A set $x$ is said to be $\in$-transitive if $\forall y$ $\forall z$($ y \in x$ and $ z \in y \Rightarrow z \in x$). A set $x$ is said to be an ordinal if $x$ and every member of $x$ is $\in$-transitive.
The following proposition is given:
Proposition: Every well ordered set is order isomorphic to a unique ordinal by a unique isomorphism.
Using this proposition how can I prove that a set which is $\in$-transitive and well ordered by $\in$ is an ordinal.
The Proposition is not necessary (or particularly useful, as far as I can tell) to prove this result.
Suppose that $a$ is $\in$-transitive and well-ordered by $\in.$ In order to show that $a$ is an ordinal (that is, an $\in$-transitive set whose elements are $\in$-transitive), it remains only to show that the elements of $a$ are $\in$-transitive. By way of contradiction, suppose that not all elements of $a$ are $\in$-transitive, meaning that there is some non-$\in$-transitive $x\in a.$ Since $x$ is not $\in$-transitive, then there exist $y\in x$ and $z\in y$ such that $z\notin x.$ Since $a$ is $\in$-transitive, then $y\in x\in a$ implies that $y\in a,$ and since $z\in y,$ we likewise have $z\in a.$
So, at this point, we know that $x,y,z\in a$ with $z\in y,$ $y\in x,$ and $z\notin x.$ Do you see where the contradiction comes in? (What property of $a$ hasn't been used yet?)