pure-lang-users

Re: [pure-lang-users] equations in Q vs Pure

[Eddie R. <er...@bm...>]

Hi guys!

In Q we could write:

(X+A)*(X+B) = X^2+(A+B)*X+(A*B);

as an equation and Q would happily reduce 

(X+2)*(X+4);

to

X^2+6*X+8

which was nice. But in Pure, since we cannot repeat a variable on the
left hand side, we cannot do any of these nice equations:

X*Y+X*Z = X*(Y+Z);
X/X = 1 if X != 0;
X+N*X = (N+1)*X;

etc...

Albert, I know you made an === operator but how are we suppose to use it
to handle this sort of stuff?

Ok. I've figured I need to do stuff like 

(a*b)+(c*d) = (a+c)*b if b === d;
a*b=a^2 if a===b;
(a+b)*(c+d) = a*c+a*d+b*c+b*d;
a*b = b*a if numberp b;

Then
  > (x+4)*(x+3)
  x^2+3*x+4*x+12

Why doesn't the rule '(a*b)+(c*d) = (a+c)*b if b === d;' reduce it
further to 

x^2+7x+12

since 
  > 3*x+4*x

reduces to
  7*x

e.r.

Re: [pure-lang-users] equations in Q vs Pure

[Eddie R. <er...@bm...>]

On Sat 26/07/08  2:49 AM , Albert Graef Dr....@t-... sent:
> Because that rule doesn't apply. Note that x^2+3*x+4*x+12 === 
> (x^2+3*x)+4*x+12 !== x^2+(3*x+4*x)+12. Pure doesn't do any rewriting 
> modulo AC (associativity+commutativity). Neither does Q, it will return
> exactly the same result with your equations.
> 
> What you can do to make this example work is add the following rule:
> 
> x+(a*b)+(c*d) = x+(a+c)*b if b === d;

Ah! Ok, the reason why I was getting the answer in Q is because I also had the associative rules of multiplication and addition.
(a+b)+c = a+(b+c);

It works but x^2+(7x+12) is kind of annoying to look at. The rule you give above makes the answer look nice.

Thanks,

e.r.