0

I was reading this question by user107952 and trying to come up with an example of a theory with no independent axiomatization or at least no independent recursive axiomatization.

I've come up with the following infinite theory with only nullary predicates, but I'm having trouble proving or disproving whether it has a (recursive) independent axiomatization.

Here's the theory, which can also be thought of as a theory in classical propositional calculus.

We are using a whole mess of nullary predicates $R_\alpha$ for each $\alpha$ in $\omega + 1$.

Let the axiomatization of our theory $T$ be the following. I'll write $R_\alpha$ instead of $R_\alpha()$.

  1. $R_\alpha \to R_\beta$ whenever $\alpha < \beta$.

Models of this theory can be thought of encodings of monotone functions $(\omega + 1) \to 2$.

I've tried for a little bit to show that we can't have an independent axiomatization of $T$. I tried to derive a contradiction regarding the handling of $R_\omega$, but I haven't manage to actually produce one. As of right now, I don't know whether it has an independent axiomatization or not.

My question is threefold:

  1. Does $T$ have an independent axiomatization?
  2. If (1), does $T$ have a recursive independent axiomatization?
  3. If (1), is there an independent axiomatization of $T$ such that there exists a $k \in \mathbb{N}$ such that for all sentences $\varphi$ in $T$, it holds that $|\varphi| \le k$?
Greg Nisbet
  • 11,657

1 Answers1

4

Every first-order theory has an independent axiomatization; while tricky to prove in general (Reznikoff), the result for countable theories is straightforward. See this old answer of mine. In particular, the strategy presented there gives a recursive independent axiomatization of this theory; the only potential point where non-recursiveness enters the picture in that strategy is the problem of determining when one sentence in (the "initial" axiomatization of) $T$ is provable from a given set of others, but in this case that's easy.

Noah Schweber
  • 245,398