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Let us check these lists.

(2) = {{},{0},{1},{0,1}} = 2^2 = 4P

and also can be represented as:

00

01

10

11

(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8P

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 ofand get some output result, we find that this result is already somewhere in the list.2^power_value

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 theinduction, on theZF Axiom of infinityside of our combinations list (by changingleft).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:

where by2^aleph0 - aleph0 = Ewe mean that there areEpossible combinations, which are out of the range of Cantor's Diagonalization function, where one of these combinations, is Cantor's Diagonalization function result.E

Therefore Cantor's Diagonalization function result is not a new combination.

Because thelong Cantor's Diagonalization function result cannot cover thealeph0list, it means that2^aleph0, but we can define a map between any unique combination and some natural number, therefore2^aleph0 > aleph0

.2^aleph0 = aleph0

Therefore(2^aleph0 >= aleph0) ={}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects.

Organic

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# Cantor's Diagonalization function on the combinations list

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