Talk:Order-Extension Principle

Alternate (better?) proofs
One option is Wikipedia's proof at Szpilrajn extension theorem. --Dfeuer (talk) 20:20, 13 December 2012 (UTC)

I moved the mess to One-Point Extension of Ordering is Ordering. --Dfeuer (talk) 21:41, 13 December 2012 (UTC)

It would be very nice to add a proof from BPIT or an equivalent axiom. --Dfeuer (talk) 21:41, 13 December 2012 (UTC)

notation
Enlighten me: how can:
 * $\forall a, b \in S: a \preceq b \implies a \le b$

be interpreted in any other way than:
 * $\forall a, b \in S: \left({a \preceq b \implies a \le b}\right)$

?


 * It must be syntactically unambiguous before semantic analysis and filtering for sensibility. It could be interpreted as the result of changing variable names in the statement $(\forall x, y \in S: x \preceq y) \implies a \le b$. --Dfeuer (talk) 06:28, 17 May 2013 (UTC)


 * To clarify: it is much easier and faster to reach an understanding of what the statement says if it is parenthesized properly, though one can generally puzzle it out otherwise. --Dfeuer (talk) 06:31, 17 May 2013 (UTC)


 * A separate issue: the version of the finite set lemma that I put up is not intended to be transcluded on the front page&mdash;it covers strict orderings. Probably the thing to do is to rename it, and make a separate page for the strict version, which Proof 2 will then use in a trivial fashion as a lemma. --Dfeuer (talk) 06:33, 17 May 2013 (UTC)