[r6rs-discuss] Comparison procedures' number of arguments

[Thomas Lord]

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