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Define an order $\leq$ on strong enough consistent logical theories by $T \leq U$ if $U, \text{Con}(U) \vdash \text{Con}(T)$.

What does this order look like?

  1. Is it linear (for any $T$ and $U$, either $T \leq U$ or $U \leq T$)?
  2. If it's not linear, does it at least have finite upper bounds (for any $T$ and $U$ there is a $V$ such that $T \leq V$ and $U \leq V$)?
  3. Is it well-founded, or can we have an infinite chain of theories $T_0 > T_1\gt\ldots$, where $T\lt U$ means $T \leq U$ but not $U \leq T$?
  • What is "strong enough"? Because otherwise 2 would be trivial: the empty theory is always consistent and thus at the top of this order. – Mark Kamsma Apr 19 '20 at 13:05
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    Relevant MathOverflow question: https://mathoverflow.net/questions/59717/non-linearity-of-the-consistency-strength-ordering-in-zf – Mark Kamsma Apr 19 '20 at 13:08
  • By strong enough I mean a theory in which it's possible to encode the statement $\text{Con}(T)$. I don't see why the empty (or minimally strong enough) theory would be at the top, it would be at the bottom. –  Apr 19 '20 at 13:12
  • Oops, I read the order the wrong way around. You're right, it would be at the bottom. – Mark Kamsma Apr 19 '20 at 13:45

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