If the sets $S$ and $T$ considered are empty or singletons, the theorem holds vacuously or trivially. Thus assume $S$ and $T$ each contain at least two elements. Let $U = \struct {S, \preceq_S} \cup \struct {T, \preceq_T}$ Define the following relation $\preceq$ on $U$: :$\forall x, y \in U: x \preceq y$ {{iff}}: :$x, y \in S: x \preceq_S y$ or: :$x, y \in T: x \preceq_T y$ or: :$x \in S, y \in T$ We claim that $\preceq$ is a well-ordering. First, we show it is a total ordering. Checking in turn each of the criteria for a total ordering: === Reflexivity === If $x = y$, they're necessarily both in $S$ or $T$ simultaneously. Reflexivity then follows from $\preceq_S$ and $x \preceq_T$ being reflexive, as they are both orderings. === Transitivity === Let $x, y, z \in U$. If $x, y, z \in S$ or $x, y, z \in T$ simultaneously, then $\preceq$ is transitive by the transitivity of $\preceq_S$ and $\preceq_T$. Suppose $x, y \in S$ and $z \in T$. Let $x \preceq y$ and $y \preceq z$. Then $x \preceq z$ because $x \in S$ and $y \in T$. Suppose $x \in S$ and $y, z \in T$. Then $x \preceq z$ also because $x \in S$ and $y \in T$. Thus $\preceq$ is transitive. === Antisymmetry === Let $x \preceq y$ and $y \preceq x$. If $x, y \in S$ then $x = y$ by the antisymmetry of $\preceq_S$. Likewise if $x, y \in T$. If $x \in S$ and $y \in T$, then $y \in S$ and $x \in T$ as well. Thus $x = y$ from the antisymmetry of $\preceq_S$ or $\preceq T$. Conclude that $\preceq$ is a total ordering. To show $\preceq$ is a well-ordering, consider a non-empty set $X \subseteq U$. Then either: :$X \cap S = \O$ or: :$X \cap T = \O$ or: :$X \cap S$ is non-empty and $X \cap T$ is non-empty. In the first case, $X \subseteq T$, by Intersection with Complement is Empty iff Subset. Then $X$ has a smallest element defined by $\preceq_T$. In the second case, $X \subseteq S$, also by Intersection with Complement is Empty iff Subset. Then $X$ has a smallest element defined by $\preceq_S$. In the third case, the smallest element of $X \setminus T$ is an element of $S$. Thus it precedes any element of $T$ by the definition of $\preceq$. The smallest element of $X \setminus T$, which is a subset of $S$, is guaranteed to exist by the well-ordering on $S$. This smallest element is then also the smallest element of $\left({X \setminus T}\right) \cup T = X$. Thus $\preceq$ is a well-ordering on $S \cup T$. {{qed|lemma}} Consider the mapping: :$k: \struct {T, \preceq_T} \to \struct {U, \preceq}$: ::$\map k \alpha = \alpha$ Then $k$ is strictly increasing, by the construction of $\preceq$. Thus there is a strictly increasing mapping from $T$ to $U$. From Strictly Increasing Mapping Between Wosets Implies Order Isomorphism, $T$ is order isomorphic to $U$ or an initial segment of $U$. Let $\II_x$ denote the initial segment in $U$ determined by $x$, according to $k$. Note that $\II_x = \II_{\map k x}$, because $\map k x = x$. Suppose $x \in S$. Then $\II_x \subseteq S$ because $\preceq$ is a well-ordering. Thus there is an order isomorphism from $T$ to $\II_x$ in $S$. Suppose $x = \min T$, the smallest element of $T$. Then every element of $S$ strictly precedes $x$, as $x$ is in $T$. Also, $x$ precedes every element of $T$, so  $\II_x \ne T$. Thus there is an order isomorphism from $T$ to all of $S$. Suppose $x \in T$ and $x \ne \min T$. Then $\II_x$ defines an initial segment in $T$. Also, every element of $S$ strictly precedes $x$, as $x$ is in $T$. Thus there is an order isomorphism from an initial segment of $T$ to all of $S$. The cases are distinct by No Isomorphism from Woset to Initial Segment. {{qed}}