Talk:Equidistance is Independent of Betweenness

I can't wait to see a proof of this; it appears that induction on the form of the formula containing $\mathsf B$ which is used to try to define $\equiv$ cannot possibly be a fruitful track to pursue. It is, however, for now the only way I know to prove things like this. --Lord_Farin 13:43, 25 January 2012 (EST)
 * It's too advanced for me to tackle by myself at this point, but I'll quote Givant (p. 200)

This can be proved by means of the well-known method of Padoa, indeed by exhibiting, for any given system, two relational structions $\langle S_1,\mathsf{B}_1,\equiv_1 \rangle$ and $\langle S_2,\mathsf{B}_2,\equiv_2 \rangle$ which are both models of the given system, and in which $S_1$ and $\mathsf{B}_1$ respectively coincide with $S_2$ and $\mathsf{B}_2$, while $\equiv_1$ and $\equiv_2$ do not coincide, i.e., for some $x,y,z,u$, one of the formulas $xy \equiv_1 zu$ and $xy \equiv_2 zu$ holds and the other fails. - Givant

Then he gives such models on pp. 201-202, which I don't feel comfortable putting up. Perhaps after I finish Linear Algebra (I just had my first class today). Feel free to get to it before I do. --GFauxPas 17:29, 25 January 2012 (EST)


 * That sounds rather crafted, but it is effective (as is obvious). Thanks for posting it. --Lord_Farin 17:49, 25 January 2012 (EST)