Re: Is it possible with Mathematica?big problem.thank you
- To: mathgroup at smc.vnet.net
- Subject: [mg102635] Re: [mg102602] Is it possible with Mathematica?big problem.thank you
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Mon, 17 Aug 2009 04:05:36 -0400 (EDT)
- References: <200908161041.GAA01276@smc.vnet.net>
- Reply-to: drmajorbob at bigfoot.com
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
- References:
- Is it possible with Mathematica?big problem.thank you
- From: olfa <olfa.mraihi@yahoo.fr>
- Is it possible with Mathematica?big problem.thank you