Simplifications
----------------

Simplification levels:
0 Do not simplify.
1 add real only to real   and  int only to int.
2 Same as 1, but integers are added to reals. (real1=int+real2)
3 Convert all integers to real, and then do 1.

SimplifyConstants:
If Mode=0: only check integrity


- Evaluates all real constants, including things like Sin(5.0)
- Evaluates Real


                                  CO
                                /   \
                               /     \
                             A         B
                           /    \      /  \
                         /       \   /     \
                        C         D  E      F


Node types:

        Ci constant             (nodetype=iconstnode)
        Cr real constant,       (nodetype=constnode)
        Cy expression           (ExprIsConstant IN Flags, things like Sin(5) or even 4 DIV 6 if integer to real is off)
        Cn is any of the three above constant types
        CO= Commutative Operator (mul, add)
        X  Any other expression,


        Constants always have to be arranged Ci<Cr<Cy<

if A <> CO then (C and D have no relevance)
if B <> CO then (E and F have no relevance)

          A   C    D   B    E   F
action    Cn  -    -   CO   Cn  Cn               (killed in SimplifyConstants}
          Cn  -    -   CO   X   Cn               (Changed to E=C, F=X in killed in SimplifyConstants)
          Cn  -    -   CO   Cn   X               (if A=Cx and E=Ci then swap(A,E))
          X   -    -   CO

----------------
(from an older version of this doc:)

 A           D                  Action
Xcomm    <>Xcomm          Process [b c d]
Xcomm      Xcomm          Process [b c e f]
<>Xcomm    Xcomm          Process [a e f]
<>Xcomm  <>Xcoom          Process [a d]


How to process:

If Simplicationlevel<>0 then
 begin
   if (Simplicationlevel=3) or ((simplicationlevel=2) and (Cr in [])) then
    {convert all Ci to Cr}
   If more than one Ci in [] then
    {addall Ci's to one Ci}
   If more than one Cr in [] then
    {addall Crs to one Cr}
 end;
{determine how many elements in set left. (in practice kept track of in code
above)}
If we have only one xcomm on the right, xchg right and left}

{Rearrange set so that
Ci < Cr < Cx < X

#nodes     nodes filled:
0          Not possible.
1          Root node only, but not possible (cases that could lead to this
            are covered by the standard simplifications)
2          A, D
3          A, e f
4          b c e f

TreeType:            0    A d
                     1    b c d
                     2    A e f
                     3    b c e f

}