Instead, I am predicting, from the vantage point of the original study, the probability that a future replication study result would again be significant at the
0.05 level.
That would be
P(b_\text{repl}/\sqrt{v_2} > 1.96 \mid b,v_1,v_2) = \Phi\left( \frac{b - 1.96\sqrt{v_2}}{\sqrt{v_1+v_2}} \right)
as I’ve explained many times. See also Goodman (1992).
Your b_\text{repl}/\sqrt{v_1+v_2} is not a useful statistic. It is misleading to refer to the (conditional) probability that it exceeds 1.96 as the “replication success probability”. By continuing to do so, you are essentially forcing me to keep correcting you. You are abusing this platform, and you need to stop it!