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Under Which Algebra?

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


 
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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

 
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Enkidu
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      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
 
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bok
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      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?
 
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bok
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      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'.



 
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Lawrence D'Oliveiro
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      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 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.
 
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Enkidu
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      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 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
 
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