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Re: Is it possible with Mathematica?big problem.thank you


Mathematica can do this:

Simplify[z + x*y == zP + xP*yP, z == zP && x*y == xP*yP]

True

Yet this gets us nowhere:

Simplify[(z == zP && x*y == xP*yP) || (z + x*y == zP + xP*yP)]

(z == zP && x y == xP yP) || x y + z == xP yP + zP

I think that's not surprising.

Bobby

On Sun, 16 Aug 2009 05:41:45 -0500, olfa <olfa.mraihi at yahoo.fr> wrote:

> hello mathematica community,
> is it possible with mathematica to simplify this expression:
> [(z == zP and x*y == xP*yP)or (z + x*y == zP + xP*yP)]
> into:
> (z + x*y == zP + xP*yP)
>
> in fact something in mathematica should detect automatically that (z
> == zP and x*y == xP*yP) implies (z + x*y == zP + xP*yP)and then
> simplify the expression above based on a theorem or rule that says
> when [a implies b] then [a or b] is simplified into b .
>
> hope my explanation is clear and thank you very much for your help.
>



-- 
DrMajorBob at bigfoot.com


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