obama used a statement taken from Psalms (46:11), which says:
"הרפו ודעו, כי אנכי אלהים" which means "Let be, and know than I am God".
"הרפו" means "Release", "Relax", "Let be", "Be still", "Be calm" etc.
In other words, one can't know God as long as his\her mind is not aware of the calm and non-subjective state, which is the natural source of any possible expression, whether it is mental or physical.
epix, your
statement is an actual example of the lack of "הרפו" of your awareness.epix said:
(Your noisy mind can't get http://www.internationalskeptics.com/forums/showpost.php?p=7255966&postcount=15594 as can be seen by you reply in http://www.internationalskeptics.com/forums/showpost.php?p=7256470&postcount=15597, which actually can't comprehend http://www.internationalskeptics.com/forums/showpost.php?p=7258971&postcount=15606).
The Man and jsfisher are also actual examples of the lack of "הרפו" of their awareness.
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So the prez couldn't even come up with his own way to publicly declare himself God and had to resort to plagiarism.
Hey, Doron, do you remember if you were born in the USA? Can you look it up in your birth certificate? Well, the presidential election is coming up the next year...
No one who hopes in you will ever be put to shame, but shame will come on those who are treacherous without cause.
Psalm 25:3
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doronshadmi
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So the prez couldn't even come up with his own way to publicly declare himself God and had to resort to plagiarism.
Hey, Doron, do you remember if you were born in the USA? Can you look it up in your birth certificate? Well, the presidential election is coming up the next year...
No one who hopes in you will ever be put to shame, but shame will come on those who are treacherous without cause.
Psalm 25:3
Gently? That diagram of yours is a mutilated version of Cantor set
epix, where is the straight (calm) line of this diagram?
Can your noisy mind actually gets http://www.internationalskeptics.com/forums/showpost.php?p=7255966&postcount=15594 ?
Can you actually be at "הרפו" state of mind?
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doronshadmi
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My reply to your next-to-last post disappeared. I'm very closely watched now, as my experiment with the plasma vortex brings the scientific evidence of the existence of God uncomfortably close.
Which one is your birth certificate that I've asked for?
doronshadmi
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Translation: Use only your verbal_symbolic skills.
Translation: Satisfaction is achieved only by using verbal_symbolic skills.
Translation: "AB" has to be defined in terms of "A,B" or "A" or "B", otherwise you will not get your reply read.
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doronshadmi
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Here is a part taken from H. M. Hubey book “The diagonal infinity: problems of multiple scales” (http://books.google.com/books?id=wD...m=6&sqi=2&ved=0CD0Q6AEwBQ#v=onepage&q&f=false ) page 297:
“A large part of the human brain is devoted to spatial computation, which in the language of serial logic becomes mere intuition. But the only way in which we can actually comprehend the most primitive concepts of all sciences, including mathematics of course, is in terms of these basic intuitions. If so, then what purpose does it serve to split the formation into syntax and semantics? What is obviously true by what some call intuition is the working of the parallel-visual-spatial system of the brain. Rigor seems to consist of turning these truths into words. Either this is done so that we can then learn from this to solve those problems that cannot be visualized or it’s done to satisfy those that cannot visualize. In no case is it necessary to stamp and certify only serial symbols as constituting rigor. If anything it is our very capability of spatial and parallel processing that even allowed us to entertain the possibility of a language for expressing truths. If anything language is a tool that allows us to partially reconstruct what we can see or seen to those that haven’t. If it were not so, only animals capable of speech (i.e. humans) would be capable of intelligence, and clearly it is not so.”
I would add that verbal_symbolic AND visual_spatial skills complement each other into a valuable framework.
“A large part of the human brain is devoted to spatial computation, which in the language of serial logic becomes mere intuition. But the only way in which we can actually comprehend the most primitive concepts of all sciences, including mathematics of course, is in terms of these basic intuitions. If so, then what purpose does it serve to split the formation into syntax and semantics? What is obviously true by what some call intuition is the working of the parallel-visual-spatial system of the brain. Rigor seems to consist of turning these truths into words. Either this is done so that we can then learn from this to solve those problems that cannot be visualized or it’s done to satisfy those that cannot visualize. In no case is it necessary to stamp and certify only serial symbols as constituting rigor. If anything it is our very capability of spatial and parallel processing that even allowed us to entertain the possibility of a language for expressing truths. If anything language is a tool that allows us to partially reconstruct what we can see or seen to those that haven’t. If it were not so, only animals capable of speech (i.e. humans) would be capable of intelligence, and clearly it is not so.”
I would add that verbal_symbolic AND visual_spatial skills complement each other into a valuable framework.
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doronshadmi
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The axiom of an infinite collection:
C is an infinite collection if for all x in C, x is arbitrarily the first object of C no more than once.
Example:
Given collection Z of points, any arbitrary chosen point is the first object of collection Z no more than once.
All the collections that do not satisfy the given axiom, are not infinite collections.
Example:
Given collection Z of points, any arbitrary chosen point is the first object of collection Z more than once.
Here is an example of finite Z:
If Z is infinite, then the diagram above does not hold.
C is an infinite collection if for all x in C, x is arbitrarily the first object of C no more than once.
Example:
Given collection Z of points, any arbitrary chosen point is the first object of collection Z no more than once.
All the collections that do not satisfy the given axiom, are not infinite collections.
Example:
Given collection Z of points, any arbitrary chosen point is the first object of collection Z more than once.
Here is an example of finite Z:
If Z is infinite, then the diagram above does not hold.
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jsfisher
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Oh, let's see how this works.
The set, {A, B, D, E}, as a collection. The set is unordered, so for all members of our set, none are "the first object."
We have satisfied the conditional. Therefore, according to Doronetics' Axiom of an Infinite Collection, the set, {A, B, D, E}, is an infinite collection.
How useful.
doronshadmi
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Wrong, all members of {A, B, D, E} are arbitrarily first objects of {A, B, D, E} exactly because order has no significance (the serial symbolic order has no significance).
Furthermore, since {A, B, D, E} is a finite collection, the first arbitrarily picked member can be picked again as the first member, since all the rest members are arbitrarily picked (which is not the case if the collection has infinitely many members).
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jsfisher
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Umm, no. Unordered means there isn't an ordering. There is no first, no last, not second, none of that. Now, if you want to impose an ordering (arbitrary or otherwise) on the set, you may, but then you will have one and only one first object, and it will be first exactly once. {A, B, D, E} would still be an infinite collection according to Doronetics' latest axiom of nonsense.
The words you used do not mean what you think they mean.
Be that as it may, if I can pick member D twice to be the first object (contradicting the whole notion of first, but whatever) from the set {A, B, D, E}, then I can also pick the member 17 twice to be the first object from the set of integers.
So, the set of integers must not be infinite.
doronshadmi
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You are right if "Order has no significance" is the same as "Order does not exist".
In that case there is a non-empty collection (of distinct members, in this case) but non of its members can be picked and used, because any attempt to pick something from this collection must be the first pick.
Furthermore, how can one define a collection of distinct members by not being able to get them both particularly (locally) and globally (non-locally)?
In other words, if order does not exist among sets, then their members are not distinct.
Since this is not the case, then "Order has no significance among sets" is the right one.
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doronshadmi
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Since members from a given set are used, they are picked, and being picked has first, second, etc ... picks that are done right from the first picked member.
There is another error that has to be fixed in my Axiom, which is: "no member is picked twice unless the rest of the members are picked".
It means that the "picking" follows the property of Distinction, which is essential to sets.
It is obvious that given an infinite set, "being picked twice" is not satisfied, by following Distinction (one can't return to the first pick).
There is another error that has to be fixed in my Axiom, which is: "no member is picked twice unless the rest of the members are picked".
It means that the "picking" follows the property of Distinction, which is essential to sets.
It is obvious that given an infinite set, "being picked twice" is not satisfied, by following Distinction (one can't return to the first pick).
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doronshadmi
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The axiom of an infinite collection:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.
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jsfisher
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That is not an axiom by any stretch of the imagination. Please stop trying to say it is.
It tries to be a definition, but it fails. It looks to be circular, but you'd need to define what you mean by the verb, to pick, first, before we can be certain.
As it stands, though, the set of integers would qualify as finite for the same reason I gave before.
doronshadmi
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Edit:
It is under construction.
The axiom of infinite set:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.
There is no "first" and no "arbitrary" in this version.
"Picked" is "chosen by distinction".
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jsfisher
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"Order has no significance" is your pet phrase, not ours. Unordered sets are, well, unordered. No ordering exists for such a set. Make of that what you will.
You really need to define what you mean by this picking thing. However, selecting an arbitrary member of a set (is that what you mean by "picking") does not require the set have any ordering. You selecting a member, no big deal. The set remains unchanged.
"Distinct members" -- there's another of your pet phrases. Sets have no such requirement. {A, A, B, A, B, B, B} is a perfectly fine set. It happens to be the same set as {B, A}, but so what?
Your premises were invalid, but your conclusion still did not follow from them. If A is an element of some set, S, and B is an element of S, it is possible for A to be the same as B.
Nope. Unordered.
jsfisher
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Construct all you like. It is still not an axiom.
Ah! The much heralded Doronetics word shift, in which the need to define one word or phrase is side-stepped by jumping to a different word or phrase in need of definition.
"Chosen by distinction" means what, exactly?
And are we heading towards your definition involving a sequence of steps in one-to-one correspondence with the positive integers? That will complete the circle.
doronshadmi
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No. You reduced redundancy in order to get {A,B} out of {A, A, B, A, B, B, B}.
If you do not do that then {A, A, B, A, B, B, B} is a multiset and {A,B} is a set.
I am talking about members of the same set.
Do you mean that A=3, B=3 and S={A,B} (where S is not a mutiset)?
doronshadmi
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Please define the difference between and axiom and a definition,
Do you understand what "under construction" means?
Means that one picks members according to their distinct properties.
It is exactly not complete the circle, since no distinct member can be picked twice, if the set is infinite.
jsfisher
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No, I did not. The two sets I presented are identical according to set theory, uncontaminated by unnecessary Doronetics concepts.
There is a multiset that looks like that, but I was clear in calling it a set, and the two sets, {A, A, B, A, B, B, B} and {A,B}, are indistinguishable (and are therefore the same set).
No. I mean A \in S and B \in S where S = {3,7,33}, for example. (Substitute the normal set membership symbol for \in.) A and B can both be 7.
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doronshadmi
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In that case, for example, |{A,A}| = |{A}| by based on, what you call, set theory.
As for no order, for example, (AB,AB) really has no order exactly because the considered framework is under superposition of identities of 2-Uncertainty x 2-Redundancy Distinction Tree.
According to this example (without a loss of generality) sets are based on Distinction State (A,B) = (B,A) under F(1,1).
Please be aware of the fact that (A,B) is just a particular case of Frame (1,1) under 2-Redundancy x 2-Uncertainty Distinction Tree, as follows:
Code:
(AB,AB) (AB,A) (AB,B) (AB) (A,A) (B,B) (A,B) (A) (B) ()
A * * A * * A * . A * . A * * A . . A * . A * . A . . A . .
| | | | | | | | | | | | | | | | | | | |
B *_* B *_. B *_* B *_. B ._. B *_* B ._* B ._. B *_. B ._.
(2,2) = (AB,AB)
(2,1) = (AB,A),(AB,B)
(2,0)= (AB)
[COLOR="magenta"][B](1,1)[/B][/COLOR] = (A,A),(B,B), [COLOR="magenta"][B](A,B)[/B][/COLOR]
(1,0)= (A),(B)
(0,0)= ()Since the members of {A,B} are based on the particular case of DS (A,B) under F (1,1), and since {A,B} has 0-Uncertainy and 0-Redundancy, then its members are pick-able no matter if {A,B} = {B,A}, and being picked has unconditionally first pick, second pick, etc ...
In other words {A,B} = {B,A} means that "order has no significance" (which is not the same as "order does not exist", where "order does not exist" holds, by this example, at DS (AB,AB)) such that the members of this set are picked as follows:
The first pick can be any member.
The second pick is any member that is not the first picked member.
(By going beyond {A,B} case, the third pick is any member that is not the first or the second picked members.
...
The n pick is any member that is not any of the previously picked members.
...
etc... ad infinitum, such that no member can be picked twice, unless the amount of the members is finite).
In other words:
C is a set.
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.
You are indeed clear about nonessential decelerations, in this case, which is probably contaminated by unnecessary jsfisheretics concepts.
In that case A or B are not used as members in terms of {A,B}, but they are used as different variables of the same S member.
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Doron really came up with a peculiar definition and not an axiom, which is a statement mostly establishing a relationship. He would have to present a theorem and prove it with the statement that he regards as an axiom. Since he has shown that his understanding of giving an example differes from what one can normally expect, any inquiry into the meaning of his wild constructions is likely to fail, if he gives an "example" as a form of an explanation. Then there is only a short step to the notorious "strict box."
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And passing page 410... wow, good luck with this!
doronshadmi
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C is a set.
The axiom of infinite set:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.
The axiom of completeness:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is incomplete.
Theorem: If C is infinite, then C is incomplete.
Proof: derived directly from the axioms.
doronshadmi
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Alternatively, you are using the ZF Axiom of extensionality, which sates:
( http://en.wikipedia.org/wiki/Zermelo–Fraenkel_set_theory )
jsfisher
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Yes, exactly. Two sets are equal if they have the same elements. Is this a revelation to you? And since the set {1, 1, 1, 2, 2, 2} has the exact same members as {2, 1}, they are the same set.
The Man
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Thus by your own requirement of what constitutes your current (nC) “pick” you can never pick any member previously picked (an nP pick). So your purported “infinite” and “completeness” are not aspects of any given set but simply of the limitation you have placed on what you defined as the members you can currently pick from (never those that have been picked already). By your own requirement no member of, say, the set {A,B,C} can be both pick nC and pick nP as simply being pick nP excludes it from consideration as pick nC. So set {A,B,C} must be “infinite” and “complete” by your requirement (as already noted by jsfisher). Be careful when you’re “Under construction” that you don’t just blindly bulldoze yourself into claiming clearly finite sets are infinite.
Oh and making your definition of “complete” the same definition (by essentially simply replacing the word “infinite” with “complete”) as your definition of “infinite” makes your definition of “complete” superfluous and a contrivance to assert what we already understand. That you only want to consider “infinite” sets as “complete”. The simple statement that ‘only “infinite” sets are “complete”” in your notions states clearly and explicitly what you mean without the pretence of thinking you’re defining a different aspect with the same definition you use for “infinite”.
You have two statements going where the property for_all_x in C is identical, but there are two different "axioms." What does prevent Doronetics to conditionally define set C as
For all x in C, if all x in C are picked AND no x can be picked twice, then C is an incomplete infinite set.
Your "theorem" wouldn't materialize, would it?
The real axiom of completeness states that every non-empty subset S of R that has an upper bound in R has a least upper bound, or supremum, in R.
And I say that it is not true -- at least one of non-empty subsets... doesn't have supremum.
Can I prove it?
No, there are no axioms available for the task, and so the original statement is not theorem but axiom.
The axioms usually start with "there exists...," but in the set theory they can be constructed different way, so knock yourself off.
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Example.
C = {1, 2, 3, 4}
Ordered selection under the condition: {2, 3, 4}, {3, 4}, {4}, { }.
In which way the statement "C is a set" precludes C being any set and therefore a finite set?
When you open a jigsaw puzzle box and there is nothing in there, you call it an incomplete box?
doronshadmi
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There is no requirement to not pick again the first picked object, if all the objects are picked.
For example, it can be done in the case of a non-empty finite set.
But it can't be done in the case of a non-empty infinite set.
This is exactly what the axiom of infinite set asserts:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite.
The Man, you simply replied only to a picked part of what I wrote, by ignoring the next part, here it is again:
doronshadmi said:
and you ignored:
doronshadmi said:
Next time please read the whole post before you reply.
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doronshadmi
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Is this a revelation to you that {1, 1, 1, 2, 2, 2} and {2, 1} actually have the same elements in terms of type?
Being the same in terms of type is based on the ability to distinguish between them, where redundancy and uncertainty have 0 value.
So according to your reasoning "the same" is actually limited to 0-Uncertainy x 0-Rerundancy case.
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doronshadmi
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I made a typo mistake in the name of the second axiom, here is the right one:
The axiom of incompleteness:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is incomplete.
No problem, it can done as follows:
C is a set.
Axiom:
For all x in C, if all x in C are picked AND no x can be picked twice, then C is infinite AND incomplete.
Theorem: If C is infinite, then C is incomplete.
Proof: derived directly from the axiom.
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