Deeper than primes

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I have no idea what Magority means, if you mean ‘Majority’ then I have no idea why you would associate that in the first place only to then try to dissociate it from exclusive.

Thanks anyway for giving me credit and I do appreciate the indications of edits and will naturally read and comment on them. Unfortunately the contents of that edit were effectively covered by my previous post.

http://www.internationalskeptics.com/forums/showthread.php?postid=4221544#post4221544

You require the symmetry by claiming that you “use "=" as self identity”. Have you decided which you want to give up your claim of asymmetry or your claim that you “use "=" as self identity”?
 
jsfisher said:
First, the word being used is "equal to", not "compare". You even provided a reference for your usage, and the reference is clear. "Equal to" is a symmetric relation.

Second, if you goal is to obtuse and confuse the discussion, you are doing an excellent job. Continue to abuse common usage and hide behind secret meanings. Well done!

I was looking for communication. My bad.
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Again

x = .

y = ___

x = y only from x point of view (for example: .__ , _._ , __. )

y >and= or <and> or <and= x olny from y point of view (for example: .__ , _._ , __. )

So as you see, the same interactions are understood differently from different points of view.
 
The Man said:
I have no idea what Magority means, if you mean ‘Majority’ then I have no idea why you would associate that in the first place only to then try to dissociate it from exclusive.

Thanks anyway for giving me credit and I do appreciate the indications of edits and will naturally read and comment on them. Unfortunately the contents of that edit were effectively covered by my previous post.

http://www.internationalskeptics.com/forums/showthread.php?postid=4221544#post4221544

You require the symmetry by claiming that you “use "=" as self identity”. Have you decided which you want to give up your claim of asymmetry or your claim that you “use "=" as self identity”?
Click to expand...

x = .

y = ___

By self identity x = x , y = y , etc...

It does not mean that if x = y from x point of view (example: _.__) then y = x from y point of view (Y < and > x from y point of view).

x,y relations are non-commutative
 
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doronshadmi said:
Again...
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Again, we were discussing your failed Definition #4 and the definitions that precede it.

According to your most recent version of Definition #4, everything is local.

If you are relying on some private meaning for the equal and not-equal relations, then please provide those meanings. Perhaps you can salvage Definition #4 yet.
 
jsfisher said:
Again, we were discussing your failed Definition #4 and the definitions that precede it.

According to your most recent version of Definition #4, everything is local.

If you are relying on some private meaning for the equal and not-equal relations, then please provide those meanings. Perhaps you can salvage Definition #4 yet.
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Definition 4: If object x = xor ≠ (where ≠ is < xor >) w.r.t object y, then object x is called Local.



Definition 5: If object x = and ≠ (where ≠ is < xor >) or < and > w.r.t object y, then object x is called Non-Local.
 
doronshadmi said:
By self identity x = x , y = y , etc...

It does not mean that if x = y from x ... then y = x
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Gee, according to the wikipedia link you yourself provided, that's exactly what it means.

x,y relations are non-commutative
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Huh? x and y (as you used them) are objects, not relations.
 
doronshadmi said:
Definition 4: If object x = xor ≠ (where ≠ is < xor >) w.r.t object y, then object x is called Local.
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Ah! A brand new revision. I see you have reverted back to less than and greater than relations. You also are being a bit sloppy with your formula notation, so I need you to be a bit more precise:

You state "≠ is < xor >". So, in "A ≠ B" I can translate that to "A < xor > B", right? Well, "A < xor > B" is not a valid formula. Did you really mean that "A ≠ B is (A<B) xor (A>B)"?

May I similarly assume that your "= xor ≠" notation was meant as binary relation, call it %, such that A % B is equivalent to (A = B) xor (A ≠ B) ?

If not, what did you mean?
 
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jsfisher said:
Ah! A brand new revision. I see you have reverted back to less than and greater than relations. You also are being a bit sloppy with your formula notation, so I need you to be a bit more precise:

You state "≠ is < xor >". So, in "A ≠ B" I can translate that to "A < xor > B", right? Well, "A < xor > B" is not a valid formula. Did you really mean that "A ≠ B is (A<B) xor (A>B)"?

May I similarly assume that your "= xor ≠" notation was meant as binary relation, call it %, such that A % B is equivalent to (A = B) xor (A ≠ B) ?

If not, what did you mean?
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I mean that x is considered as local wr.t y if for example:

x = .

y = ___

x = y (.__ , _._ , __.)

x ≠ y ( in this case x < y ( . __) or x > y( __ .))

EDIT:

In general:

If only a one relation between x and y satisfies the obseravtion of y through x , then x is local.

If only more than a one relation between x and y satisfies the obseravtion of y through x , then x is non-local.
 
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doronshadmi said:
I mean that x is considered as local wr.t y if:

x = .

y = ___

x = y (.__ , _._ , __.)

x ≠ y ( in this case x < y ( . __) or x > y( __ .))
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You didn't understand my question, did you? So rather than explain your notation, even when I provided you the most likely interpretation, you dash off into your world of lines and dots.

Let's take this quoted post as your final word on local. According to you, for one thing to be local with respect to another thing, the first thing must be a dot and the other thing must be a line.

So, only the dot is local, and only with respect to a line.

Is that really what you wanted to say? How about answering my question about your notation, using words, without asides nor irrelevant symbols.


ETA: I see you've edited your original post. Doesn't matter. It still doesn't address the original question.
 
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doronshadmi said:
x = .

y = ___

By self identity x = x , y = y , etc...


It does not mean that if x = y from x point of view (example: _.__) then y = x from y point of view (Y < and > x from y point of view).

x,y relations are non-commutative
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That is exactly what it means, that they are the same. If as you say "Y < and > x from y point of view" then Y ≠ X and X ≠ Y. Your problem seems to be you conflating a point within a line as being equal to or by some strange application in your thinking a “self identity” with that line from the perspective of the point. However that is easily dismissed as you assert that a line is not a collection of points, making your above augment irrelevant. If the line Y does not contain points, one of which being X, then X can never be part of Y. So what do you want to give up now your claim that a line is not a collection of points or your bizarre dot and dash diagrams intended to shown circumstances where point X is part of line Y?


ETA: Even if your were to concede that a line Y is a collection of points one of which being X that only means that Y contains X and not that X=Y from any point of view.
 
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jsfisher said:
ETA: I see you've edited your original post. Doesn't matter. It still doesn't address the original question.
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Do you mean to this part?
doronshadmi said:
In general:

If only a one relation between x and y satisfies the obseravtion of y through x , then x is local.

If only more than a one relation between x and y satisfies the obseravtion of y through x , then x is non-local.
Click to expand...

If you get the idea, then please address it by your style.
 
doronshadmi said:
Do you mean to this part?
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No. That's not even a part of the post I was quoting.

Here's my question again:

You make use of what I'll call compound operators, like "< XOR >". I simply want to know what you explicitly mean by this. Since you tend to be inconsistent in your notation, and you conflate concepts, it is only reasonable I insist on a little precision, here.

If I use this compound operator with a pair of operands, A and B, I get:

A < xor > B​

That is not a well-formed formula. It doesn't have meaning. However, the way I think you meant it, the way I'd interpret it coming from anyone with any sort of consistency in there notation would be:

(A < B) xor (A > B)​

Is that what you would mean by it?
 
jsfisher said:
No. That's not even a part of the post I was quoting.

Here's my question again:

You make use of what I'll call compound operators, like "< XOR >". I simply want to know what you explicitly mean by this. Since you tend to be inconsistent in your notation, and you conflate concepts, it is only reasonable I insist on a little precision, here.

If I use this compound operator with a pair of operands, A and B, I get:

A < xor > B​

That is not a well-formed formula. It doesn't have meaning. However, the way I think you meant it, the way I'd interpret it coming from anyone with any sort of consistency in there notation would be:

(A < B) xor (A > B)​

Is that what you would mean by it?
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Of course, you're spot-on. And, as you undoubtedly realize, you'll never get a satisfactory answer from doron as his mathematical notation is consistently wrong out of deliberate ignorance (deliberate because he refuses to enroll in college courses) and his English his deliberately obtuse (deliberate because he's been called often enough on it).

The concept you describe is called lifting - at least, it was when I used to dabble in SquiggolWP. You lift a function/operator which operates on simple things to a higher-order function - one which operates on functions - which applies the original function to the results of the two function arguments. In a functional programming definition, lifting would be the function:
Code: 
lift :: (c -> d -> e) -> (a -> b -> c) -> (a -> b -> d) -> (a -> b -> e)
lift f g h x y = f (g x y) (h x y)
Of course, in Squiggol we'd write lifting with some squiggly symbol over the operator being lifted. Doron, wallowing in his ignorance, probably doesn't even see the difference between the lifted and the not-lifted operator. Not that he'd understand my or your comments in the first place.
 
doronshadmi said:
Yes.
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Great! Progress. Here's Definition 4, revision 3 again:

Definition 4: If object x = xor ≠ (where ≠ is < xor >) w.r.t object y, then object x is called Local.​

Let's parse the IF-part:

object x = xor ≠ w.r.t object y ==>
(object x = w.r.t. object y) xor (object x ≠ w.r.t object y)​

The not-equal part has it's own set of questions. I'd rather not ask them now and create more distractions, so I'll focus on just the equal part.

(object x = w.r.t. object y) ==>
Object x is the same as with respect to object y.​

Doron, do you see how your use of "w.r.t." makes nonsense out of that sentence? Either object x is the same as object y or it is not. So, now the next question that needs an answer: What do you actually mean by "object x = w.r.t. object y"?
 
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