Theorem: A partially ordered set is totally ordered if it obeys the law of trichotomy.
Things I know:
- A relation on some set $A$ is said to be a partially ordered set if the relation is reflexive, anti-symmetric and transitive.
- The law of trichotomy states: for all $a$ and $b$ in the poset $A$, the law of trichotomy holds if exactly one of the following is true: $a < b$, $a = b$ or $b < a$.
- A partially ordered set is said to be totally ordered if given any two elements in A, either $a \leq b$ (less than/equal to) or $b < a$.
I restructured the theorem to match the form of $P$ implies $Q$. If a partially ordered set $A$ obeys the law of trichotomy, then it is totally ordered.
Proceed by direct proof. Let $A$ be any arbitrary but fixed partially ordered set that follows the law of trichotomy. Let $a$ and $b$ be ABF elements in $A$, such that exactly one of the following is true: $a < b$, $a = b$ or $b < a$. By the definition of the partially ordered set, the relation on $A$ is reflexive, anti-symmetric, and transitive.
How do I continue?