Alan Bawden wrote:
> For `=' it is indeed not bogus. Unfortunately, substituting `<' for `=' in > the above equivalence results in something that -is- bogus. So while you > may be able to use this example to argue that `(=)' should return #T, you > can not use it to argue that `(<)' should return #T. > > Well, the transformation isn't supposed to definitively prove anything it's just an example that illustrates how the 0 and 1 arity cases can simplify reasoning. It seems like it would be instructive, given an externally supplied observation like "this pattern doesn't work for <", to creatively search for something that does work for < or, better, that works for both. And then there's <=. And all kinds of interesting questions arise about the logic of these things. So, you wanted a starting place for explaining things to the guy with "high school math" and maybe in this little eddie of the thread there is something.... -t _______________________________________________ r6rs-discuss mailing list [email protected] http://lists.r6rs.org/cgi-bin/mailman/listinfo/r6rs-discuss
