#497 f(i)==f(j) where i==j (is/equal fails)

closed
nobody
5
2006-12-19
2004-01-27
No

As a shorthand in this note, I'll use a==b to mean is
(equal(a,b)).

assume(equal(i,j))

x[i] == x[j] => Unknown NO!
f(i) == f(j) => Unknown NO!
sin(i) == sin(j) => Unknown NO!
sinh(i) == sinh(j) => True OK!!!!

It turns out that it has to do with the Increasing
property, even though of course that is completely
irrelevant:

declare(g,increasing)
g(i) == g(j) => true OK

Strangely, Decreasing doesn't work:

declare(h,decreasing)
h(i) == h(j) => Unknown ???

I guess that technically this is an enhancement request
(Unknown is always a valid way to give up), but....

Discussion

  • Robert Dodier

    Robert Dodier - 2006-07-22
    • labels: --> Lisp Core - Assume
     
  • Robert Dodier

    Robert Dodier - 2006-12-19

    Logged In: YES
    user_id=501686
    Originator: NO

    Fixed by r1.16 src/compar.lisp (by Barton Willis).

     
  • Robert Dodier

    Robert Dodier - 2006-12-19
    • status: open --> closed
     

Log in to post a comment.

Get latest updates about Open Source Projects, Conferences and News.

Sign up for the SourceForge newsletter:





No, thanks