First, note that since $\prec$ is a strict ordering, $\preceq$ is an ordering by Reflexive Closure of Strict Ordering is Ordering. === $a$ and $b$ are $\preceq$-incomparable === Suppose that $a \preceq b$. By the definition of reflexive closure, either $a \prec b$ or $a = b$. Each possibility contradicts one of the premises, so $a \not \preceq b$. For the same reasons, $b \not \preceq a$. {{qed|lemma}} === $\prec'^+$ is antireflexive === Let $p \in S$. Then by the definition of strict ordering: $p \not\prec p$. Suppose that $p \preceq a$ and $b \preceq p$. Then since $\preceq$ is transitive, $b \preceq a$, contradicting the fact that $a$ and $b$ are $\preceq$-incomparable. Since neither $p \prec p$ nor $\paren {p \preceq a \land \paren {b \preceq p}$ holds: :$p \not \prec'^+ p$. Since this is the case for all $p \in S$, $\prec'^+$ is antireflexive. {{qed|lemma}} === $\prec'^+$ is transitive === Let $p \prec'^+ q$ and $q \prec'^+ r$. Then there are three possibilities: $(1): \quad p \prec q$ and $p \prec r$ Because $\prec$ is transitive: :$p \prec r$ Thus: :$p \prec'^+ r$ $(2): \quad p \prec q$, $q \preceq a$, and $b \preceq r$ By Extended Transitivity, $p \prec a$. Since $p \prec a$ and $b \prec r$: :$p \prec'^+ r$ $(3): \quad p \preceq a$, $b \preceq q$, and $q \prec r$ By Extended Transitivity: :$b \prec r$ Therefore: :$b \preceq r$ Since $p \preceq a$ and $b \preceq r$: : $p \prec'^+ r$ Note that it is impossible to have $p \preceq a$, $b \preceq q$, $q \preceq a$ and $b \preceq r$. If that were so, $b \preceq q$ and $q \preceq a$ together would imply by transitivity that $b \preceq a$. But this contradicts the fact that $a$ and $b$ are $\preceq$-incomparable. Thus in all cases, $p \prec'^+ q$ and $q \prec'^+ r$ imply $p \prec'^+ r$, so $\prec'^+$ is transitive. {{qed|lemma}} Since $\prec'^+$ is transitive and antireflexive, it is by definition a strict ordering. === $\prec'^+$ is the transitive closure of $\prec'$ === First note that $\prec'$ is a subset of $\prec'^+$: If $p \prec' q$ then either :$p \prec q$ or :$p = a$ and $q = b$ If $p \prec q$ then $p \prec'^+ q$ by definition. If $p = a$ and $q = b$, then $p \preceq a$ and $q \preceq b$ by the definition of $\preceq$. Thus $p \prec'^+ q$ by the definition of $\preceq'^+$. Let $\mathcal R$ be a transitive relation containing $\prec'$. Let $p, q \in S$ and let $p \prec'^+ q$. Then either: :$p \prec q$ or :$p \preceq a$ and $b \preceq q$ If $p \prec q$, then by the definition of subset, $p \mathrel {\mathcal R} q$. Suppose instead that $p \preceq a$ and $b \preceq q$. By the definition of $\preceq'$, $a \preceq' b$. Since Reflexive Closure is Closure Operator, it is increasing, so $\preceq \subseteq \mathcal R^=$. Thus $p \mathrel {\mathcal R}^= a$, $a \mathrel {\mathcal R}^= b$, and $b \mathrel {\mathcal R}^= q$. Since $\mathcal R$ is transitive, $\mathcal R^=$ is as well, by Reflexive Closure of Transitive Relation is Transitive. Thus $p \mathrel {\mathcal R}^= q$. Since $\preceq'^+$ is antireflexive and we assumed $p \preceq'^+ q$, we conclude that $p \ne q$. Thus by the definition of reflexive closure: :$p \mathrel {\mathcal R} q$ Thus we have shown that $\preceq'^+$ is a subset of $\mathcal R$. Since this holds when $\mathcal R$ is any transitive superset of $\preceq'$, $\preceq'^+$ is the transitive closure of $\preceq'$. {{qed}}