MathGroup Archive 2006

[Date Index] [Thread Index] [Author Index]

Search the Archive

Boolean algebra

  • To: mathgroup at
  • Subject: [mg69373] Boolean algebra
  • From: "Menace" <menace232323 at>
  • Date: Sat, 9 Sep 2006 03:26:54 -0400 (EDT)

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?

  • Prev by Date: Re: Fit rectangle to set of points
  • Next by Date: General--Series Command uses 100% of CPU
  • Previous by thread: Re: Exporting text to (or in) vector graphics as "sentences"?
  • Next by thread: Re: Boolean algebra