# Under Which Algebra?

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

1. ### Lawrence D'OliveiroGuest

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

Lawrence D'Oliveiro, Nov 2, 2010

2. ### Patrick FitzGeraldGuest

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

3. ### bokGuest

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
4. ### Lawrence D'OliveiroGuest

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
5. ### EnkiduGuest

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
6. ### bokGuest

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
7. ### bokGuest

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
8. ### Lawrence D'OliveiroGuest

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 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.

Lawrence D'Oliveiro, Nov 8, 2010
9. ### EnkiduGuest

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 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 - NiccolÃ² di Bernardo dei Machiavelli.

The end excuses any evil - Sophocles

Enkidu, Nov 9, 2010