Proving $0$ is an ordinal In Introduction to Set Theory by J. Donald Monk, he defines ordinal as follows.

Definition (1): $A$ is an ordinal iff $A$ is $\in$-transitive and each member of $A$ is $\in$-transitive. $A$ is $\in$-transitive iff for all $x$ and $y$ , $x \in y \in A \implies x \in A$.

And using definition (1), I have a problem of showing  $0$ is an ordinal.
My solution: Let $x \in y \in 0$, then show that $x \in 0$, but in the theorem shown before this, there cannot exist an $x \in 0$ for any $x$, for if $x \in 0$, then $x$ is a set and $x \neq x$, which is absurd. Hence  there cannot be an $x$ such that $x\in 0$ .But if this is the case, we cannot shown the above theorem by definition. Any ideas?
thanks for your help.
Edit: Initially I have 2 question to ask. But later on, I have found the answer to my first question. That is why I deleted it and the question post above is my second question.
 A: Seoral: A statement of the form "for all $x\in A$ blah" literally means "for all $x$, if it is the case that $x\in A$, then blah".
Now: As you mentioned, $y\in 0$ is false for any $y$. We need to show that if $x\in y\in 0$ then $x\in 0$. More precisely, this means: For all $x$ and $y$, if $y\in 0$ and $x\in y$, then $x\in 0$. Since $y\in 0$ is false, we need to show that for all $x$ and $y$, (false and $x\in y$) implies $x\in 0$. Recall that "false and blah" is false, so this is just: For all $x$ and $y$, false implies $x\in 0$.
But false implies anything! (More precisely, statements $p\to q$ are true whenever $p$ is false. They are also true when $q$ is true, but that doesn't matter here.) So, we have "for all $x$ and $y$, true", which is the same as "true". I.e., you have proved what you needed.
One usually doesn't express situations as the one above this way; instead, people talk of statements being "vacuously true", as pointed out in a comment above.
Let's review: 0 is transitive, because any element of 0 is a subset. (Precisely, because there are no elements of 0.)
Similarly, any element of 0 is transitive. Again, because there are no elements of 0.
This is somewhat strange the first time one encounters it, since it is also the case that any element of 0 fails to be transitive (and that any element of 0 is blue, etc).
