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Organic
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PLEASE READ THIS POST UNTIL ITS LAST WORD, BEFORE YOU REPLY.
THANK YOU.
Let us check these lists.
P(2) = {{},{0},{1},{0,1}} = 2^2 = 4
and also can be represented as:
00
01
10
11
P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8
and also can be represented as:
000
001
010
011
100
101
110
111
Let us call any 0,1 list, combinations list.
When we use Cantor's Diagonalization function on the combinations list of 2^power_value and get some output result, we find that this result is already somewhere in the list.
The formula that gives us the number of combinations , which are out of the range of Cantor's Diagonalization function, is:
2^n - n
Combinations are first of all structural changes, based on at least two parameters:
a)The number of different notations.
b)The number of places that have been given to permute these notations.
We get our list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list (by changing power_value).
When we have infinitely many places to combine our two different notations, then the number of combinations, which are out of the range of Cantor's Diagonalization function is:
2^aleph0 - aleph0 = E where by E we mean that there are E possible combinations, which are out of the range of Cantor's Diagonalization function, where one of these combinations, is Cantor's Diagonalization function result.
Therefore Cantor's Diagonalization function result is not a new combination.
Because the aleph0 long Cantor's Diagonalization function result cannot cover the 2^aleph0 list, it means that 2^aleph0 > aleph0, but we can define a map between any unique combination and some natural number, therefore
2^aleph0 = aleph0.
Therefore (2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects.
Organic
THANK YOU.
Let us check these lists.
P(2) = {{},{0},{1},{0,1}} = 2^2 = 4
and also can be represented as:
00
01
10
11
P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8
and also can be represented as:
000
001
010
011
100
101
110
111
Let us call any 0,1 list, combinations list.
When we use Cantor's Diagonalization function on the combinations list of 2^power_value and get some output result, we find that this result is already somewhere in the list.
The formula that gives us the number of combinations , which are out of the range of Cantor's Diagonalization function, is:
2^n - n
Combinations are first of all structural changes, based on at least two parameters:
a)The number of different notations.
b)The number of places that have been given to permute these notations.
We get our list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list (by changing power_value).
When we have infinitely many places to combine our two different notations, then the number of combinations, which are out of the range of Cantor's Diagonalization function is:
2^aleph0 - aleph0 = E where by E we mean that there are E possible combinations, which are out of the range of Cantor's Diagonalization function, where one of these combinations, is Cantor's Diagonalization function result.
Therefore Cantor's Diagonalization function result is not a new combination.
Because the aleph0 long Cantor's Diagonalization function result cannot cover the 2^aleph0 list, it means that 2^aleph0 > aleph0, but we can define a map between any unique combination and some natural number, therefore
2^aleph0 = aleph0.
Therefore (2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects.
Organic
Last edited: