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