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I am trying to understand this definition https://plato.stanford.edu/entries/independence-large-cardinals/#IntHie of Interpretability Hierarchy and how it relates to the concept of Consistency Strength. For example, I have found on some lecture notes that $ZFC+CH \equiv ZFC \equiv ZFC + \neg CH $ but this claim comes without proof. Now, I know that $Con(ZFC) \leftrightarrow Con(ZFC + CH) $ and that $ Con(ZFC) \leftrightarrow Con(ZFC + \neg CH) $. Moreover, the two notions seem to be related somehow, so I wonder what the exact connection is.

Matteo Casarosa
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    Consistency strength calculations are generally proved as corollaries of sharper results. E.g. the proof that $\mathsf{ZFC}$ and $\mathsf{ZFC+\neg CH}$ are equiconsistent follows by exhibiting an interpretation of the latter in the former (and verifying that in (a subtheory of) $\mathsf{ZFC}$ itself). Note that if $T$ interprets $S$ then the consistency of $T$ implies the consistency of $S$ (since consistency is equivalent to satisfiability), and this is provable within an appropriate "background theory" ($\mathsf{ZFC}$ being vast overkill here). – Noah Schweber Dec 15 '21 at 18:56
  • Kurt Godel in the 1930's showed that his "constructible class" L (defined in ZF) satisfies ZFC+GCH. Hence Con(ZF) implies Con(ZFC+GCH). In the 1960's Paul Cohen invented Forcing to show that Con(ZF) implies Con (ZF+$\neg$ AC) and that Con (ZFC) implies Con(ZFC+$\neg$ CH). At least a couple of courses in advanced set theory are needed to understand the proofs of these "claims". – DanielWainfleet Jun 01 '22 at 10:45
  • Related: How to define "interpretation" for general formal systems. – user21820 Jun 10 '22 at 15:18

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Belatedly turning my comment into an answer:

Consistency strength calculations are generally proved as corollaries of sharper results. E.g. the proof that $\mathsf{ZFC}$ and $\mathsf{ZFC+\neg CH}$ are equiconsistent (relative to $\mathsf{PA}$, say) follows by exhibiting an interpretation of the latter in the former.

The point is that if $T$ interprets $S$ and $S$ is inconsistent, then $T$ must also be inconsistent; contrapositively, if $T$ is consistent, then so is $S$. So interpretability results yield relative consistency results, and the former are indeed the chief means of attaining the latter.

In a sense, interpretability is a better notion than relative consistency. Remember that relative consistency is both dependent on a choice of "base theory" and somewhat capricious; for example, if we take as our base theory $\mathsf{PA}$, then the linear order axioms and the group axioms are boringly equiconsistent since $\mathsf{PA}$ proves that each is consistent outright.

Noah Schweber
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