Does the simplicity of the zeta zeros imply the Riemann hypothesis?

Question

If the Riemann zeros are proved to be simple, that is, have multiplicity ${m_f (\alpha)}=1$ such that their Taylor series about the point $\alpha$ has the form \begin{equation} f (t) = c (t - \alpha)^{m_f (\alpha)} + (...) \end{equation} where $c \neq 0$ and $m \geqslant 1$, does it follow that the Riemann hypothesis is true?

Answer 1

It is not known to follow. Conversely, it is not known that the Riemann Hypothesis would imply the simplicity of the zeroes. [And for that matter, I suspect that they are independent, and that the Riemann Hypothesis would be solved before one shows that the zeroes are simple].

The only line of results in either direction stem from a line of inquiry from Montgomery, who showed that a large percentage of the zeroes are simple if the Riemann Hypothesis is true.