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MathGroup Archive 2006

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Re: Boolean algebra

  • To: mathgroup at smc.vnet.net
  • Subject: [mg69395] Re: Boolean algebra
  • From: p-valko at tamu.edu
  • Date: Sun, 10 Sep 2006 07:20:23 -0400 (EDT)
  • References: <edtrbb$lcb$1@smc.vnet.net>

I am not sure Simplify is up to the task, but for not too large
problems you may find useful the following:

prelims = {(a1 && a2) == False, (a1 || a2) == True};
f = (b1 || a1 || c1) && (b2 || a2 || a1);

vars = DeleteCases[DeleteCases[Union[Cases[prelims, x_Symbol,
Infinity]], True], False];
ins = FindInstance[prelims, vars, Booleans, 2^30];
{f /. ins, ins} // TableForm

The result is:
{
{True,   {a1->True,a2->False}},
{b1||c1, {a1->False,a2->True}}
}

so you can conclude that
either      a1||b1||c1      or     Not[a2] || b1 || c1
is a good representation.

Regards
Peter


Menace wrote:
> I'd like to solve the following problem in Mathematica,
>
> Given are the following preliminaries:
> (a1 || a2)=True
> (a1 && a2) = False
>
> Then, given the following conjunctive normal form:
>
> (b1 || a1) && (b1 || a2)
>
> This can be simpliefied to: b1 || (a1 && a2)
>
> Given the prelininaries I'd like Mathematica to simplify this to: b1.
> However, I cant figure out how to do this. I tried the following:
>
> prelims := {(a1 && a2) -> False, (a1 || a2) -> True};
> f1 := (b1 || a2) && (b1 || a1);
> Simplify[f1] /. prelims
>
> and this seems to work. However
>
> f2 := (b1 || a1 || c1) && (b2 || a2 || a1);
> Simplify[f2] /. prelims
>
> evaluates to: (b1 || a1 || c1) && (b2 || a2 || a1) instead of b1 || a1
> || c1.
> The reason of this seems to be the order in which a2 and a1 occur in
> f2. How can I make sure it works regardless of the order of a1 and a2?


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