Proving Blank-halt to be recursively enumerable through reduction to Halting problem
If you reduce $X$ to&nbsp;$Y$, you're showing that $X$&nbsp;is no harder than&nbsp;$Y$ and that $Y$&nbsp;is at least as hard as&nbsp;$X$.

So, by showing that some problem&nbsp;$X$ reduces to the halting problem, you're showing that $X$&nbsp;is no harder than the halting problem, i.e., that it's recursively enumerable. Remember that we normally define problem classes to include all lower classes, so showing that $X$&nbsp;is RE doesn't necessarily imply that it's not recursive, for example. In this case, we can separately prove that the blank halting problem isn't recursive, by reducing the halting problem to it, which shows it's at least as hard as the halting problem, as above.