Cont: Deeper than primes - Continuation 2

[jsfisher]

Why am I restricted to a diagonal argument?

Because you are the one who claimed (in http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965) about 2-valued unbounded logical tree that "all dyadic boolean functions can be enumerated"

Multiple fails on your part, Doronshadmi.
(1) Nothing in what you wrote would require diagonal proof arguments.
(2) I said nothing about "unbounded logical trees".
(3) Dyadic boolean functions can be completely enumerated.

(And lest there be ambiguity, I am using 'enumerate' in its common definition of to name one after another in a finite or countable infinite list.)

 

[doronshadmi]

Multiple fails on your part, Doronshadmi.
(1) Nothing in what you wrote would require diagonal proof arguments.
(2) I said nothing about "unbounded logical trees".
(3) Dyadic boolean functions can be completely enumerated.

(And lest there be ambiguity, I am using 'enumerate' in its common definition of to name one after another in a finite or countable infinite list.)

(1) Wrong jsfisher, it requires diagonal proof argument because (2) In http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965 you explicitly replied about unbounded logical trees (3) by using an argument about them, which is based on the claim that Dyadic boolean functions can be completely enumerated (or in other words, they are at most countably infinite).

(And lest there be ambiguity, I am using 'enumerate' in its common definition of to name one after another in a finite or countable infinite list.)

It is trivially clear that the 2-vauled unbounded logical tree is not finite, so all is left for you is to use the diagonal argument on the the 2-vauled unbounded logical tree in order to prove that it is at most countably infinite, by simply demonstrate an unbounded path that is not included in it.

It is logically guaranteed that by any attempt to do that you realize that both unbounded complement paths are already included in that tree.

 

[jsfisher]

(1) Wrong jsfisher, it requires diagonal proof argument because (2) In http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965 you explicitly replied about unbounded logical trees (3) by using an argument about them, which is based on the claim that Dyadic boolean functions can be completely enumerated (or in other words, they are at most countably infinite).

You really need to understand the terms you argument so vehemently about. The dyadic boolean functions can be completely enumerated. There are a total of sixteen of them.

 

[doronshadmi]

You really need to understand the terms you argument so vehemently about. The dyadic boolean functions can be completely enumerated. There are a total of sixteen of them.

http://www.internationalskeptics.com/forums/showpost.php?p=11374991&postcount=1962

In other words, you have no argument about the issue at hand.

Moreover, http://www.internationalskeptics.com/forums/showpost.php?p=11393612&postcount=2036 which is your reply to http://www.internationalskeptics.com/forums/showpost.php?p=11393247&postcount=2035, clearly demonstrates the weakness of your currently used standard reasoning.

 

[jsfisher]

...
In other words, you have no argument about the issue at hand.

You clearly didn't understand a particular mathematical term; you made yourself yet again appear the fool by arguing vehemently about something you didn't understand; and now you attempt to dismiss your failures by shifting topics.

No. You were wrong, doronshadmi. You are wrong now.

Let's stick to the current subject of this long and twisted departure from reason. Here are your recent failures again:

Multiple fails on your part, Doronshadmi.
(1) Nothing in what you wrote would require diagonal proof arguments.
(2) I said nothing about "unbounded logical trees".
(3) Dyadic boolean functions can be completely enumerated.

 

[doronshadmi]

You clearly didn't understand a particular mathematical term; you made yourself yet again appear the fool by arguing vehemently about something you didn't understand; and now you attempt to dismiss your failures by shifting topics.

No. You were wrong, doronshadmi. You are wrong now.

Let's stick to the current subject of this long and twisted departure from reason. Here are your recent failures again:

Multiple fails on your part, Doronshadmi.
(1) Nothing in what you wrote would require diagonal proof arguments.
(2) I said nothing about "unbounded logical trees".
(3) Dyadic boolean functions can be completely enumerated.

http://www.internationalskeptics.com/forums/showpost.php?p=11395283&postcount=2042 and http://www.internationalskeptics.com/forums/showpost.php?p=11396004&postcount=2044 clearly demonstrate that your last reply (quoted above) is no more than hands waving that has nothing to do with the issue at hand, which is exactly unbounded logical trees.

 

[doronshadmi]

So can the triadic tri-state functions.

(http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965)

jsfisher, try to google "triadic tri-state functions" in order to realize how you use Standard Mathematics (also called by you real mathematics).

 

[jsfisher]

http://www.internationalskeptics.com/forums/showpost.php?p=11395283&postcount=2042 and http://www.internationalskeptics.com/forums/showpost.php?p=11396004&postcount=2044 clearly demonstrate that your last reply (quoted above) is no more than hands waving that has nothing to do with the issue at hand, which is exactly unbounded logical trees.

Those two posts of yours you linked to do nothing to address the specifics of your failures.

(1) Nothing in what you wrote would require diagonal proof arguments.
You asserted that it was required; you have provided no justification for the requirement. And that doesn't even get to the additional failure that you were asking for a proof of something for which I have no responsibility.

(2) I said nothing about "unbounded logical trees".
Had I said something about "unbounded logical trees", I would likely have made some mention of them. I did not.

(3) Dyadic boolean functions can be completely enumerated.
Apparently, you finally tired to look up those words and may have figured out it means a true/false-valued function of two true-false arguments.

I would have guessed we could assume your failure regarding the countability of dyadic boolean functions has been put to rest, except now you are tripped up by triadic tri-state functions.

This does not bode well for you.

 

[doronshadmi]

Those two posts of yours you linked to do nothing to address the specifics of your failures.

(1) Nothing in what you wrote would require diagonal proof arguments.
You asserted that it was required; you have provided no justification for the requirement. And that doesn't even get to the additional failure that you were asking for a proof of something for which I have no responsibility.

(2) I said nothing about "unbounded logical trees".
Had I said something about "unbounded logical trees", I would likely have made some mention of them. I did not.

(3) Dyadic boolean functions can be completely enumerated.
Apparently, you finally tired to look up those words and may have figured out it means a true/false-valued function of two true-false arguments.

I would have guessed we could assume your failure regarding the countability of dyadic boolean functions has been put to rest, except now you are tripped up by triadic tri-state functions.

This does not bode well for you.

In other words, you actually admit that your responses to http://www.internationalskeptics.com/forums/showpost.php?p=11380724&postcount=1964 form http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965 forward have nothing to do with the issue at hand, which is exactly unbounded logical trees without free variables.

When you actually dealing with unbounded logical trees without free variables, please let me know.

Here is some concrete example of the logical failure of your "real mathematics":

By using your "real mathematics" you logically can't comprehend that, for example, in the particular case of 2-valued logic unbounded tree 0.111... and 1.000... are logically complements of each other and therefore are logically not the same thing (it is logically proven that 0.111...₂ ≠ 1.000...₂).

No, that would be a bare assertion, the same assertion you have made multiple times, always without proof.

It is directly logically proven without any need of free variables, but since your "real mathematics" can't work without using free variables in the considered case, your reasoning is simply blind to any direct logical proof.

Indeed by being consistently blind to the issue at hand, one concludes that all there is is "a bare assertion" ... "always without proof".

 

[jsfisher]

In other words, you actually admit that your responses to http://www.internationalskeptics.com/forums/showpost.php?p=11380724&postcount=1964 form http://www.internationalskeptics.com/forums/showpost.php?p=11380745&postcount=1965 forward have nothing to do with the issue at hand, which is exactly unbounded logical trees without free variables.

Nope. I was addressing the three posts by you that preceded my post. The comments were entirely responsive to those posts.

However, you, as you have done so many times in the past, assumed you understood what the posts meant and responded accordingly. It is obvious to all, including even you at this point, you hadn't a clue what they meant and responded accordingly.

It is also clear to all (except you) from your fascination with the phrase 'free variable' that it is another thing you don't really understand.

 

[jsfisher]

Here is some concrete example of the logical failure of your "real mathematics":

By using your "real mathematics" you logically can't comprehend that, for example, in the particular case of 2-valued logic unbounded tree 0.111... and 1.000... are logically complements of each other and therefore are logically not the same thing (it is logically proven that 0.111...₂ ≠ 1.000...₂).

Real Mathematics recognizes your "2-valued logic unbounded tree" as simply a binary tree of infinite height (or depth, depending on your point of view). You have labeled the branches with 0's and 1's in a conventional way.

As such, a sequence of 0's and 1's can be interpreted as a path down the tree from the root. For some reason, you have chosen to insert a radix point in that sequence, but it is of no consequence in defining the path.

Sequences of 0's and 1's for paths correspond to numeric values expressed in binary. Well, sort of, but not completely. Take for example 0.01 and 0.01000000. They are clearly different paths down the tree even though they represent the same numeric value in base-two.

Only a fool would use the tree to conclude that 0.01₂ and 0.01000000₂ were different numbers.

Similarly, 1.00000... and 0.11111... are different paths in the tree, but the tree provides no basis to conclude 1.00000...₂ and 0.11111...₂ be different numbers. The tree encodes paths, not numeric values.

Doronshadmi, would you like to argue that 0.01₂ and 0.01000000₂ are different values?

 

[doronshadmi]

Nope. I was addressing the three posts by you that preceded my post. The comments were entirely responsive to those posts.

However, you, as you have done so many times in the past, assumed you understood what the posts meant and responded accordingly. It is obvious to all, including even you at this point, you hadn't a clue what they meant and responded accordingly.

It is also clear to all (except you) from your fascination with the phrase 'free variable' that it is another thing you don't really understand.

Indeed I wrongly gave you a credit that you do not diverse, by asserting that you use enumerated in terms of infinite paths.

 

[doronshadmi]

Take for example 0.01 and 0.01000000. They are clearly different paths down the tree even though they represent the same numeric value in base-two.

Wrong, jsfisher.

0.01₂ is some bounded case along the 2-valued unbounded logical tree, where 0.01000...₂ is one of the infinitely many possible unbounded paths that emerge form 0.01₂

Standard Math simply uses a default rule that determines x000... as the unbounded path of any bounded case along any unbounded logical tree (including the 2-valued unbounded logical tree).

Standard Math default rule does not work in case of 0.111... or in other words (in case of the 2-valued unbounded logical tree)
0.111...₂ ≠ 1.000...₂ , 0.0111...₂ ≠ 0.1000...₂ etc.

The following 2-valued unbounded logical tree directly logically proves that Standard Math default rule does not work along the unbounded paths (without loss of generality):

Unity
|\
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
|         \
|          \
|           \
|            \
|             \
|              \
0---------------1---------------Integers
|\              |\
| \             | \
|  \            |  \
|   \           |   \
|    \          |    \
|     \         |     \         Fractions
|      \        |      \
0       1       0       1
|\      |\      |\      |\
| \     | \     | \     | \
|  \    |  \    |  \    |  \
0   1   0   1   0   1   0   1
|\  |\  |\  |\  |\  |\  |\  |\
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

              ...

 

[doronshadmi]

Nope. I was addressing the three posts by you that preceded my post. The comments were entirely responsive to those posts.

However, you, as you have done so many times in the past, assumed you understood what the posts meant and responded accordingly. It is obvious to all, including even you at this point, you hadn't a clue what they meant and responded accordingly.

It is also clear to all (except you) from your fascination with the phrase 'free variable' that it is another thing you don't really understand.

Indeed I wrongly gave you a credit that you do not diverse, by asserting that you use enumerated in terms of infinite paths.

It has to be:

Indeed I wrongly gave you a credit that you do not deserve, by asserting that you use enumerated in terms of infinite paths.

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

Any way, since you are fond of finitely many logical functions, then please write down all the logical functions of the 3-valued bounded logical tree (there are only 7,625,597,484,987 functions as clearly seen in http://www.internationalskeptics.com/forums/showpost.php?p=11375190&postcount=1963, so it is an easy task for you).

 

[doronshadmi]

For some reason, you have chosen to insert a radix point in that sequence, but it is of no consequence in defining the path.

Wrong jsfishr.

Any finite amount of logical states (bits, trits, etc.) "above" the radix point is some integer, which is an integral (bounded) part of any given unbounded logical tree.

Take, for example, the exact logical syntax of a radix point along the 2-valued unbounded logical tree:

Unity
|\
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
|         \
|          \
|           \
|            \
|             \
|              \
0---------------1---------------Integers
|\              |\
| \             | \
|  \            |  \
|   \           |   \
|    \          |    \
|     \         |     \         Fractions
|      \        |      \
0       1       0       1
|\      |\      |\      |\
| \     | \     | \     | \
|  \    |  \    |  \    |  \
0   1   0   1   0   1   0   1
|\  |\  |\  |\  |\  |\  |\  |\
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

              ...

Unity
|\
| \
|  \
|   \
|    \
|     \
|      \
|       \
|        \
|         \
|          \
|           \
|            \
|             \
|              \
0               1               
|\              |\
| \             | \
|  \            |  \
|   \           |   \
|    \          |    \
|     \         |     \         
|      \        |      \
0-------1-------0-------1---------Integers
|\      |\      |\      |\
| \     | \     | \     | \       
|  \    |  \    |  \    |  \      Fractions
0   1   0   1   0   1   0   1
|\  |\  |\  |\  |\  |\  |\  |\
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

              ...

etc.

 

[doronshadmi]

Only a fool would use the tree to conclude that 0.01₂ and 0.01000000₂ were different numbers.

Only a fool does not understand how numbers are logically directly defined along an unbounded logical tree (the needed details are already given in http://www.internationalskeptics.com/forums/showpost.php?p=11397847&postcount=2053 and http://www.internationalskeptics.com/forums/showpost.php?p=11398093&postcount=2055).

 

[jsfisher]

Indeed I wrongly gave you a credit that you do not diverse, by asserting that you use enumerated in terms of infinite paths.

You did no such thing. This is a bare-faced lie.

You didn't understand what I wrote because you didn't understand the math term I used. You ridiculed what I wrote under a bogus interpretation. As far as "giving credit", you did that other thing -- the opposite of giving credit.

You don't understand math. You make stuff up to pretend you do. As a result, you get just about everything wrong.

...and you blame everyone else for your failures.

 

[jsfisher]

Take for example 0.01 and 0.01000000. They are clearly different paths down the tree even though they represent the same numeric value in base-two.

Wrong, jsfisher.

Well, isn't that interesting.

I made two statements, (1) that 0.01 and 0.01000000 when interpreted as paths from the root of your binary tree were different and (2) that 0.01₂ and 0.01000000₂ as base-2 values represented the same number.

Which of those two were wrong?

 

[doronshadmi]

You did no such thing. This is a bare-faced lie.

jsfisher, you simply chose to avoid http://www.internationalskeptics.com/forums/showpost.php?p=11394670&postcount=2039, no less no more.

 

[doronshadmi]

Well, isn't that interesting.

I made two statements, (1) that 0.01 and 0.01000000 when interpreted as paths from the root of your binary tree were different and (2) that 0.01₂ and 0.01000000₂ as base-2 values represented the same number.

Which of those two were wrong?

Both.

The answers are given in the rest of http://www.internationalskeptics.com/forums/showpost.php?p=11397847&postcount=2053.

Edit:

(some typo correction: instead of "emerge form 0.01₂" it has to be "emerge from 0.01₂")