Show that if a node in a binary search tree has two children, then its successor has no left child and its predecessor has no right child.
Let's say we have a tree $T$ with left subtree $L$ and right subtree $R$ and parent $P$. Let's also assume distinct keys.
Having two children implies that both the successor and the predecessor is going to be among them. To illustrate this, let's consider the successor. It can't be to the right of an ancestor $A$, because this makes $T$ on the left of $A$, therefore making $A$ a better candidate as $T < A$. It can't be that ancestor either, because $T < R$ and $R < A$, making $R$ a better candidate than $A$. Furthermore, if $R$ exists, it's going to contain the (it can't be in $L$, as those elements are smaller).
Furthermore, if the successor (somewhere in $R$) had a left child, that would be a better candidate for a successor, since it's smaller than $R$ but still bigger than $T$.
Same logic applies for the predecessor.