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

[Albert G. <Dr....@t-...>]

Eddie Rucker wrote:
> 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;

Yeah, that's exactly how to do it. (In fact, the Q interpreter does 
exactly the same for non-linearities, it's only done automatically 
behind the scenes.)

> (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

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;

Cheers,
Albert

-- 
Dr. Albert Gr"af
Dept. of Music-Informatics, University of Mainz, Germany
Email:  Dr....@t-..., ag...@mu...
WWW:    http://www.musikinformatik.uni-mainz.de/ag