1

Let $D$ be an effectively generated set of recursive ordinals that contains $0$ and that is closed under an effectively defined ordinal successor function "$+1$" and that have a well defined relation of ordinal strict smaller than relation $<$ on it, such that no two distinct elements of $D$ have the same order type. We'll simply label such a set as an "effective set of recursive ordinals"

Define recursively

  • $T_0 = T$,
  • $T_{i+1} = T_i + \operatorname{Con}(T_i)$ for every $i \in D$,
  • $T_j = \bigcup\{T_i\mid i<j\}$ for every limit ordinal $j \in D $.

Where $T$ is an extension of $PA$.

Say that $T$ is $D$_ultraconsistent iff $\forall i \in D \ ( \operatorname{Con}(T_i)) $

Say that $T$ is ultraconsistent iff

$\forall D (D$ is an effective set of recursive ordinals$ \implies T $ is $D$_ultraconsistent$)$

Do the following statements follow?

  1. If $T,H$ are ultraconsistent then $T \cup H$ is ultraconsistent.

  2. If $T$ is ultraconsistent and $G$ is a fragment of $T$ then $G$ is ultraconsistent.

  3. Can we have a theory $T$ that is ultraconsistent and yet $FALSE$?

    Of relevance to the last question is the following quote:

"Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's O, but could just prove a theorem for Π1 sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "n-Reflection" statements for every n each time one extends the theory then there are paths along which every true statement of arithmetic is proven". (emphasis added)

From the following page of Mathoverflow:

Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.?

Now this points to completeness of such paths for theories formulated in classical logic, and so ultra-consistency seems to parallel truth here, since no ultra-consistent theory can be false since it would be disproved along the hierarchy of theories on top of it proving its negation. I'm not really sure of this, that's why I'm asking, if such a quote goes in favor with what I'm at here.

Note: This post modified the earlier approach before the [HOLD] which was erroneous in using indefinable ordinal indices, here this is fixed to index the theories after an ordinal notation that is effectively recursively generated.

Community
  • 1
Zuhair Al-Johar
  • 9,152
  • 2
    What a horror... – Andrés E. Caicedo Dec 18 '17 at 19:28
  • 2
    I tried to change the format into something palatable. You may want to explain what basic requirements you are imposing on your theories $T$. Is $T_0$ a subtheory (or an extension) of $\mathsf{PA}$? Is it a theory in an arbitrary language but capable of coding enough recursion to make sense of the $T_i$ and of consistency statements? If the latter, then you need to specify what the natural model is that you have in mind, to make sense of the notion of "false". – Andrés E. Caicedo Dec 18 '17 at 19:38
  • Thanks. I've responded to the basic requirements of T, it should be an extension of PA. – Zuhair Al-Johar Dec 18 '17 at 20:17
  • 15
    This is something of a FAQ (and pretty much everyone makes this mistake at some point): $\mathrm{Con}(T)$ does not depend merely on the set of axioms of $T$ but on the way they are enumerated. So $T_i$ is not well-defined, unless you let $i$ be a recursive ordinal notation and not merely a recursive ordinal (but then you run into other problems). – Gro-Tsen Dec 18 '17 at 20:18
  • @Gro-Tsen, it is well defined in the second order language of arithmetic. – Zuhair Al-Johar Dec 18 '17 at 20:22
  • 2
    If you work in second-order logic (with full models), then I don't know what $\mathrm{Con}(T)$ means, but $\mathrm{PA}^2$ is categorical (meaning, has a single model, the intended model) from the start, so adding axioms to it is unlikely to produce anything interesting. – Gro-Tsen Dec 18 '17 at 20:30
  • 1
    @Zuhair Are you saying that you have a way of circumventing the issues pointed out in https://mathoverflow.net/a/67237/6085 ? – Andrés E. Caicedo Dec 18 '17 at 20:33
  • 4
    (@Zuhair More likely, I think you did not understand the problem Gro-Tsen mentioned.) – Andrés E. Caicedo Dec 18 '17 at 20:34
  • 5
    Now that you've edited the question to iterate along a recursive notation system, it makes sense, so I nominated for reopening. But it would have been better if you had made it clear in your edit that your original question was wrong and that you fixed it, so people can see it and vote for reopening. – Gro-Tsen Dec 20 '17 at 14:38
  • 1
    Also, I don't know the answers to your (modified) questions, but I suspect "ultraconsistency" is strongly tied to $\Sigma_2$-soundness. A candidate for a theory that would be "ultraconsistent" but not sound (I suppose "FALSE" means unsound) might be the theory $\mathrm{PA}$+"$\mathrm{PA}$ is $\Sigma_2$-unsound", which I think is $\Sigma_2$-sound (but obviously not sound), and I suspect "ultraconsistent". – Gro-Tsen Dec 20 '17 at 14:41
  • @Gro-Tsen: Your last comment makes me wonder whether you perhaps know the answer to my question here? =) – user21820 Dec 20 '17 at 16:20
  • @Gro-Tsen Thanks for drawing my attention to making the explicit note. I made that note now. – Zuhair Al-Johar Dec 20 '17 at 17:07
  • 3
    Any $\Sigma_1$-sound theory is ultraconsistent. (This is likely if and only if.) Thus, 1. there are ultraconsistent theories T, H whose union is (plain) inconsistent, and 3. there is an ultraconsistent theory that's not even $\Sigma_2$-sound. 2. is likely easily shown true if you fix issues with the formalization mentioned in Gro-Tsen's comment. – Emil Jeřábek Dec 23 '17 at 16:08

0 Answers0