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How to prove, in modal logic, that \square $A\to A$ is valid iff $R$ is reflexive? (shouldn't this be the definition of $T$ axiom in modal logic)?

(NOTE: The question was edited because I made a mistake by misplacing the necessity sign. Former formulation does not make sense at all)

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  • $\begingroup$ The formula defines the pathetic relations, i.e. those relations $R$ with $xRy$ iff $x = y$ $\endgroup$
    – sequitur
    Commented Oct 24, 2021 at 21:34
  • $\begingroup$ You've already asked the updated question in this post and fully accepted an answer without any further question and comment there, so what do you still doubt? $\endgroup$
    – cinch
    Commented Oct 28, 2021 at 2:26

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$T$ axiom is not defined as such, instead it's defined in most modal logics the othere way around as below:

T, Reflexivity Axiom: □p → p (If p is necessary, then p is the case.)

T holds in most but not all modal logics. Zeman (1973) describes a few exceptions, such as $S1^0$.

$A \to \square A$ is an inference rule (not an axiom) in the weakest normal modal logic $K$ only when $A$ is a theorem:

Necessitation rule: $\vdash A$ implies $\vdash \square A$.

Obviously in general if A is a mere contingent proposition, we cannot assert A must be necessarily true in our world even if our world is reflexive normal modal system $T$ accessible to itself.

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