Tertium non datur and Inductive Subset of a Well Ordered Set Theorem: *Let (S,) be a well-ordered set. Let T⊆S be an inductive subset of S, that is such that: ∀s∈S:(∀t∈S:t≺st∈T)s∈T (induction hypothesis) Then T=S.* Using reductio ad absurdum the proof is very simple. If S\T is nonempty, then it has a lowest element x, which satisfies the induction hypothesis. As such x∈T, contradiction, q.e.d. My question is whether _tertium non datur/reductio ad absurdum_ can be avoided in this proof.

Essentially you're proving the contrapositive: if $T\subsetneq S$ then $T$ is not inductive. Suppose $T\subsetneq S$. Then $S\setminus T\
e \emptyset$, so it has a least element $x$. For any $s\in S$, if $sotin T$. Therefore, $$ \exists s\in S\,((\forall t\in S\,(totin T), $$ which says "T is not inductive".