#505 assume conflates programming and math vrbls

open
nobody
5
2006-04-10
2004-02-01
No

Assume/is is not consistent in its treatment of
programming vs. mathematical variables

x:1$
assume(x<0) => Inconsistent
OK, this presumably is equivalent to assume(1<0)
assume('x<0) => Inconsistent
Questionable. Should 'x really refer to the
*programming
variable* x here even though it's quoted?

but

assume(y<0)$
y:1$ => no error
This is inconsistent with the situation above. But I
certainly don't want every variable assignment to be
checking the Assume database!
is(y<0) => false
OK, 1<0 is false.
is('y<0) => true
OK if 'y is treated as a *mathematical variable* and
uses the assume database, not the programming variable
y. But above, assume('x<0) treats them as the same.

Discussion

  • Stavros Macrakis

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    user_id=588346

    Clearer demonstration:

    assume(equal('r,0))
    r:10

    is(equal('r,0)) => true
    assume(equal('r,0)) => [inconsistent]
    assume(not(equal('r,0))) => [redundant]
    facts() => [ equal(r,0) ]

    is(equal('r,10)) => false
    assume(equal('r,10)) => [redundant]
    is(not(equal('r,10))) => true

     
  • Robert Dodier

    Robert Dodier - 2006-04-10
    • labels: --> Lisp Core - Assume
     
  • mikolg

    mikolg - 2007-11-03

    Logged In: YES
    user_id=1902378
    Originator: NO

    Maxima version: 5.13.0cvs
    Maxima build date: 17:3 11/2/2007
    host type: i686-redhat-linux-gnu
    lisp-implementation-type: GNU Common Lisp (GCL)
    lisp-implementation-version: GCL 2.6.7

    Original issue is already fixed by a someone. See code below.
    But sample code from the note "2004-02-03 03:43" gives the following error:

    is(equal('r,10)) => true
    is(not(equal('r,10))) => true

    The following code shows that origianl issue is fixed:
    (%i1) assume(y<0);
    (%o1) [y < 0]
    (%i2) y:1$
    (%i3) is(y<0);
    (%o3) false
    (%i4) is('y<0);
    (%o4) false

    The following code from comment "2004-02-03 03:43" shows that problem is changed a little

    (%i1) assume(equal('r,0));
    (%o1) [equal(r, 0)]
    (%i2) r:10;
    (%o2) 10
    (%i3) is(equal('r,0));
    (%o3) false
    (%i4) assume(equal('r,0));
    (%o4) [inconsistent]
    (%i5) assume(not(equal('r,0)));
    (%o5) [redundant]
    (%i6) facts();
    (%o6) [equal(r, 0)]
    (%i7) is(equal('r,10));
    (%o7) true
    (%i8) assume(equal('r,10));
    (%o8) [redundant]
    (%i9) is(not(equal('r,10)));
    (%o9) true

     
  • Stavros Macrakis

    Logged In: YES
    user_id=588346
    Originator: YES

    The last comment claims that "the original issue is fixed". But the example given shows that it is *not* fixed. The second case, is('y<0), should return true; instead, the 'y is being evaluated twice, giving 1. It takes double quoting to get it right: is('('y)<0) => true. (Two single quotes in a row (''y) are parsed differently, hence the parentheses.)

    The situation is still a mess, and the issue has not been fixed.

     

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