# Proving that □A → A is valid iff R is reflexive?

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)

• The formula defines the pathetic relations, i.e. those relations $R$ with $xRy$ iff $x = y$ Commented Oct 24, 2021 at 21:34
• 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? Commented Oct 28, 2021 at 2:26

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