Is it possible to prove that the Grz axiom is valid in a modal frame iff the frame is reflexive and transitive?

We need to prove (or disprove?) that $\square (\square (A \rightarrow \square A) \rightarrow A) \rightarrow A$ is valid in the Kripke modal frame $F = <S, R>$ iff R is transitive and reflexive.

I think that more is required of R for this to be correct. Am I right?

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Consider following structure: $U=\{ w_{1},w_{2},w_{3}\}$, $R$ is transitive, reflexive plus $(w_{3},w_{2}) \in R$ and following valuation $v(w_{1},A)=0, v(w_{2},A)=1, v(w_{3},A)=0$. It's easy to check that Grz axiom is false in this structure with respect to given valuation. In general this http://en.wikipedia.org/wiki/Method_of_analytic_tableaux method is helpful for such issues.