# I don't understand the case 4 of red-black tree deletion

I don't know why case 4 will resolve the issue of the double black of $$x$$ described in Introduction to algorithm p.329. I know case 1 is transformed into one of {2,3,4} case, and case 2 re-point $$x$$ to its parent node so it's approaching the root, and case 3 will be transformed into case 4, but both case {3,4} will not change the pointer $$x$$, anyway so how case 4 work?

Is it because for paths include $$C,E$$ there is no difference about the number of blacks while for that about $$\alpha,\beta$$ an extra black node $$B$$ is passed?

Let us check case 4. Although the parent pointer of node $$x$$ is not changed, i.e., the parent node of $$x$$ is still $$B$$, the color of $$B$$ is set to black and node $$D$$ is lifted to become $$B$$'s parent with $$B$$'s original color.
So, going from the graph on the left hand side of case 4 on the image in the question to the graph on the right hand side, we have added that "extra" black node that was supposed to attached to $$x$$, which is, in your words, an extra black node $$B$$ is passed for the paths going through $$\alpha,\beta$$.
Furthermore, for that graph on the right hand side, we can check that all simple paths from the node to descendant leaves that do not pass through $$A$$ contain the same number of black nodes as before. Since all properties of black-red tree have been restored, case 4 ends the while loop, which ends RB-DELETE-FIXUP$$(T, x)$$, which ends RB-DELETE$$(T, z)$$.