Definition talk:Strict Ordering

One can derive the asymmetric result from irreflexivity and transitivity alone:

$\neg a R a$, and $( a R b \land b R a ) \implies a R a$, so by modus tollens, $\neg ( a R b \land b R a )$ and thus $( a R b \implies \neg b R a )$ asalmon

Many thanks. The proof has been added. --prime mover 12:19, 25 November 2011 (CST)

Because the converse of above statement also holds (we have by asymmetry $a R a \implies \neg a R a$, so $\neg a R a$), I have added an alternative definition part. The page Equivalence of Definitions of Strict Ordering can probably supersede Strict Ordering is Asymmetric. --Lord_Farin 12:45, 25 November 2011 (CST)