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I am currently trying to learn about the fundations of mathematical logic, and the incompleteness theorem. I was curious to know if there's a way, given some given axioms, to analyze all the possible theories that are compatible with them.

By "analyze", I mean "say anything that could be interesting", for instance counting how many such theories there are, or automatically generating statements that are true in one of the theories, but false in the other etc...

For instance, given the first 4 axioms of planar geometry, can we automatically deduce that only 3 geometries will be compatible with those axioms (euclidean, spherical and hyperbolic), generate automatically the additional axioms needed to define those geometries, and generate statements that are true in one of the geometries but not the other ?

Even further, given a starting set of axioms, do the allowed theories generated have some sort of structure ? For instance, intuitively I would imagine the possible theories could be organized with a spanning tree, where we branch out by adding different alternative axioms on top of the growing stack of axioms...

By automatically I literally mean using a computer for instance,

I'm just trying to get more intuitions on this area, (at the same time I'm learning more formally to get more into the details of the proofs, formal systems, godel numbering etc)

I'd be really glad to read about any reference about the subject! Particularly with concrete examples using classical axioms systems (like geometry, arithmetic etc)

Thanks a lot,

  • A computer generally cannot decide if two theories are equivalent since you always say add the Riemann hypothesis an axiom and then what are you supposed to do? –  May 25 '21 at 14:52
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    This is a very good question, but not research-level, so I’m voting for migration to math.stackexchange. – Peter LeFanu Lumsdaine May 25 '21 at 15:25

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Your idea that there is a computable tree of possible extensions of a given theory is completely right, in the case of arithmetic.

If we start with any consistent computably axiomatizable theory $T$ of arithmetic (extending a certain sufficient weak theory), then it is a consequence of the Rosser variation of the Gödel incompleteness theorem that there must be a sentence $\rho_T$, the Rosser sentence, that is neither provable nor refutable in the theory. Thus, we may extend the theory in two ways, either by adding the sentence or its negation $$T_0=T+\neg\rho_T\qquad\qquad T_1=T+\rho_T.$$ Each of these theories is a consistent computably axiomatizable theory of arithmetic, a finite extension of the original theory $T$. These new theories $T_s$ therefore have their own Rosser sentences $\rho_s$, and so we may extend to a tree of theories $$T_{s^\frown 0}=T_s+\neg\rho_s\qquad\qquad T_{s^\frown 1}=T_s+\rho_s.$$ In this way, for every binary sequence $s$ we get a theory $T_s$ forming altogether a binary-branching tree of distinct pairwise incompatible consistent theories of arithmetic, each a finite extension of the initial theory $T$. Furthermore, this tree is computable, in the sense that we can compute from any finite sequence $s$ what are the sentences $\rho_{s\upharpoonright n}$ that we had added to form $T_s$.

Every infinite binary sequence $x$ determines a branch through this tree $$T_x=\bigcup_n T_{x\upharpoonright n}$$ and any two such theories will disagree. Thus, there are continuum many distinct consistent theories extending $T$. And the theory $T_x$ is computably axiomatizable from oracle $x$.

You can see an application of this idea on my blog post Every function can be computable!.

JDH
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    Meanwhile, I have voted to transfer the question to math.stackexchange, which I think is a better fit for the question, and I think the transfer will just take this answer along with it. – JDH May 25 '21 at 17:50
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    What is also amazing is that even though there are continuum-many branches of the "Rosser tree" (in particular many branches that are not even arithmetically definable, let alone computably enumerable), there is a single arithmetical sentence that is independent of all the branches. This can be proved by appealing to a strengthening of Rosser's theorem, due to Mostowski, that says that given an r.e. sequence of consistent theories T_n (extending PA, say) there is a sentence independent of all of them. The relevant T_n here is the set of finite extensions of T built by the Rosser tree. – Ali Enayat May 26 '21 at 17:15
  • Thank you very much for your answer @JDH, this is very interesting, also thanks for the reference on your blog which is very interesting ! – dorikolmo May 28 '21 at 19:02
  • @Ali Enayat thanks for your comment, that’s really interesting ! – dorikolmo May 28 '21 at 19:03