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# Under Which Algebra?

Lawrence D'Oliveiro
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 11-02-2010
A + BC = (A + B)(A + C)

Patrick FitzGerald
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 11-02-2010
On Tue, 02 Nov 2010 23:53:37 +1300, Lawrence D'Oliveiro
<(E-Mail Removed)_zealand> wrote:

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

The modified Gyzorcki-Tchesofen system

Patrick

bok
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Posts: n/a

 11-07-2010
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

Lawrence D'Oliveiro
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 11-07-2010
In message <ib5hjj\$u81\$(E-Mail Removed)-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

Enkidu
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Posts: n/a

 11-07-2010
On 07/11/10 23:13, Lawrence D'Oliveiro wrote:
> In message<ib5hjj\$u81\$(E-Mail Removed)-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 - Niccolň di Bernardo dei Machiavelli.

The end excuses any evil - Sophocles

bok
Guest
Posts: n/a

 11-08-2010
On 7/11/2010 11:13 p.m., Lawrence D'Oliveiro wrote:
> In message<ib5hjj\$u81\$(E-Mail Removed)-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
Guest
Posts: n/a

 11-08-2010
On 7/11/2010 11:13 p.m., Lawrence D'Oliveiro wrote:
> In message<ib5hjj\$u81\$(E-Mail Removed)-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'.

Lawrence D'Oliveiro
Guest
Posts: n/a

 11-08-2010
In message <ib86de\$cnr\$(E-Mail Removed)-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 itâ€™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.

Enkidu
Guest
Posts: n/a

 11-09-2010
On 08/11/10 20:54, Lawrence D'Oliveiro wrote:
> In message<ib86de\$cnr\$(E-Mail Removed)-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 itâ€™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.
>

theorems. Of course, he didn't explicitly state his theorems.

Cheers,

Cliff

--

The ends justifies the means - NiccolĂ˛ di Bernardo dei Machiavelli.

The end excuses any evil - Sophocles