Theories and indiscernible propositions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:50:05Z http://mathoverflow.net/feeds/question/90860 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90860/theories-and-indiscernible-propositions Theories and indiscernible propositions Everett Piper 2012-03-11T01:09:10Z 2012-03-11T01:44:42Z <p>Are there known examples of statements which are strong from a proof-theoretic standpoint but which are indistinguishable by one set of axioms (or proof system) yet distinct according to a stronger set of axioms?</p> <p>More specifically, I'm wondering if there are examples of the following kind: </p> <p>Let $T_1$ and $T_2$ be theorems in some formal language $\mathcal{L}$. Let $A_1$ and $A_2$ be two distinct sets of axioms in $\mathcal{L}$ but which are not (obviously) incompatible. Are there known examples of $T_1$ and $T_2$ where $A_1$ proves "$T_1$ is equivalent to $T_2$" but $A_2$ proves "$T_1$ is strictly stronger than $T_2$"? </p> <p>At first glance, this notion of "stronger" conflicts with the received notion of "stronger" as "proving the same and more theorems" so the notion of "stronger theory" I'm asking about rules out characterizations like "the collection of formulas deducible from $A_1$ is properly contained in the collection of formulas deducible from $A_2$". I'm wondering if there is a sense in which two theories can disagree on the equivalence of two propositions because the weaker theory views the propositions as the same in some sense and the stronger theory witnesses some kind of first-order (or higher?) distinction between the propositions. Is this a useful notion in general? Or would this require some kind of axiom like "indiscernible objects are identical" to make precise and/or useful? </p> http://mathoverflow.net/questions/90860/theories-and-indiscernible-propositions/90866#90866 Answer by Andreas Blass for Theories and indiscernible propositions Andreas Blass 2012-03-11T01:44:42Z 2012-03-11T01:44:42Z <p>When I compare the question with the example in the comment, I infer that you regard "in $L[\mu]$" as weak, presumably because there's "only" a measurable cardinal there. But that theory is strong in another direction, namely by saying that the universe is only $L[\mu]$ rather than something bigger. </p> <p>In a similar vein, V=L is a strong theory. But if you want to regard it as weak because it doesn't allow very impressive large cardinals, then another example of what you asked for would be that V=L proves that the cardinal of the continuum and the first uncountable cardinal are the same.</p>