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    $\begingroup$ Adding LEM doesn't increase the strength. This is seen by the double negation translation. $\endgroup$
    – Trebor ♦
    Commented Mar 29, 2022 at 23:42
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    $\begingroup$ @Trebor there is no such thing as a double negation for CIC. If you do it on a proof relevant sort, it will break dependent elimination. If you do it on SProp, it works but SProp can only be used to prove dead branches so you don't increase you expressivity in Type. $\endgroup$
    – Pierre-Marie Pédrot
    Commented Mar 30, 2022 at 9:39
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    $\begingroup$ @Trebor: No, double-negation translation works for only certain systems, such as pure FOL to pure IFOL, or PA to HA. Even for FOL theories, it does not imply that adding LEM does not change the strength. Over the base theory IFOL we have ( A+LEM ⊢ Q ) iff ( DN(A) ⊢ DN(Q) ) where DN is the double-negation translation, which is not the same as ( A ⊢ DN(Q) ). In fact, CZF is quite weak compared to ZFC but CZF+LEM = ZFC. $\endgroup$
    – user21820
    Commented Mar 30, 2022 at 10:05
  • $\begingroup$ Of course it depends on what you call LEM. If you interpret it naively, then HoTT + LEM = Falso, which is the most dramatic strength increase you can name. In practice we mean some version of truncated LEM, which is fine. $\endgroup$
    – Trebor ♦
    Commented Mar 30, 2022 at 10:16
  • $\begingroup$ @Trebor: I meant full LEM, which is not known to be inconsistent with MLTT and CIC. And I am not interested in HoTT in this post. Either way, I don't understand why you don't acknowledge that there is no double-negation translation for MLTT/CIC, which makes your initial comment wrong. $\endgroup$
    – user21820
    Commented Mar 30, 2022 at 12:46