Exercise 13.4.2

Argue that if in RB-DELETE-FIXUP both $x$ and $x.p$ are red, then property 4 is restored by the call to RB-DELETE-FIXUP(T, x)

Similar to the previous exercise, just observe that RB-DELETE-FIXUP(T, x) will immediately color x black if it was red. Thus, if both $x$ and $x.p$ are red, $x$ will become black and the property will be retained (as $x.p$ is red and it's other child was unchanged, it's bound to be black).