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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?

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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.

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    It's pretty easy to show they are simple, see https://fs23.formsite.com/viXra/files/f-1-2-9265569_LTyulPJf_ZSimpleZeros.pdf and according to Wikipedia, https://en.wikipedia.org/wiki/Riemann_hypothesis, Speiser (1934) proved that the Riemann hypothesis is equivalent to the statement that ζ ′ ( s ), the derivative of ζ(s), has no zeros in the strip 0 < ℜ ( s ) < 0.5 . That ζ has only simple zeros on the critical line is equivalent to its derivative having no zeros on the critical line. – 2017-02-27
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    I cannot find a copy of Speiser, Andreas (1934), "Geometrisches zur Riemannschen Zetafunktion", Mathematische Annalen, 110: 514–521, doi:10.1007/BF01448042, JFM 60.0272.04 on the web, nor on sci-hub, but I would like to check that out – 2017-02-27
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    small typo in definition 3. the 3rd equation shouldnt be there. that is lamba_N_f(t) not lambda_f(t). doesn't change the results though – 2017-02-27
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    http://projecteuclid.org/euclid.acta/1485889821 – 2017-02-27