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Wikipedia says:

Theory is complete if it is a maximal consistent set of sentences.

Than it says:

Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete.

But completeness definition says that theory is consistent so above definition says that

Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and (maximal and consistent).

So it should be enough to say:

Any effectively generated theory capable of expressing elementary arithmetic cannot be complete.

Am I right or am I missing something?

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2 Answers 2

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Your statement of the Incompleteness Theorem is fine, provided that you make clear your definition of a complete theory.

The fact of the matter is that it is not uncommon for people to define a complete theory as a theory that for each sentence $\phi$ in its language contains (at least one of) $\phi$ or $\lnot\phi$. The only difference between this definition and Wikipedia's is that this one makes any inconsistent theory complete.

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  • $\begingroup$ So I suppose wiki author who described incompleteness theorem thought about version of completeness that you provided. $\endgroup$ Aug 23, 2013 at 8:31
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Your first quote is from Complete Theory, I presume, and the second from Gödel's Incompleteness Theorems, so the first problem we encounter is that of consistency within Wikipedia. However, if you carry on reading Complete Theory, you will notice that different notions of completeness exist and Gödel is about the other.

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    $\begingroup$ Actually (as opposed to many other questions we've had here on this topic) this is the appropriate notion of completeness for the incompleteness theorems. It's the Completeness Theorem that confuses people. $\endgroup$ Aug 22, 2013 at 20:56
  • $\begingroup$ The other version provided by wiki article is also consistent because is says that theory must contain either $\phi$ or $\neg \phi$. So for my understanding of English language it may not contain both so it is consistent. $\endgroup$ Aug 23, 2013 at 8:32

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