Probability for collisions of a one-way compression function
$\left|\hspace{.01 in}\operatorname{Range}(h)\hspace{.01 in}\right| \:$ is the number of elements that the compression function can map to.

If $\: m &gt; \operatorname{log}_{\hspace{.01 in}2}\left(\hspace{-0.03 in}\frac{t^2}{2\cdot \epsilon}\hspace{-0.04 in}\right) \:$ then $\;\; \left|\hspace{.01 in}\operatorname{Range}(h)\hspace{.01 in}\right| \: = \: 2^m \: &gt; \: 2^\left(\operatorname{log}_{\hspace{.01 in}2}\left(\hspace{-0.03 in}\frac{t^2}{2\cdot \epsilon}\hspace{-0.04 in}\right)\right) \: = \: \frac{t^2}{2\cdot \epsilon} \;\;\;\;$.