First, define S as the set of all sets that are members of themselves. That is, self-referential. The S set is well defined, and easy to see that it exists.

I disagree: your definition of S is insufficient for me to deduce whether S contains S. It contains exactly the same hole as the definition of R, except that in the case of S it is consistent either way.

More precisely, I read your definition as:

  S := A \in A => A \in S

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.. and evaluating that for A = S gives the uninformative:
  S \in S => S \in S

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.. and analogously:
  R \notin R => R \in R

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I'd say therefore that these sets are not well-defined unless they additionally specify self-membership:

  S := (A = S | A \in A) => A \in S
  or S := (A != S & A \in A) => A \in S
  R := (A = R | A \notin A) => A \in R
  or R := (A != R & A \notin A) => A \in R

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.. and then everything is simultaneously hunky and dory.

Hugo

In reply to Re^4: BlooP and FlooP and GlooP: Turing Equivalence, Lazy Evaluation, and Perl6 by hv
in thread BlooP and FlooP and GlooP: Turing Equivalence, Lazy Evaluation, and Perl6 by jonadab

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