       Re: Boolean algebra

• To: mathgroup at smc.vnet.net
• Subject: [mg69396] Re: Boolean algebra
• From: Peter Pein <petsie at dordos.net>
• Date: Sun, 10 Sep 2006 07:20:28 -0400 (EDT)
• References: <edtrbb\$lcb\$1@smc.vnet.net>

```Menace schrieb:
> 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?
>

Hi,

Simplify allows to set some Information as 2nd parameter:

Simplify[(b1 || a1) && (b1 || a2), ! (a1 && a2)]
evaluates to b1 as desired.

Regards,
Peter

```

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