Re: Boolean functions' invariant properties.

From: conesetter (conesetter_at_btopenworld.com)
Date: 05/29/04

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    Date: 29 May 2004 04:30:26 -0700

    "Mikito Harakiri" <mikharakiri@iahu.com> wrote in message news:<W1Mtc.19$qk1.182@news.oracle.com>...
    > "conesetter" <conesetter@btopenworld.com> wrote in message
    > news:8083d02b.0405280912.729b90bc@posting.google.com...
    > > An example: ((wAx)Uy)Az. The tabular form is
    > > w x y z
    > > w A U A
    > > x A U A
    > > y U U A
    > > z A A A
    > > Complete sets under A are w.x,z and y,z giving the DNF
    > > (wAxAz)U(yAz).
    > > Complete sets under U are w,y and x,y and z giving the CNF
    > > (wUy)A(xUy)Az.
    >
    > What is tabular form for
    >
    > (xAy)U(yAz)U(zAx)
    >
    > ?
      Your formula has repeated variables which have to be replaced as
    explained in my post. So it becomes say (xAs)U(yAt)U(zAr) which leads
    without difficulty to a six-by-six table with one A and four U's in
    each row and column. Function-structure is separated from conditions
    on variables which in this case are x=r, y=s, z=t.

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