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[PrimeNumbers] Logic and Sets.

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  • Juan Pedro Hecht
    Hello Paul, Thanks, you are right with the comment about the Boolean operator (E) belongs to , but it is not hard to obtain it from the primitive Boolean
    Message 1 of 5 , Nov 1, 2001
      Hello Paul,

      Thanks, you are right with the comment about the Boolean operator (E)
      'belongs to', but it is not hard to obtain it from the primitive Boolean
      operators:

      Let a, b, c, 0, 1 be objects 'belonging to' the K Boolean Set. All the
      properties for 'belongs to' are:

      (a E a)
      if (a E b) and (b E a) then a=b
      if (a E b) and (b E c) then (a E c)
      (a E 1) and (0 E a)
      (a E (a+b)) and ((a.b) E a)
      if (a E b) then ((not b) E (not a))

      but Boolean Algebra don't need that operator, and of course Morgan's Law's
      are easy to derive from the original Huntington set (1904).

      Thanks again for your comments!

      Juan
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