-3

Zhang Hong recently asked "is there a paradox lurking in Godel's 1931 incompleteness proof" (paraphrase)? This can be answered in two ways: First, by proving quite generally that there are no paradoxes in mathematical logic. Or, second, by actually producing a paradoxical result within mathematical logic.

The following is putatively an answer of the second kind, to Hong's question. I'm posting here to ask, "is the following a paradox, or just a mistake?"

  1. Let #Q# be the Godel number of the expression, Q.
  2. Let @r@ be the expression corresponding to the Godel number, r.
  3. Now let n = #@n@ --> P#, with P being arbitrary.
  4. Then P.

(Note that Q = @#Q#@ and r = #@r@#, for all expressions, Q, and all Godel numbers, r.)

The proof from 3 to 4 goes this way: Suppose @n@. Then by applying @..@ quotes to both sides of 3, it follows that @n@ --> P. So it follows that P (by modus ponens). Suppose ~@n@, contrariwise. Then @n@ --> P is false. So the antecedent is true and the conclusion is false. Thus it follows that @n@ (since this is the antecedent). And it follows from this that @n@ --> P. So we have it again (by modus ponens) that P. So, whether @n@ is true or false, it yeilds a proof that P, even though P could be anything.

Edit: It was suggested that I add a link to Zhang Hong's question. So here it is: Hong's question.

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  • Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Philosophy Meta, or in Philosophy Chat. Comments continuing discussion may be removed.
    – Philip Klöcking
    Commented Jun 12 at 15:03
  • I want to register my alarm at the way this question has been treated. Beside the objections of user58697 and Bumble, which are based on a use/mention conflation, there have been no other critical notes. And yet the question is downvoted to -3. More, the question has today been closed with the declaration that it is off-topic. But according to the definition displayed, off-topic questions "push a personal philosopy" -- my question does not do this. Further, my question is answerable.
    – James King
    Commented Jun 12 at 17:23
  • I'm sure nobody intends this, but I feel like I'm being censored in order to hide a secret. Lol.
    – James King
    Commented Jun 12 at 17:24
  • @JamesKing If you need help understanding why the question is off-topic, or believe it's on-topic and would like it to be reopened, you can ask on Meta.
    – F1Krazy
    Commented Jun 13 at 10:05

1 Answer 1

2

@n@ is an expression which Godel number is n. @n@ --> P is a different expression, so its Godel number is different. Therefore, there is no such n for which the equation n = #@n@ --> P# may hold.

3
  • I have three contraries to present: (1) Within the equation @n@ --> P, we can regard @n@ simply as a string of the basic symbols of the language, and not as anything like a metalinguistic function which must be resolved before the overall equation can be seen. Seeing it this way, the equation involves no circularity. Now, @n@ is an expression which resolves_to (not 'is') the expression with Godel number n. @n@ is like a name for an expression; it is not the expression itself. '@n@ --> P', conceived syntactically is another name.
    – James King
    Commented Jun 10 at 22:44
  • If that doesn't confound you, consider this: (2) A very similar solution can be given just in consideration that a numeral is not it's number. (3) Also: The fixed point lemma seems to provide that such an n must exist -- after all, both the @..@ and #..# functions are computable.
    – James King
    Commented Jun 10 at 22:52
  • You see, whatever mysteries there are concerning what @n@ may be, '@n@' is just an expression. We can compute a value for #@n@ --> P#. And we can set n to take that value. So then it follows that n = #@n@ --> P#. What's the problem?
    – James King
    Commented Jun 10 at 23:06

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