Deeper than primes

Status
Not open for further replies.
Once again Doron the commutative property of an operator or connective is not dependent on the variables or that some Truth\False value is indeterminate.
Let's see how you support you claim.

Please provide the result of A AND B, such that A's value is indeterminate because it is in Truth/False superposition, or B's value is indeterminate because it is in Truth/False superposition.
 
It also can’t be obtained by just spouting self-contradictory nonsensical gibberish for 20 years.
A typical reply of context-dependent-only reasoner professional in the box waiter.
Doron, we’ll be waiting for you whenever you decide you want to catch up.
You are already a professional in the box waiter, and the result is: you can't catch up OM.
 
Last edited:
Let's see how you support you claim.

Please provide the result of A AND B, such that A's value is indeterminate because it is in Truth/False superposition, or B's value is indeterminate because it is in Truth/False superposition.

Perhaps if you actually read and understood what The Man wrote, you wouldn't have posted something so irrelevent. For your convenient reference, here's The Man's statement again:

Once again Doron the commutative property of an operator or connective is not dependent on the variables or that some Truth\False value is indeterminate.
 
http://www.internationalskeptics.com/forums/showpost.php?p=7292565&postcount=15714 is tangible result that can't be comprehended by using context-dependent-only reasoning.

Assumptions aren't results.

From the Peano axioms comes all of Arithmetic with its myriad of results and applications, including the rather practical ability to balance one's check book. Number Theory has sired some incredible developments in cryptography. Basic principles in Topology led to the proof of the Four-Color Theorem. Even something as mundane as the Fibonacci Sequence finds application in something equally mundate, polyphase sorting.

Got anything like that? Or are you still stuck on claiming everyone and everything else is wrong while making no progress of your own?
 
Let's see how you support you claim.

Please provide the result of A AND B, such that A's value is indeterminate because it is in Truth/False superposition, or B's value is indeterminate because it is in Truth/False superposition.


A AND B = C

Once again you seem to be deliberately ignoring that a result can also be a, well, variable.


Also as noted by jsfisher the actual claim was that A AND B = C = B AND A, which is also supported by, well, mathematics.
 
A typical reply of context-dependent-only reasoner professional in the box waiter.

Well since spouting your self-contradictory nonsensical gibberish for 20 years hasn't gotten you your “paradigm-shift” or even a rudimentary understanding of mathematics. So the validity of the assertion is self-evident.

You are already a professional in the box waiter, and the result is: you can't catch up OM.

Once again hardly compelling or meaningful as your “OM” still has no results and, evidently from your latest posts, that’s just the way you’ve contrived it

Oh and stop simply trying to posit aspects of your own failed reasoning onto others
 
Perhaps if you actually read and understood what The Man wrote, you wouldn't have posted something so irrelevent. For your convenient reference, here's The Man's statement again:
In http://www.internationalskeptics.com/forums/showpost.php?p=7271242&postcount=15638 I used AND connective in terms of simultaneous existence of the considered inputs, which has no impact on the results.

Here it is again with again:

doronshadmi said:
For The Man and other posters:

A = (that has no predecessor) AND (that has no successor)

B = (that has no successor) AND (that has no predecessor)

((A AND B) is the same as (B AND A)) is unordered.

((A AND B) is different than (B AND A)) is ordered.
EDIT:

In the first case we determine that the inputs are in superposition, such that we can't determine the exact result exactly because of the symmetry of superposition between A;B inputs.

In the second case we determine that the inputs are not in superposition, such that we can determine the exact result exactly because of the non-symmetry of A;B strict input values.

The commutativity of AND connective has no impact on the indeterminate or determinate result of the considered cases.

Unfortunately, The Man ignored the deference between the results that are derived from the properties of the inputs, and focused on the commutativity of AND connective, that has no impact on the different results.
 
Last edited:
A AND B = C
C has strict result only if (A AND B) OR (B AND A) are strict inputs.

Otherwise C does not have strict result, and the commutativity of AND connective has no impact on C different results.
 
Last edited:
In http://www.internationalskeptics.com/forums/showpost.php?p=7271242&postcount=15638 I used AND connective in terms of simultaneous existence of the considered inputs, which has no impact on the results.

Here it is again with again:


EDIT:

In the first case we determine that the inputs are in superposition, such that
we can't determine the exact result exactly because of the symmetry of superposition between A;B inputs.

In the second case we determine that the inputs are not in superposition, such that we can determine the exact result exactly because of the non-symmetry of A;B strict input values.

The commutativity of AND connective has no impact on the indeterminate or determinate result of the considered cases.

Unfortunately, The Man ignored the deference between the results that are derived from the properties of the inputs, and focused on the commutativity of AND connective, that has no impact on the different results.


Doron once again…



“(that has no predecessor) AND (that has no successor)”

Is the same as…

“(that has no successor) AND (that has no predecessor)

Because “AND” is commutative.

So all you have is “A” = “B”.

“((A AND B) is the same as (B AND A))” because “AND” is commutative which is again why you ended up setting “A” = “B” before.

“((A AND B) is different than (B AND A)) is ordered.” Is again just nonsense as “((A AND B) is the same as (B AND A))” because “AND” is commutative” and you set “A” = “B” before. Also “(A AND B)” is ordered as well as "(B AND A)", that they are ordered differently again can not change the results because “AND” is commutative. Again the results are not different and can not be different because “AND” is commutative. Once again since you keep missing it the “impact” on your “different results” is that they can not be different because “AND” is commutative.
 
Doron once again…



“(that has no predecessor) AND (that has no successor)”

Is the same as…

“(that has no successor) AND (that has no predecessor)

Because “AND” is commutative.

The Man once again ...

A = “(that has no predecessor) AND (that has no successor)”

B = “(that has no successor) AND (that has no predecessor)"

The statement “((A AND B) is the same as (B AND A)) is unordered” means that A;B inputs are in superposition, and as a result C is indeterminate, and the commutativity of AND connective has no impact of C indeterminate result.

The statement “((A AND B) is different than (B AND A)) is ordered” means that A;B inputs are strict and different, and as a result C is strict, and the commutativity of AND connective has no impact of C strict result.
 
Last edited:
C has strict result only if (A AND B) OR (B AND A) are strict inputs.

Otherwise C does not have strict result, and the commutativity of AND connective has no impact on C different results.

Nope, “C” is strictly equal to “B AND A” which is strictly equal to “A AND B” as well as being strictly not equal to “A XOR B” and strictly not equal to “B XOR A”. That you simply prefer no results (strict or otherwise) is no ones problem but yours. Once again the impact on the results from the fact that AND is commutative is that “A AND B” can not give "different results" from “B AND A”.

To try to put it more succinctly for you, what C= B AND A does is it defines the variable “C” as the logical conjunction of the variable “B” with the variable “A”. What those variables (“A”, “B’) represent is irrelevant to that definition which strictly eliminates certain other definitions for “C” (like C= A XOR B) from consideration.
 
Nope, “C” is strictly equal to “B AND A” which is strictly equal to “A AND B” as well as being strictly not equal to “A XOR B” and strictly not equal to “B XOR A”.
The Man, once again, please provide C strict result, such that A or B inputs are indeterminate.
 
The Man once again ...

A = “(that has no predecessor) AND (that has no successor)”

B = “(that has no successor) AND (that has no predecessor)"

The statement “((A AND B) is the same as (B AND A)) is unordered” means that A;B inputs are in superposition, and as a result C is indeterminate, and the commutativity of AND connective has no impact of C indeterminate result.

The statement “((A AND B) is different than (B AND A)) is ordered” means that A;B inputs are strict and different, and as a result C is strict, and the commutativity of AND connective has no impact of C strict result.

Again "The statement “((A AND B) is different than (B AND A)) is” just wrong. That “AND” is commutative means that “A” = “B” as well as ““(A AND B) is the same as (B AND A)” and your “unordered”, “ordered”, “indeterminate” and “strict” nonsense can not change that. That is the impact of the commutative property of “AND” that you simply want to deliberately ignore. It makes all of your “unordered”, “ordered”, “indeterminate” and “strict” nonsense definitively and strictly irrelevant.
 
Again "The statement “((A AND B) is different than (B AND A)) is” just wrong.

I understand your problem.

You take what I wrote as a logical statements, but my statements are not logical statements, but they are simply statements which show that A;B inputs are strict and different ( “((A AND B) is different than (B AND A)) is ordered” ), or in superposition ( “((A AND B) is the same as (B AND A)) is unordered” ).

EDIT:

You have missed it because you ignored the "is ordered" or "is unordered” parts of these statements.
 
Last edited:
I understand your problem.

You take what I wrote as a logical statements, but my statements are not logical statements, but they are simply statements which show that A;B inputs are strict and different ( “((A AND B) is different than (B AND A)) is ordered” ), or in superposition ( “((A AND B) is the same as (B AND A)) is unordered” ).

EDIT:

You have missed it because you ignored the "is ordered" or "is unordered” parts of these statements.

Nope, I took them (without ‘missing’ or ‘ignoring’ any “parts”) as clearly illogical statements as that problem remains simply yours. Once again you evidently simply do not understand ordering.
 
Nope, I took them (without ‘missing’ or ‘ignoring’ any “parts”) as clearly illogical statements as that problem remains simply yours. Once again you evidently simply do not understand ordering.
EDIT:

The Man I am not a participator of your "running after my own tail" closed game.

A = “(that has no predecessor) AND (that has no successor)”

B = “(that has no successor) AND (that has no predecessor)"

Every reasonable person, but you, that sees a statement like:

“((A AND B) is different than (B AND A)) is ordered” immediately understands that this statement is about the ability to strictly distinguish between A and B inputs under different orders, where AND connective is used here as simultaneous existence of A;B strict inputs, which provides a strict result. The commutativity of AND connective has no impact on the fact that the result is strict, and the result is strict because A;B inputs are strict.

Also every reasonable person, but you, that sees a statement like:

“((A AND B) is the same as (B AND A)) is unordered” immediately understands that this statement is about the inability to strictly distinguish between A and B inputs under different orders (because A;B values are in superposition, which prevent their strict ids), where AND connective is used here as simultaneous existence of A;B non-strict inputs, which provides a non-strict result. The commutativity of AND connective has no impact on the fact that the result is non-strict, and the result is non-strict because A;B inputs are non-strict.

Let's take, for example, the 2-Uncertainty x 2-Redundancy Distinction-Tree:

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)
(1,1) = (A,A),(B,B),(A,B)
(1,0)=  (A),(B)
(0,0)=  ()

Any appearance of that tree is called Distinction State (DS), where any DS is under a structure called Frame (F), for example: (AB ,B ) is a DS that is under (2,1) F. The order in each DS or F has no significance (similar to {a,b}={b,a}) but any DS is the basis of any possible order (similar to the concept of Set as being the basis of permutations).

--------------------------

Some example:

A = True
B = False

AB AND AB has non-strict result because the input is non-strict.

AB AND A or A AND AB has non-strict result because the input is non-strict (the commutativity of AND connective has no impact on the non-strict result).

AB AND B or B AND AB has non-strict result because the input is non-strict (the commutativity of AND connective has no impact on the non-strict result).

A AND B or B AND A is strictly False (the commutativity of AND connective has no impact on the strict result).

--------------------------

The current use of AND connective inputs is limited only to DS (A,A),(B,B),(A,B) under F (1,1), and by this limitation we get only strict results, where the commutativity of AND connective has no impact on the strict property of the results.

Organic Mathematics is not limited only to strict inputs, and without this limitation we get also non-strict results, where also in this case the commutativity of AND connective has no impact on the non-strict property of the results.

In other words The Man, your "running after my own tail" closed game, does not even scratch my arguments.

Once again you evidently simply do not understand ordering.
Once again you evidently simply do not understand symmetrical and non-strict values (which are naturally unordered) exactly because your reasoning is limited only to DS (A,A),(B,B),(A,B) under F (1,1). You simply can't get anything beyond it.

-------------------------

Furthermore, you have asked about OM results, so one of OM results is the exposure of the limitations of your reasoning to DS (A,A),(B,B),(A,B) under F (1,1). You simply can't get anything beyond it and as a result you don't understand that the commutativity of AND connective has no impact on the result, whether it is strict or non-strict.
 
Last edited:
The Man I am not a participator of your "running after my own tail" closed game.

A = “(that has no predecessor) AND (that has no successor)”

B = “(that has no successor) AND (that has no predecessor)"

Every reasonable person, but you, that sees a statement like:

“((A AND B) is different than (B AND A)) is ordered” immediately understands that this statement is about the ability to strictly distinguish between A and B inputs under different orders, where AND connective is used here as simultaneous existence of A;B strict inputs, which provides a strict result. The commutativity of AND connective has no impact on the fact that the result is strict because A;B inputs are strict.

Also every reasonable person, but you, that sees a statement like:

“((A AND B) is the same as (B AND A)) is unordered” immediately understands that this statement is about the inability to strictly distinguish between A and B inputs under different orders (because A;B values are in superposition, which prevent their strict ids), where AND connective is used here as simultaneous existence of A;B non-strict inputs, which provides a non-strict result. The commutativity of AND connective has no impact on the fact that the result is non-strict because A;B inputs are non-strict.
No.
 


The most convincing way to prove your gibberish has any value whatsoever would be to demonstrate a single, real result. Just one.

You can't, because there aren't any.

No one else is to blame; it is not due to any failed understanding on our part. It is all your doing. After 20+ years of contradiction and muddle, you still have nothing to show.
 
The most convincing way to prove your gibberish has any value whatsoever would be to demonstrate a single, real result. Just one.

You can't, because there aren't any.

No one else is to blame; it is not due to any failed understanding on our part. It is all your doing. After 20+ years of contradiction and muddle, you still have nothing to show.

The results are there: http://www.internationalskeptics.com/forums/showpost.php?p=7294838&postcount=15744 in front of your blind mind, jsfisher.
 
A real result, Doron. Not just some nonsense you assume must be true, then declare it a result.
They are real results, unlike your nonsense reasoning that can't comprehend them.

--------------------------

Some example:

A = True
B = False

AB AND AB has non-strict result because the input is non-strict.

AB AND A or A AND AB has non-strict result because the input is non-strict (the commutativity of AND connective has no impact on the non-strict result).

AB AND B or B AND AB has non-strict result because the input is non-strict (the commutativity of AND connective has no impact on the non-strict result).

A AND B or B AND A is strictly False (the commutativity of AND connective has no impact on the strict result).

--------------------------

You and The Man simply can't get that the commutativity of AND connective has no impact on these results.

--------------------------




--------------------------

Furthermore, the suggested framework has no entropy exactly because no collection of lower dimensional spaces has the magnitude of higher dimensional space. For example |R| < |1-dimensional space|, where in this case R members are 0-dimensional spaces (known as points).

The following diagram illustrates |R| < |1-dimensional space| as follows:

5736095487_99cb0b393a_b.jpg


As can be seen, the cardinality of the intersection points along the line segments with different lengths is the same, whether it is finite or infinite cardinality. So only the sets of points with the same cardinality can't provide the solution for the existence of 1-dim elements with different lengths.

This fundamental fact is exactly the incompleteness of R w.r.t 1-dimensional space, which preserves the openness of R and enables its endless further development.

The claim of Traditional Mathematics about 0-dimensional R members that completely cover a 1-dimensional space, is actually "death by entropy" of the considered framework.

--------------------------

Actually You and The Man simply can't provide a logical solution of the fact that line segments have different lengths even if the cardinality of the the set of points (where a point is the smallest existing element) along them is the same, where by your assertion this set of points (notated as R) completely covers these 1-dimensional elements.
 
Last edited:
They are real results, unlike your nonsense reasoning that can't comprehend them.


You are being intentionally obtuse, it would seem. You are still focused on your own assumptions and special meanings and claiming they're conclusions. We are looking for something that goes beyond your self-inconsistent assumptions.

That S(0) is 1 is not a "result" that can be credited to Peano. It is just a definition and part of Peano's foundation. Arithmetic, on the other hand, is derivable from the Peano axioms and rightly is a signficant result.

You, Doron, on the other hand, provide nothing constructive. You simply deny what you don't understand, and you imagine things that aren't real. None of it has foundation and none of it conforms to any valid line of reasoning.

You have been wallowing in nonsense for over 20 years, Doron. Time to move on.
 
You are being intentionally obtuse, it would seem. You are still focused on your own assumptions and special meanings and claiming they're conclusions. We are looking for something that goes beyond your self-inconsistent assumptions.

That S(0) is 1 is not a "result" that can be credited to Peano. It is just a definition and part of Peano's foundation. Arithmetic, on the other hand, is derivable from the Peano axioms and rightly is a signficant result.

You, Doron, on the other hand, provide nothing constructive. You simply deny what you don't understand, and you imagine things that aren't real. None of it has foundation and none of it conforms to any valid line of reasoning.

You have been wallowing in nonsense for over 20 years, Doron. Time to move on.

Evasion of http://www.internationalskeptics.com/forums/showpost.php?p=7296289&postcount=15750 was noted.

Because of this evasion you can't get the result of the following expression: 1 - 0.999...[base 10] = 0.000...1[base 10] (where "...1" of 0.000...1 expression is the non-locality of the real-line that is simultaneously < AND = 1, which is a property that no point (locality) along the real-line has).

Please try again.

Start, for example, by providing a logical solution of the fact that line segments have different lengths even if the cardinality of the the set of points (where a point is the smallest existing element) along them is the same, where by your assertion this set of points (notated as R) completely covers these 1-dimensional elements.
 
Last edited:
No evasion; there was no result.
The result in http://www.internationalskeptics.com/forums/showpost.php?p=7296289&postcount=15750, was, for example, the conclusion that the commutativity of AND connective has no impact on the property of the output (which can be strict or non-strict).

Now evasion of http://www.internationalskeptics.com/forums/showpost.php?p=7297036&postcount=15755 was noted, by ignoring a result that is based on expressions with non-local numbers.

In order to get the validity of non-local numbers please this time start, for example, by providing a logical solution of the fact that line segments have different lengths even if the cardinality of the the set of points (where a point is the smallest existing element) along them is the same, where by your assertion this set of points (taken as R members) completely covers these 1-dimensional elements.

Furthermore by the following diagram, there are no proper subsets for R members exactly because they are results of common intersections, marked by the black lines:

5736095487_99cb0b393a_b.jpg
 
Last edited:
The result in http://www.internationalskeptics.com/forums/showpost.php?p=7296289&postcount=15750, was, for example, the conclusion that the commutativity of AND connective has no impact on the property of the output (which can be strict or non-strict).

That is not a result. That is just you pretending AND means something different than it really does.

Even if you were to invent a brand-new boolean operator unique to OM (which would be a perfectly ok thing to do, but it would require something so far proven to be beyond your capability, namely providing an actually definition for something), it would still not be a result since it would not follow from but instead would have been given to OM.
 
Start, for example, by providing a logical solution of the fact that line segments have different lengths even if the cardinality of the the set of points (where a point is the smallest existing element) along them is the same, where by your assertion this set of points (notated as R) completely covers these 1-dimensional elements.
Aleph0 and Aleph1 are not the same as A and B defined by particular quantities. Imagine points as sub-atomic particles. Some astrophysicists see the origin of Big Bang in something called "Big Crunch," which is a scenario where all the matter in the pre-existing universe collapses under heavy gravitational influences into one point called "singularity." That means all the particles in the pre-existing universe were squeezed into a mass the size of our sun, for example. Then the contraction continued and the size of the ancestor universe was the one of the earth. That went on and the size shrank to such dimensions that the scientists must be given various awards to elevate their social image, otherwise folks would call them nuts: How can you squeeze all the matter in the universe into something smaller than a tennis ball?

So you can see a similarity between the same number of particles and the dimension of the objects that contain them, be it a sphere or your line segment:

the sun: __________________________________________________

the earth: ___________

the tennis ball: __

Since points are zero-dimensional, the "squeezing" goes much better and it's much safer as well, coz a line segment whose length approaches zero never explodes to become a very long line segment U_________E
 
Last edited:
Status
Not open for further replies.

Back
Top Bottom