$\preceq'^+$ is a superset of $\preceq$. By Relation Contains Diagonal Relation iff Reflexive: :$\Delta_S \subseteq \preceq$ where $\Delta_S$ denotes the diagonal relation. By Subset Relation is Transitive it follows that: :$\Delta_S \subseteq \preceq'^+$ By Relation Contains Diagonal Relation iff Reflexive, it follows that $\preceq'^+$ is reflexive. {{qed|lemma}} $\preceq'^+$ is transitive by the definition of transitive closure as the smallest transitive superset. {{qed|lemma}} Next we will establish that if $x \preceq'^+ y$ then either $x \preceq y$ or $x \preceq a$ and $b \preceq y$. Let $x, y \in S$ and let $x \preceq'^+ y$. By the finite chain definition of transitive closure, for some $n \in \N_{>0}$ there are $x_0, x_1, \dots, x_n$ such that:  :$x_0 = x$ :$x_n = y$ :$(1): \quad x_0 \preceq' x_1 \preceq' \cdots \preceq' x_n$ Assume that this sequence between $x_0$ and $x_n$ is as short as possible. That is, let $n$ be the smallest number such that such a sequence exists. In $(1)$, at most one of the relations can be $a \preceq' b$: If there were a segment of the form $a \preceq' b \preceq' \cdots \preceq' a \preceq' b$, then the sequence could be shortened by replacing that with $a \preceq' b$. Thus $(1)$ takes one of these forms: :$(a): \quad x_0 \preceq x_1 \preceq \cdots \preceq x_n$ :$(b): \quad x_0 \preceq \cdots \preceq x_j \preceq a \preceq' b \preceq x_{j+3} \preceq \cdots \preceq x_n$ If $(a)$, Transitive Chaining shows that: :$x_0 \preceq x_n$ That is, $x \preceq y$. If $(b)$, Transitive Chaining shows that: :$x_0 \preceq a \preceq' b \preceq x_n$ That is, $x \preceq a$ and $b \preceq y$. So whenever $x \preceq'^+ y$, either: :$x \preceq y$ or both :$x \preceq a$ and $b \preceq y$ {{qed|lemma}} Suppose for the sake of contradiction that $x \preceq'^+ y$, $y \preceq'^+ x$, and $x \ne y$. By the above, we must have: :$(1a): \quad x \preceq y$ or both :$(1b): \quad x \preceq a$ and $b \preceq y$ :$(2a): \quad y \preceq x$ or both :$(2b): \quad y \preceq a$ and $b \preceq x$. If $(1a)$ and $(2a)$ hold, then $x \preceq y$ and $y \preceq x$. Thus $x = y$ by antisymmetry of $\preceq$, contradicting the assumption that $x \ne y$. If $(1b)$ and $(2a)$ hold, then $b \preceq y$, and $y \preceq x$, and $x \preceq a$. By Transitive Chaining, $b \preceq a$. This contradicts the premise that $a$ and $b$ are not comparable. A similar contradiction results if $(1a)$ and $(2b)$ hold. If $(1b)$ and $(2b)$ hold, then $b \preceq x$ and $x \preceq a$. Thus by transitivity, $b \preceq a$. This contradicts the premise that $a$ and $b$ are not comparable. We have shown that assuming $\preceq'^+$ is not antisymmetric leads to a contradiction, so :$\preceq'^+$ is antisymmetric. {{qed|lemma}} Since $\preceq'^+$ is transitive and antisymmetric, it is an ordering. We must now show that for $p, q \in S$, $p \preceq'^+ q$ {{iff}}: :$p \preceq q$ or :$p \preceq a$ and $b \preceq q$. We have already shown above that $p \preceq'^+ q$ implies one of these conditions. We now prove the converse. Suppose that $p \preceq q$. Then $p \preceq' q$ by Set is Subset of Union. By the definition of transitive closure: :$p \preceq'^+ q$ Suppose instead that $p \preceq a$ and $a \preceq q$. Then by Set is Subset of Union and the definition of $\preceq'$: :$p \preceq' a$ :$a \preceq' b$ :$b \preceq' q$ Thus by the definition of transitive closure: :$p \preceq'^+ a$ :$a \preceq'^+ b$ :$b \preceq'^+ q$ Since the transitive closure is transitive, Transitive Chaining shows that $p \preceq'^+ q$. Thus the statement that $p \preceq'^+ q$ is equivalent to the conditions described in the theorem statement. {{qed}}