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)

If it needs to be transcluded into the proof page, it needs to be transcluded into the main page. End of. If not, then don't transcluded it. End of end of. --prime mover (talk) 06:42, 17 May 2013 (UTC)