On the Infinity of Prime Numbers

1 points by Kabbalist ↗ HN
Let A be the set of prime numbers. Let B_1 and B_2 be subsets of A. Let C be an element of A Let F(X) signify that X is finite. Let I(X) signify that X is infinite. 1. ∃B_1 (F(B_1 )) 2. ∃B_2∃C(C∉B_2 ) 3. ∃B_1 (F(B_1 ))→∃B_2∃C(C∉B_2 ) Conclusion: ∀B_1∃B_2∃C(F(B_1 )→C∉B_2 ) Premise 1 is true because from the set of prime numbers I can create a sequence of primes that is finite. Premise 2 is true because from the set of prime numbers I can create a sequence of prime numbers which C is not a member of. Premise 3 is true because of Premises 1 and 2 The Conclusion is true because it is Premise 3 in prenex normal form. Thus, if I let B_1=B_2 then the conclusion should read: For any finite sequence of prime numbers, there exist a prime number that is not a member of it. The conclusion is also true if I let B_1=B_2=A. For since the consequent is false, the antecedent must be false too by Modus Tollens.

1 comment

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Where does the proof fail if you use

A = {the set of integers between 0 and 9} = {0,1,2,3,4,5,6,7,8,9}

instead of using

A = {the set of prime numbers}

?