# Gödel Without Tears — 8

Here’s the latest episode — in which we at last get to the First Incompleteness Theorem. (You can download the latest versions of earlier episodes here, as well as a composite PDF of all the episodes to date.)

That’s a good place to pause and get back to other stuff. The plan is for five more episodes: two more around and about the First Theorem, two on the Second Theorem, and one lightning tour of the bits from the last part of my book which are most relevant to Gödel’s Theorems. But these episodes won’t be written until lectures start again in the New Year.

### 3 thoughts on “Gödel Without Tears — 8”

1. The title of section 31.3 should presumably be: If PA is w-consistent, it can’t prove not-G.

There are two mistakes: that it said G rather than not-G, but also that it said “it is can’t” rather than “it can’t”. The “it is can’t” error also occurs in the title of 31.1.

Another point: It’s not clear to be how the “true” gets into theorems 41 and 42. (A mistake?) Note that theorem 2 doesn’t have it.

1. Thanks for the corrections, now made.

G is true, because it is unprovable, by Theorem 34 (so there’ no mistake — but I’ve added a remark as a reminder).

2. I’ve been reading through your lectures, and there was a point on the last page of Lecture 2 that jumped out at me:

Note, however, that as far as Theorem 6 is concerned, it could be that U repairs the gaps in T and proves every truth statable in T’s language, while the incompleteness has now ‘moved outwards’, so to speak, to claims involving U’s new vocabulary. G ̈odel’s result is a lot stronger: he shows that some incompleteness will always remain even in the theory’s arithmetical core.

I was thinking: What if we strengthened “T is sufficiently strong” to mean “T captures each decidable numerical property with an arithmetic wff“? This would be equivalent for any theory in the first-order language of number theory (since all wff’s are arithmetic) and would be preserved under any strengthening of T (say to U above).

Moreover, Theorem 4 would be strengthened to “No consistent, sufficiently strong, axiomatized formal theory has decidable arithmetic part”, since in the proof one could choose to enumerate only the arithmetic wff’s with free variable x. Otherwise the argument would be identical. Theorem 6 would in turn become “The arithmetic part of any consistent, sufficiently strong, axiomatized formal theory is negation-incomplete”.

We still wouldn’t have a specific undecidable arithmetic sentence, but we would at least have the strengthening mentioned above.

Thoughts? Obvious mistakes?

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