Under Which Algebra?

Discussion in 'NZ Computing' started by Lawrence D'Oliveiro, Nov 2, 2010.

  1. A + BC = (A + B)(A + C)
    Lawrence D'Oliveiro, Nov 2, 2010
    #1
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  2. On Tue, 02 Nov 2010 23:53:37 +1300, Lawrence D'Oliveiro
    <_zealand> wrote:

    >A + BC = (A + B)(A + C)



    The modified Gyzorcki-Tchesofen system

    Patrick
    Patrick FitzGerald, Nov 2, 2010
    #2
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  3. Lawrence D'Oliveiro

    bok Guest

    On 2/11/2010 11:53 p.m., Lawrence D'Oliveiro wrote:
    > A + BC = (A + B)(A + C)


    Boolean algebra.

    (A+B)(A+C) = A*A + A*(B+C) + B*C
    = A*(1+B+C) + B*C [1+B+C = 1]
    = A + BC
    bok, Nov 7, 2010
    #3
  4. In message <ib5hjj$u81$-september.org>, bok wrote:

    > On 2/11/2010 11:53 p.m., Lawrence D'Oliveiro wrote:
    >>
    >> A + BC = (A + B)(A + C)

    >
    > Boolean algebra.


    Yes, but ...

    > (A+B)(A+C) = A*A + A*(B+C) + B*C
    > = A*(1+B+C) + B*C [1+B+C = 1]
    > = A + BC


    .... why did you have to work it out in such a long way?

    A B C A + BC (A + B)(A + C)
    0 0 0 0 + 0 = 0 0.0 = 0
    0 0 1 0 + 0 = 0 0.1 = 0
    0 1 0 0 + 0 = 0 1.0 = 0
    0 1 1 0 + 1 = 1 1.1 = 1
    1 0 0 1 + 0 = 1 1.1 = 1
    1 0 1 1 + 0 = 1 1.1 = 1
    1 1 0 1 + 0 = 1 1.1 = 1
    1 1 1 1 + 1 = 1 1.1 = 1
    Lawrence D'Oliveiro, Nov 7, 2010
    #4
  5. Lawrence D'Oliveiro

    Enkidu Guest

    On 07/11/10 23:13, Lawrence D'Oliveiro wrote:
    > In message<ib5hjj$u81$-september.org>, bok wrote:
    >
    >> On 2/11/2010 11:53 p.m., Lawrence D'Oliveiro wrote:
    >>>
    >>> A + BC = (A + B)(A + C)

    >>
    >> Boolean algebra.

    >
    > Yes, but ...
    >
    >> (A+B)(A+C) = A*A + A*(B+C) + B*C
    >> = A*(1+B+C) + B*C [1+B+C = 1]
    >> = A + BC

    >
    > ... why did you have to work it out in such a long way?
    >
    > A B C A + BC (A + B)(A + C)
    > 0 0 0 0 + 0 = 0 0.0 = 0
    > 0 0 1 0 + 0 = 0 0.1 = 0
    > 0 1 0 0 + 0 = 0 1.0 = 0
    > 0 1 1 0 + 1 = 1 1.1 = 1
    > 1 0 0 1 + 0 = 1 1.1 = 1
    > 1 0 1 1 + 0 = 1 1.1 = 1
    > 1 1 0 1 + 0 = 1 1.1 = 1
    > 1 1 1 1 + 1 = 1 1.1 = 1
    >

    In which case it is more usual to write it:

    A ^ (B v C) = (A ^ B)V(A ^ C)

    Cheers,

    Cliff

    --

    The ends justifies the means - Niccolò di Bernardo dei Machiavelli.

    The end excuses any evil - Sophocles
    Enkidu, Nov 7, 2010
    #5
  6. Lawrence D'Oliveiro

    bok Guest

    On 7/11/2010 11:13 p.m., Lawrence D'Oliveiro wrote:
    > In message<ib5hjj$u81$-september.org>, bok wrote:
    >
    >> On 2/11/2010 11:53 p.m., Lawrence D'Oliveiro wrote:
    >>>
    >>> A + BC = (A + B)(A + C)

    >>
    >> Boolean algebra.

    >
    > Yes, but ...
    >
    >> (A+B)(A+C) = A*A + A*(B+C) + B*C
    >> = A*(1+B+C) + B*C [1+B+C = 1]
    >> = A + BC

    >
    > ... why did you have to work it out in such a long way?
    >
    > A B C A + BC (A + B)(A + C)
    > 0 0 0 0 + 0 = 0 0.0 = 0
    > 0 0 1 0 + 0 = 0 0.1 = 0
    > 0 1 0 0 + 0 = 0 1.0 = 0
    > 0 1 1 0 + 1 = 1 1.1 = 1
    > 1 0 0 1 + 0 = 1 1.1 = 1
    > 1 0 1 1 + 0 = 1 1.1 = 1
    > 1 1 0 1 + 0 = 1 1.1 = 1
    > 1 1 1 1 + 1 = 1 1.1 = 1
    >


    Under what algebra is 8 less than 3?
    bok, Nov 8, 2010
    #6
  7. Lawrence D'Oliveiro

    bok Guest

    On 7/11/2010 11:13 p.m., Lawrence D'Oliveiro wrote:
    > In message<ib5hjj$u81$-september.org>, bok wrote:
    >
    >> On 2/11/2010 11:53 p.m., Lawrence D'Oliveiro wrote:
    >>>
    >>> A + BC = (A + B)(A + C)

    >>
    >> Boolean algebra.

    >
    > Yes, but ...
    >
    >> (A+B)(A+C) = A*A + A*(B+C) + B*C
    >> = A*(1+B+C) + B*C [1+B+C = 1]
    >> = A + BC

    >
    > ... why did you have to work it out in such a long way?


    The steps I outlined were intended to be an informal proof loosely based
    on the laws or theorems of Boolean algebra and not how I 'worked it out'.

    For example, that [1+B+C = 1] was based on an application of the 'law of
    identity' or 'a term OR ed with 1 equals 1'.
    bok, Nov 8, 2010
    #7
  8. In message <ib86de$cnr$-september.org>, bok wrote:

    > The steps I outlined were intended to be an informal proof loosely based
    > on the laws or theorems of Boolean algebra and not how I 'worked it out'.


    But it’s interesting that you assumed distributivity of AND over OR in order
    to prove distributivity of OR over AND. What makes one more valid than the
    other? Otherwise, you were just offering a circular argument.
    Lawrence D'Oliveiro, Nov 8, 2010
    #8
  9. Lawrence D'Oliveiro

    Enkidu Guest

    On 08/11/10 20:54, Lawrence D'Oliveiro wrote:
    > In message<ib86de$cnr$-september.org>, bok wrote:
    >
    >> The steps I outlined were intended to be an informal proof loosely based
    >> on the laws or theorems of Boolean algebra and not how I 'worked it out'.

    >
    > But it’s interesting that you assumed distributivity of AND over OR in order
    > to prove distributivity of OR over AND. What makes one more valid than the
    > other? Otherwise, you were just offering a circular argument.
    >

    That's what algebra is about. Assumptions called axioms lead to
    theorems. Of course, he didn't explicitly state his theorems.

    Cheers,

    Cliff

    --

    The ends justifies the means - Niccolò di Bernardo dei Machiavelli.

    The end excuses any evil - Sophocles
    Enkidu, Nov 9, 2010
    #9
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