Deeper than primes

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epix said:
Your definitions prevents XOR to be placed non-locally within some hypothetical program
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You are wrong.

a) By Relation\Element Linkage XOR is the Relation (Non-local) aspect and the considered values are the Element (Local) aspect of the linkage.

By (a), elements values are simultaneously different, and we get bipartite, tripartite and four-partite maximally entangled states (bolded is mine) as written in Implementing a non-local xor function with quantum communication.
http://www.informaworld.com/smpp/content~db=all~content=a748066033


b) In this quote
doronshadmi said:
Wrong.

Again, if A=B or A≠B are checked by F T or T F XOR inputs, then the T result is related to the simultaneity of a one value (because XOR is a preventive relation), which is Locality in both A=B (it is exactly one value, and we have simultaneity of a one value) or A≠B (one value prevents the other value, and we have simultaneity of a one value) cases.

Simultaneity of a one value is local.

Simultaneity of more than one value is non-local.
Click to expand...
we get Non-localiy and Locality by the simultaneity of values, and XOR connective prevents the simultaneity of more than one value, which is property of Locality.


epix said:
Here is the source of the confusion: you are talking about logical connectives and nothing but that...
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Again, OM is not a context-dependent framework (Non-locality\Locality Linkage is fundamental in any given context, exactly as shown by (a) (b) cases) and as long as you miss that fact you can't get OM.
 
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doronshadmi said:
The Man said:
No, once again A AND ~A is a contradiction specificaly because the truth value of ~A depends on the truth value of A such the result is always FALSE regardless of the truth value of A.
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And the reason that such a result is always FALSE is because A is disjoint form ~A domain and any attempt to force sameness (~A AND A are the same domain) on disjoint domains like A and ~A is resulted by contradiction.
Click to expand...

The Man said:
No, once again A AND ~A is a contradiction specificaly because the truth value of ~A depends on the truth value of A such the result is always FALSE regardless of the truth value of A.
Click to expand...




doronshadmi said:
"the truth value of ~A depends on the truth value of A" is simply the result of two disjoint domains that have opposite true values w.r.t each other.
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Doron two “domains” do not have to be the same to have the same truth value, so your “two disjoint domains” is just superfluous nonsense. That they “have opposite true values w.r.t each other” is specifically the result of their mutual dependence by negation, not of being “two disjoint domains” as you put it.


doronshadmi said:
On the contrary, the line segment is included AND excluded w.r.t A or ~A exactly because it is a non-composed thing that is not fully included in A or ~A domains, which is the property of Non-locality that A and ~A do not have, exactly as shown here:
[qimg]http://farm5.static.flickr.com/4077/4789944385_7e4d198597.jpg[/qimg]
Click to expand...

“not fully included in A or ~A domains”? So it is only partially included or excluded “in A or ~A domains”. See, now was that so hard for you to admit? Once again however we see that you are one of the biggest opponents of your own notions as your “non-composed thing” is “is not fully included in A or ~A domains”.


doronshadmi said:
So the line segment is Non-local w.r.t A or ~A, and ~A and A are Locals w.r.t each other because no one of them is included AND excluded w.r.t each other, as the line segment is.
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Nope, the “line segment” by your own assertions “is not fully included in A or ~A domains” thus not fully excluded as well. A smaller line segment is “fully included in “A” and another “fully included in” “~A”. Together they compose your original line segment.

Again…
The Man said:
That you would like to choose not to partition the line is irrelevant to the fact that the “A;~A domains” do partition that line and in determining those “A;~A domains” you have in fact partitioned that space as well as any lines in that space crossing both domains.
Click to expand...


doronshadmi said:
Again, so what? A+~A=the whole space (A and ~A=the whole space) is not A AND ~A, as you by yourself wrote.
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Again see above and my post you quoted. “A and ~A=the whole space” just as the line segment “fully included in “A” and another “fully included in” “~A” compose your original line segment.

Again…
The Man said:
That you would like to choose not to partition the line is irrelevant to the fact that the “A;~A domains” do partition that line and in determining those “A;~A domains” you have in fact partitioned that space as well as any lines in that space crossing both domains.
Click to expand...
 
doronshadmi said:
Again, you use a definition of a line that is based on points.

Look at this part of the definition:

"A line segment is a part of a line …"

"Line" or "Point" are used but undefined by this definition.

On the contrary I define "Line" as the minimal representation of Non-locality and a Point as a minimal representation of Locality, and I am doing that by using logical Venn diagrams.
Click to expand...

But your definitions of line and point don't define "minimal representation", "non-locality", or "locality". See, two can play that game.

And you still haven't given a clear definition of locality/local.

doronshadmi said:
[post snipped due to usage of undefined terms]

As for the meaning of the word "understand" we get it better when we realize that it used for getting the core of things (gets what is "under" a given ("stand") thing).
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"Under" means below, "stand" is device to keep things from falling. My bicycle has a stand.
 
The Man said:
Doron two “domains” do not have to be the same to have the same truth value, so your “two disjoint domains” is just superfluous nonsense. That they “have opposite true values w.r.t each other” is specifically the result of their mutual dependence by negation, not of being “two disjoint domains” as you put it.
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The Man you support my claim, which is: A,~A is nothing but a particular case of A≠B, so the generalization is derived from the fact that A and B are disjoint domains, and therefore they are local w.r.t each other.

The Man said:
So it is only partially included or excluded “in A or ~A domains”.
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No.

The line segment (which is non-composed) is exactly included AND excluded w.r.t A or ~A domains.

A or ~A are included OR excluded w.r.t the line segment (which is non-composed), or included OR excluded w.r.t each other, as clearly seen by:
4789944385_7e4d198597.jpg



See the difference?
 
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Little 10 Toes said:
But your definitions of line and point don't define "minimal representation", "non-locality", or "locality". See, two can play that game.
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Again, I am doing that by using logical Venn diagrams.

You ignore it. It is your ignorance, not mine.

Little 10 Toes said:
"Under" means below, "stand" is device to keep things from falling.
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This is not the case of "under" and "stand", if they are combined into a one word.
 
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dlorde said:
Alas indeed!

I pretty much agree with your sentiments; any effort to expand and integrate our intuition and immediate perception with our serial, logical thought processes is to be encouraged, but this thread itself is good evidence that OM is not the advertised highway to enlightenment, but an unpaved blind alley.
Click to expand...

... down which, if it proceeds, reason is likely to be mugged...
 
zooterkin said:
... down which, if it proceeds, reason is likely to be mugged...
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Why, just because Non-locality is added to Locality under a one comprehensive framework?

Take any possible form of venn diagrams, http://en.wikipedia.org/wiki/Venn_diagram ,their domains are Local w.r.t other, and this is what we have today at the foundation of the mathematical science.

OM extends it by adding Non-locality to the game:

4789944385_7e4d198597.jpg
 
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doronshadmi said:
As for the meaning of the word "understand" we get it better when we realize that it used for getting the core of things (gets what is "under" a given ("stand") thing).
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No Doron we get it better when we actually look up that meaning.
http://en.wiktionary.org/wiki/understand

As well as the Etymology.
Etymology
Middle English understanden from Old English understandan (“to understand”) from under- "between" + standan (“to stand”), akin to O.Fris. understonda, M.Dan. understande, Dutch onderstaan
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Instead of you just making up your own.

So actually it derives from a “standing” that is “between” as in an agreement, like “The two parties have reached an understanding”. However the common usage today has also come to include to be “aware of the meaning of” as well as “To apply values (axioms)”.

You got any of that Doron, Some agreement as an “understanding” between parties, any awareness of the meanings of the words you use or even some applied axioms? So far you have displayed considerable lacking in all those regards.
 
The Man said:
No Doron we get it better when we actually look up that meaning.
http://en.wiktionary.org/wiki/understand

As well as the Etymology.


Instead of you just making up your own.

So actually it derives from a “standing” that is “between” as in an agreement, like “The two parties have reached an understanding”. However the common usage today has also come to include to be “aware of the meaning of” as well as “To apply values (axioms)”.

You got any of that Doron, Some agreement as an “understanding” between parties, any awareness of the meanings of the words you use or even some applied axioms? So far you have displayed considerable lacking in all those regards.
Click to expand...

Thank you for supporting my claim that there is a common foundation for different stands in order to get an agreement (get what is "under" given ("stands") things).

EDIT:

Furthermore, that is "under" is non-local w.r.t "stands", where the "stands" are local w.r.t each other or w.r.t "under".
 
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doronshadmi said:
Thank you for supporting my claim that there is a common foundation for different stands in order to get an agreement (get what is "under" given ("stands") things).
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No, Doron, he did not support your claim at all. Please re-read the post. What exactly do you mean by 'different stands', anyway?
 
doronshadmi said:
The Man you support my claim, which is: A,~A is nothing but a particular case of A≠B,
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What claim? Do you mean A ≠ ~A is a "particular case of A≠B" where B = ~A? You may like to think of it as 'your claim', but it is simply a trivial fact just as A ≠ ~A is a "particular case of” (double negation) A = A or a “particular case of” Z = Q where Z=A and Q = A or a "particular case of” Z ≠ W where Z=A and W ≠ A. Everything Doron is “nothing but a particular case of” it being written with different symbols resulting in the same meaning. However that wasn’t your actual “claim” was it? Your actual claim is what follows below and is simply a false assumption on your part.

doronshadmi said:
so the generalization is derived from the fact that A and B are disjoint domains, and therefore they are local w.r.t each other.
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Nope, A≠B does not indicate “A and B are disjoint domains”. So your assumption is simple false from the start.


doronshadmi said:
No.

The line segment (which is non-composed) is exactly included AND excluded w.r.t A or ~A domains.
Click to expand...

Again…

The Man said:
That you would like to choose not to partition the line is irrelevant to the fact that the “A;~A domains” do partition that line and in determining those “A;~A domains” you have in fact partitioned that space as well as any lines in that space crossing both domains.
Click to expand...



doronshadmi said:
A or ~A are included OR excluded w.r.t the line segment (which is non-composed), or included OR excluded w.r.t each other, as clearly seen by:
[qimg]http://farm5.static.flickr.com/4077/4789944385_7e4d198597.jpg[/qimg]
Click to expand...

No Doron A specifically excludes ~A as ~A specifically excludes A. Also A specifically excludes part of your original line segment as as ~A specifically excludes part of your original line segment. You are just talking nonsense again.


doronshadmi said:
See the difference?
Click to expand...

Everyone (expect you) sees the difference between you just talking nonsense and the mutually dependent mutually exclusive relation of A and ~A. That is probably why you have some much difficulty in reaching an understanding with anyone about your nonsense or the mutually dependent mutually exclusive relation of A and ~A
 
doronshadmi said:
Thank you for supporting my claim that there is a common foundation for different stands in order to get an agreement (get what is "under" given ("stands") things).

EDIT:

Furthermore, that is "under" is non-local w.r.t "stands", where the "stands" are local w.r.t each other or w.r.t "under".
Click to expand...



[latex]$$$ \rho >> 11.34 \, g \, cm^{-3} $$$[/latex]​
 
doronshadmi said:
Thank you for supporting my claim that there is a common foundation for different stands in order to get an agreement (get what is "under" given ("stands") things).
Click to expand...


Doron neither I nor the etymology of the word “understand” support your “claim”. Not to worry though, we already understand that understanding is not your strong suit.

doronshadmi said:
EDIT:

Furthermore, that is "under" is non-local w.r.t "stands", where the "stands" are local w.r.t each other or w.r.t "under".
Click to expand...

More of your meaningless “non-local” and “local” ascriptions. That you feel some compulsion or get some satisfaction out of just sticking those meaningless ascriptions onto everything is another thing that we already understand.
 
doronshadmi said:
Thank you for supporting my claim that there is a common foundation for different stands in order to get an agreement (get what is "under" given ("stands") things).

Furthermore, that is "under" is non-local w.r.t "stands", where the "stands" are local w.r.t each other or w.r.t "under".
Click to expand...
I under-stand that you've been under heavy flak and that excuses a minor mistake in your example of deterministic non-isomorfism. So, with your kind permission . . .

local: ST = ST
non-local: ?____ST = ST_____?

(Use Bonham's Theorem and swap the last two letters of "non-local.")

non-locla: ?____ST = ST____?

(The swap initiate the process of non-localization and la becomes disjoint from "non-local" to become local to the left complement.)

non-loc.. : LAST = ST____?

(Now run the equation through the logical connective AND.)

non-loc.. : LAST = STAND

(With the non-local equation solved, the global state takes over to demonstrate the property of deterministic non-isomorfism.)

global: ?------? LAST STAND

As you implied: There are many stands, and it takes an OM(G) to localize the heck out of them.
 
The Man said:
Nope, A≠B does not indicate “A and B are disjoint domains”. So your assumption is simple false from the start.
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Let us see.

a) In the case of A ≠ ~A A domain and ~A domain are defiantly disjoint (A ≠ ~A is not a "particular case of” (double negation) A = A because (A ≠ ~A) ≠ (A = ~~A) = (A=A))

b) As for A≠B, since you are talking only about T or F, then (A≠B) = (A ≠ ~A), which is exactly (a) case ( (a) case = (b) case ).

If A≠B is not limited to T or F then and only then there can be a union between A and B.

In that case A and B are not disjoint, but still we can distinguish between 3 disjoint domains, which are A-B, B-A and C (where C is non-empty if A is not fully included in B or B is not fully included in A).

Disjoint domains are local w.r.t each other.
 
doronshadmi said:
Let us see.

a) In the case of A ≠ ~A A domain and ~A domain are defiantly disjoint (A ≠ ~A is not a "particular case of” (double negation) A = A because (A ≠ ~A) ≠ (A = ~~A) = (A=A))
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Are you sure that the A domain and the ~A domain are "defiantly disjoint?" Wouldn't the adverb "fiercely" characterize the relation better?

Why don't you get out of your house and observe, instead of taking pleasures in sodomizing the Venn diagram?

Look at the diagram once again: You see a rectangle, right? What can be symbolized by a rectangle?

A room, says the architect.

Alright. We have a large room full of people. They watch a guy who is drawing a circle on the floor. When he is done, he asks the others if anyone can play piano. "If you can play piano, please step inside the circle," he asks. And that's an example that the negation Venn diagram represents. That, and nothing else.

A = people who CAN play piano
~A = people who CANNOT play piano

Here is what the Venn diagram does not represent:

A = former US president Jimmy Carter
~A = anything but former US president Jimmy Carter (and that includes a lost sock.)

Now you can have your way and say that former US presidents and lost socks are two "defiantly disjoint" domains.

Btw, how can the colors of the negation Venn diagram be a disjoint domain?

Look at the Venn diagram and remind yourself the applicable definition of the word domain:the set of elements to which a mathematical or logical variable is limited.

Your esoteric definitions and characteristics don't have any application - you can't go beyond the verbal goulash and show a real-life example of your novel approach to fixing things that aren't broken.
 
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epix said:
A = people who CAN play piano
~A = people who CANNOT play piano

...

Look at the Venn diagram and remind yourself the applicable definition of the word domain:the set of elements to which a mathematical or logical variable is limited.

Your esoteric definitions and characteristics don't have any application - you can't go beyond the verbal goulash and show a real-life example of your novel approach to fixing things that aren't broken.
Click to expand...

Look at the venn diagram and ask yourself what enables you to know ~A without loosing A, or what enables you to know A without loosing ~A (and it does not matter if ~A is limited to "not playing a piano" or ~A is the non limited state of "anything but ~A")?

If you really do that you will find that you can't avoid the line segement (Non-locality) in:

4789944385_7e4d198597.jpg
 
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The Man said:
Nope, A≠B does not indicate “A and B are disjoint domains”. So your assumption is simple false from the start.
Click to expand...
If B is "anything but A" and our framework deals with only T or F true values. then B and A are disjoint.

If the framework is extended beyond only T and F, like two sets and a union (for example: the set of prime numbers P and the set of even numbers E) then we get the following:

T=2

a) E is included (by T) AND excluded (by even numbers that are not T) w.r.t P.

b) P is included (by T) AND excluded (by prime numbers that are not T) w.r.t E.

c) PE is the set of prime AND even numbers (only inclusion is considered).

d) If F=E-(P+T), then F is disjoint w.r.t P+T.

e) If G=P-(E+T), then G is disjoint w.r.t E+T.

f) If H=PE-T, then H is disjoint w.r.t T.


By carefully observe (a) to (f) cases it is realized that each set (P,E,PE,T,F,G,H) has a common property, which is non-local w.r.t any particular given element of the given set, so Non-locality\Locality Linkage is fundamental to any given set.

Furthermore, even if there is no common property to a given collection of elements, still they are considered as many elements only by Non-locality (comparer\counter)\Locality (compared\counted) Linkage.

Again Non-locality\Locality Linkage is not limited to any particular context, and as a result we get a one organic body of mathematical knowledge, which does not exist by the current context-dependent reasoning.
 
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doronshadmi said:
Look at the venn diagram and ask yourself what enables you to know ~A without loosing A, or what enables you to know A without loosing ~A (and it does not matter if ~A is limited to "not playing a piano" or ~A is the non limited state of "anything but ~A")?

If you really do that you will find that you can't avoid the line segement (Non-locality) in:

[qimg]http://farm5.static.flickr.com/4077/4789944385_7e4d198597.jpg[/qimg]
Click to expand...
Your reply exactly confirms what I said: You are not capable of translating the doctored Venn diagram into a real example; you can only use ghost phrases such as "~A without loosing A" to justify the presence of the straight line.

What matters in the Venn diagram is the area - you see white area and red area. Can you measure the area of the circle and the area of the rectangle minus the area of the circle?

No you can't, coz areas can be only computed. So what's that line, which is not the diameter line, doing in there when it belongs to the domain of geometric figures which can be measured? Isn't that configuration of yours a classic example of two "defiantly disjoint domains?"

(Hold your hat on this one: I found a real application for your graffiti. It does somewhat relate to the A and ~A topic.)
 
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doronshadmi said:
Let us see.

a) In the case of A ≠ ~A A domain and ~A domain are defiantly disjoint (A ≠ ~A is not a "particular case of” (double negation) A = A because (A ≠ ~A) ≠ (A = ~~A) = (A=A))
Click to expand...

Wrong again, look up the distributivity of negation.

doronshadmi said:
b) As for A≠B, since you are talking only about T or F, then (A≠B) = (A ≠ ~A), which is exactly (a) case ( (a) case = (b) case ).
Click to expand...

I wasn’t “talking only about T or F” and by your venn diagram you weren’t either, so you are talking only nonsense again.

doronshadmi said:
If A≠B is not limited to T or F then and only then there can be a union between A and B.
Click to expand...

Were did you even get the ridiculous notion that you can not have a union of disjoint domains? It is an intersection that will make the domains not disjoint. Also the entire space of possible values even when “limited to T or F” is T and F, a union of those disjoint domains. You really don’t have a clue to what you are talking about, do you?

doronshadmi said:
In that case A and B are not disjoint, but still we can distinguish between 3 disjoint domains, which are A-B, B-A and C (where C is non-empty if A is not fully included in B or B is not fully included in A).
Click to expand...

A and B can very well be disjoint, the defining factor is that they share no elements (there is no non-empty intersection between them). So your assertions about “3 disjoint domains,” “(where C is non-empty if A is not fully included in B or B is not fully included in A)” is again simply nonsense. You really don’t have a clue to what you are talking about, do you?

doronshadmi said:
Disjoint domains are local w.r.t each other.
Click to expand...

Again simply your meaningless ascription




doronshadmi said:
If B is "anything but A" and our framework deals with only T or F true values. then B and A are disjoint.
Click to expand...

If B = ~A “then B and A are disjoint” because A and ~A are mutually exclusive (have no elements in common). That’s it, it doesn’t matter if “our framework deals with only T or F true values” or not. It is simply a result of the mutually dependent mutually exclusive relation of A to ~A.


doronshadmi said:
If the framework is extended beyond only T and F, like two sets and a union (for example: the set of prime numbers P and the set of even numbers E) then we get the following:
Click to expand...

Again the range of available values in a “T and F” system is a union of those disjoint domains.

doronshadmi said:
T=2

a) E is included (by T) AND excluded (by even numbers that are not T) w.r.t P.

b) P is included (by T) AND excluded (by prime numbers that are not T) w.r.t E.

c) PE is the set of prime AND even numbers (only inclusion is considered).

d) If F=E-(P+T), then F is disjoint w.r.t P+T.

e) If G=P-(E+T), then G is disjoint w.r.t E+T.

f) If H=PE-T, then H is disjoint w.r.t T.
Click to expand...

Doron, the definition of disjoint is simple, the sets have no elements in common, that’s it.

doronshadmi said:
By carefully observe (a) to (f) cases it is realized that each set (P,E,PE,T,F,G,H) has a common property, which is non-local w.r.t any particular given element of the given set, so Non-locality\Locality Linkage is fundamental to any given set.

Furthermore, even if there is no common property to a given collection of elements, still they are considered as many elements only by Non-locality (comparer\counter)\Locality (compared\counted) Linkage.

Again Non-locality\Locality Linkage is not limited to any particular context, and as a result we get a one organic body of mathematical knowledge, which does not exist by the current context-dependent reasoning.
Click to expand...

Again simply your meaningless ascriptions
 
doronshadmi said:
If B is "anything but A" and our framework deals with only T or F true values. then B and A are disjoint.
Click to expand...
Nonsense. If "anything but that" refers to undeclared domain, then A and B could be

A = knife
B = salami

for example.

A Boolean operator sift through all the items (variables) like A = sock, or A = screwdriver; B = table, or B = car, and declares the input False, until it encounters A = knife and/or B = salami. In that case it declares the input(s) True. The conclusion of those various True/False inputs depends on the type of the Boolean operator. (OR, XOR, AND . . .)

A researcher opts for the AND gate to count the instances where A=True and B=True, coz the conjunction of A and B describes a knife cutting salami. Or does it?

According to you, A and B are always disjoint under your stated conditions and there is no possibility of A AND B occurring.

You are using this false notion of disjoint state of A and B as a lemma for the (a) through (f) propositions. It's like saying don't bother.

You can fix the lemma though. Just declare the domain . . .

Domain: logic
A = reason
B = Doron

Now A and B are in the permanent disjoint state and you can form the propositions.

But given the A and B condition, those six proposition will be False as well.

Sweetie, why don't you go and do the dishes.
 
Were did you even get the ridiculous notion that you can not have a union of disjoint domains? It is an intersection that will make the domains not disjoint.[/quote]
In other words, disjoint domains are not the same as non disjoint domains, which can be written as:

A ~= ~A

where

A= disjoint domains

~A= non disjoint domains

In other words we have got two birds in a one shot:

1) Your nonsense about (A ≠ ~A) = (A = ~~A)

2) Your ridiculous claim that disjoint domains are still disjoint domains even if they are non disjoint domains.

The rest of your reply has the same "value".
 
epix said:
Your reply exactly confirms what I said: You are not capable of translating the doctored Venn diagram into a real example; you can only use ghost phrases such as "~A without loosing A" to justify the presence of the straight line.
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epix, without this straight line, ~A;A are non-comparable, simple as that.

You still can't get the universal form of Relation\Element Linkage, which is non context dependent framework.
 
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doronshadmi said:
Were did you even get the ridiculous notion that you can not have a union of disjoint domains? It is an intersection that will make the domains not disjoint.
In other words, disjoint domains are not the same as non disjoint domains, which can be written as:

A ~= ~A

where

A= disjoint domains

~A= non disjoint domains
Click to expand...

A union of A and ~A then gives you disjoint as well as “non disjoint” domains. So the answer to my question of “Were did you even get the ridiculous notion that you can not have a union of disjoint domains?” is that you just don’t know what disjoint means?

doronshadmi said:
In other words we have got two birds in a one shot:

1) Your nonsense about (A ≠ ~A) = (A = ~~A)
Click to expand...

Again look up the distributivity of negation.

doronshadmi said:
2) Your ridiculous claim that disjoint domains are still disjoint domains even if they are non disjoint domains.
Click to expand...

Nope, I never made any such claim and you obviously do not understand what disjoint means.

doronshadmi said:
The rest of your reply has the same "value".
Click to expand...

As usual, you’re reply has absolutely no value.



doronshadmi said:
The Man said:
Doron, the definition of disjoint is simple, the sets have no elements in common, that’s it.
Click to expand...

Exactly, but this is not (a) (b) non-local cases even if E ~= P.
Click to expand...

So you do know that your previous assertions are simply nonsense? Again who cares (other then you) about your meaningless ascriptions like “this is not (a) (b) non-local cases”? The fact is that you don’t know what you’re talking about and even seem to assert above that you know you don’t know what your talking about.
 
doronshadmi said:
Were did you even get the ridiculous notion that you can not have a union of disjoint domains?
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Of course you can have a union of two previously disjoint categories, but only when a "certain condition" doesn't disallow it:

Originally Posted by doronshadmi
If B is "anything but A" and our framework deals with only T or F true values, then B and A are disjoint.
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A = It is raining
B = I am cold

Each proposition can be either True or False (the "framework" deals only with T or F values, as it obviously must). According to your premise above, A and B are disjoint.

How do you know that when it's raining I can't be cold at the same time? Why would B and A be disjoint? A<>B and their truth values are in no way any limiting factors that would render B and A disjoint variables recognized by the AND gate always as False, as you think it is so.

You've built your arguments to follow on a false premise.

See, there is a difference between input truth values and output truth values, which you didn't specify in the quote.
 
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doronshadmi said:
If B is "anything but A" and our framework deals with only T or F true values. then B and A are disjoint.
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This nonsense has very likely its root in your misconception that if A=True and B=False and vice versa, then A and B are disjoint and both circles A and B do not intersect each other, coz when you decide treat B as "anything but A," then B<>A is strict. That prevents two cirles A and B to intersect each other, such as

A = fruit
B = apples

Then comes fruit NOT apples represented by two intersecting circles.
 
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doronshadmi said:
epix, without this straight line, ~A;A are non-comparable, simple as that.
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Let's be specific here. It is is you and only you who can't make the comparison. When I see a horse standing by a tree, I don't need to tie him to the tree in order to tell what is what. It is you who needs to bring the rope (line) and do the connecting to prevent yourself climbing the tree and riding nowhere.

You still can't get the universal form of Relation\Element Linkage, which is non context dependent framework.
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Well, I lost the diagram of the non-universal form. Can you paste it please?
 
doronshadmi said:
epix, without this straight line, ~A;A are non-comparable, simple as that.
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Do you mean that without some aid represented by a straight line you can't tell A apart from ~A?

A = Doron is in the bathroom
~A Doron is not in the batroom

So the straight line is a symbol for a white cane. That sucks. I'm sorry . . . But how do you manage to type?
:confused:
 
The Man said:
A union of A and ~A then gives you disjoint as well as “non disjoint” domains.
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In other words, you claim that a union of A and ~A gives you non disjoint domain (which is False) AND disjoint domain (which id True), and we get F AND T, which is always F.

So, your
The Man said:
Nope, I never made any such claim and you obviously do not understand what disjoint means.
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has no basis.
 
jsfisher said:
So, without geometry, set theory and logic could not exist? Who knew it was so important. Euclid would be so proud!
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The line in venn diagram is not limited to Logic or Geomety.

It is the non-local property of both frameworks.

Your context dependent approach can't get this.


jsfisher said:
No, I didn't. I wrote: (A ≠ B) = ~(A = B). That expresses, formally, what I already understand.
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What you already understand (before using some expression (formal or not))?
 
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epix said:
Do you mean that without some aid represented by a straight line you can't tell A apart from ~A?
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Exactly, you simply ignore your memory (represented by the straight line) as the non-local aspect, which enables you to compare A with ~A (the local aspects) and get the conclusion that they are different.

Your framework is based on Non-locality\Locality Linkage.
 
jsfisher said:
So, without geometry, set theory and logic could not exist? Who knew it was so important. Euclid would be so proud!
Click to expand...

Actually, it's quite common for mathematical cranks to consider geometry the only "real" math. Many circle-squarers, for instance, react with incomprehension to the demand to tell them what the value of Pi is according to their geometrical work. See Dudley Underwood's delightful book, Mathematical Cranks.

Clearly Doron is of this variety. He keeps giving us pictures (which he considers to be geometry), and reacts with disdain that we ask him to explain with numbers or logical symbols -- such inferior, artificial creations! -- what the pictures mean.

De Morgan, the famous logician, once said about an especially incomprehensible and insistent circle-squarer that he is a special man. The usual crank -- "paradoxer", in De Morgan's terms -- reasons correctly, but from false premises. Mr. X reasons badly based on no premises at all.

That's our Doron.
 
Skeptic said:
Actually, it's quite common for mathematical cranks to consider geometry the only "real" math. Many circle-squarers, for instance, react with incomprehension to the demand to tell them what the value of Pi is according to their geometrical work. See Dudley Underwood's delightful book, Mathematical Cranks.

Clearly Doron is of this variety. He keeps giving us pictures (which he considers to be geometry), and reacts with disdain that we ask him to explain with numbers or logical symbols -- such inferior, artificial creations! -- what the pictures mean.

De Morgan, the famous logician, once said about an especially incomprehensible and insistent circle-squarer that he is a special man. The usual crank -- "paradoxer", in De Morgan's terms -- reasons correctly, but from false premises. Mr. X reasons badly based on no premises at all.

That's our Doron.
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Skeptic, you are talking without any connection to OM.

See http://www.internationalskeptics.com/forums/showpost.php?p=6134247&postcount=10713.
 
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doronshadmi said:
In other words, you claim that a union of A and ~A gives you non disjoint domain (which is False) AND disjoint domain (which id True), and we get F AND T, which is always F.
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Nope, your other words fail you again. Two domains (you will note the plural) can be disjoint or “non-disjoint”. A single domain can not be disjoint with itself, as it always has elements in common with itself. However, it can be a union of disjoint and/or non-disjoint domains, depending on how those sub domains (sub sets) are defined. In fact some of those sub domains can be disjoint from certain other sub domains while still having elements in common with some other sub-domains. You really don’t have a clue what you are talking about, do you?




doronshadmi said:
So, your

has no basis.
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Show where I made such a claim as you remarked to, otherwise it is you, as usual, that "has no basis".

doronshadmi said:
The Man, (A ~= ~A) ~= (A = A), no matter how many maneuvers you do.
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No maneuvers Doron, I’ll leave them to you, however you only maneuver in circles. So you are going to purposely and deliberately remain ignorant by not looking up the distributiviy of negation. Not surprising as it has been brought to your attention before and you still just want to go around in circles claiming the same nonsense (like “(A ~= ~A) ~= (A = A)”) as if we haven’t addressed it before.
 
doronshadmi said:
Exactly, you simply ignore your memory (represented by the straight line) as the non-local aspect, which enables you to compare A with ~A (the local aspects) and get the conclusion that they are different.

Your framework is based on Non-locality\Locality Linkage.
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Doron the relation between A and ~A is explicit. You’re the only one who requires some ‘comparison’ between them. So even if…



doronshadmi said:
without this straight line, ~A;A are non-comparable, simple as that.
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that is simply your problem. We have been over this before as well,
 
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