Proof of an $\iff$ statement on binary trees 
Let $x$ and $y$ be two nodes of a binary tree $B$.
Prove that $x$ is an ancestor of $y$ $\iff$ $x$ stands before $y$ in the pre-order traversal of $B$ and $x$ stands  after $y$ in the post-order traversal of $B$.

I can do the $\implies$ part without trouble, but I can't deal with the converse with enough rigour.
Thanks for any kind of rigorous proof.

For the converse I would proceed by contrapositive with the following ideas.
(1) Assume $x$ and $y$ are vertices for which neither is an ancestor of the other.
(2) Argue that there is a lowest common ancestor (I am picturing the tree drawn with the root at the top and going downwards) of $x$ and $y$, say $a$.
(3) Argue that one of $x$ and $y$, say $x$ is a left descendant of $a$ and the other is a right descendant of $a$.  (If both are on the same side of $a$, then there is a lower common ancestor.)
(4) Then in preorder traversal, these three vertices will be in order $a$...$x$...$y$, and in postorder traversal $x$...$y$...$a$.  
