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Following Koellner in http://plato.stanford.edu/entries/independence-large-cardinals/, "a theory $T_1$ is interpretable in $T_2$ ($T_1 \leq T_2$) when, roughly speaking, there is a translation $\tau$ from the language of $T_1$ to the language of $T_2$ such that, for each sentence $\phi$ of the language of $T_1$, if $T_1\vdash \phi$ then $T_2\vdash \tau (\phi)$." Now take a theory $T$ and a statement $\phi$ independent from $T$, then for sure we have $T\leq T+\phi$ (with $\tau$ being trivial). Additionally, Koellner says, we can have either $T+\phi\not\leq T$ or $T+\phi\leq T$. But how is the second case possible, since the trivial proof $T+\phi\vdash\phi$ cannot be translated into a proof $T\vdash\phi$?

I know there are examples for the second case (like $T+\neg Con(T)\equiv T$, $T+CH\equiv T$, $T+\neg CH\equiv T$, where $\equiv$ means both $\leq$ and $\geq$), but how are they possible?

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    $\begingroup$ Your "rough" description of interpretability is not accurate, since we could take $\tau(\phi)$ to be any tautology and have your property. $\endgroup$ Jul 9, 2016 at 14:02
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    $\begingroup$ In addition to Joel's correction, notice that the end of the first paragraph of the question, "cannot be translated into a proof $T\vdash\phi$" is irrelevant. What's required for interpretability is not $T\vdash\phi$ but $T\vdash\tau(\phi)$. $\endgroup$ Jul 9, 2016 at 16:31
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    $\begingroup$ The phenomenon is very common. For example, we can let $T$ be the theory of groups and let $\phi$ say that every element equals the identity. Then $T + \phi$ is interpretable in $T$, but $\phi$ is independent of $T$. $\endgroup$ Jul 10, 2016 at 1:44
  • $\begingroup$ Thank you all for your answers! I thought that $\tau$ should be the identity on any provable formulae of $T$ (is that right so far?) and that this identity would somehow expand to the unprovable formulae (which is not the case, I guess?) $\endgroup$
    – Henning H
    Jul 10, 2016 at 15:09

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For example, $\newcommand\ZFC{\text{ZFC}}\ZFC+V=L$ is mutually interpretable with $\ZFC$ because inside any model of $\ZFC$ we may find its version of the constructible universe $L$, which is a model of $\ZFC+V=L$. So we have a definable way to interpret $\ZFC+V=L$ inside any model of $\ZFC$. The same argument works with $\ZFC+\text{GCH}\equiv\ZFC$.

Meanwhile, a theory like $\ZFC+\exists$ inaccessible cardinal is not mutually interpretable with $\ZFC$, because on consistency strength grounds we are not able uniformly to define a model of the former theory inside any model of the latter theory.

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