Help with understing how this boolean expression was sinplif

[Maxim Vexler]

Please help me understanding how this boolean expression was simplified

1. ab + ab' + a'b =
2. a(b+b') + a'b =
3. a*1 + a'b =
4. a + a'b + 0 =
5. a + a'b + aa' =
6. a + b(a + a') =
7. a + b*1 =
8. a + b

I did a truth table on both function [1] & [8], they produce the same
result so they are the same function, but I can't seem to understand
using what theorem / postulate can move from line 5 to 6 ?

Appreciate the help,
Maxim.

Example was taken from "Art Of Assembly (DOS-16bit), Chapter 02, Page 48"

 

[Active8]

On Wed, 22 Dec 2004 22:19:14 +0200, Maxim Vexlerat012 dotnet dot
wrote:

Please help me understanding how this boolean expression was simplified

1. ab + ab' + a'b =
2. a(b+b') + a'b =
3. a*1 + a'b =
4. a + a'b + 0 =
5. a + a'b + aa' =
6. a + b(a + a') =
7. a + b*1 =
8. a + b

I did a truth table on both function [1] & [8], they produce the same
result so they are the same function, but I can't seem to understand
using what theorem / postulate can move from line 5 to 6 ?

since aa' = 0,

then baa' = 0

so you put the b there so you can use that, wtf?, associative
property? cute.


--
Best Regards,
Mike

 

[Active8]

On Thu, 23 Dec 2004 00:19:28 +0200, Maxim Vexlerat012 dotnet dot
wrote:

Active8 wrote:
On Wed, 22 Dec 2004 22:19:14 +0200, Maxim Vexlerat012 dotnet dot
wrote:

since aa' = 0,
then baa' = 0

so you put the b there so you can use that, wtf?, associative
property? cute.

My point exactly!

I'm wrong, well, I was right, but it doesn't justify the wtf? step.

a'b + aa' =
a'b + baa' =
b(a' + aa') not b(a' + a) or b(a + a')

don't be surprised if there was a misprint. Doesn't make sense,
though. Somebody will be along to either straighten this out or
confirm wtf?

Look :
P4: ab+ac=a(b+c)
P5: a+a'=1, a*a'=0
Th3: a+0=a
Th4: a*1=a

F=ab + ab' + a'b
[by P4] = a(b+b') + a'b
[by P5] = a*1 + a'b
[by Th4] = a + a'b
[by Th3] = a + a'b + 0
[by P5] = a + a'b + aa'
[by wtf?] a + b(a + a')
[by P5] a + b*1 =
[by Th4] = a + b

--
Best Regards,
Mike

 

[Rather Play Pinball]

Can't help you, too much eggnog to focus...




> Please help me understanding how this boolean expression was simplified

 

[Fred Bloggs]

Maxim Vexler wrote:

5. a + a'b + aa' =
6. a + b(a + a') =

I did a truth table on both function [1] & [8], they produce the same
result so they are the same function, but I can't seem to understand
using what theorem / postulate can move from line 5 to 6 ?

He must be relying on some fundamental results like a=a+ab and a=a+aa',
which are fairly self-evident and require little reasoning beyond
definition manipulation. Then 5) becomes: a+ ab + a'b + aa' when those
substitutions are made, and 6) obviously follows. There is no error,
typo, or illogic- it was a departure from one simple substitution per
line that threw you.

 

[Fred Bloggs]

Maxim Vexler wrote:

Please help me understanding how this boolean expression was simplified

1. ab + ab' + a'b =

Any term in the summation can be replicated - so replicate ab.

This is rewritten as:

2. ab + ab' + ab + a'b

and then

3. a(b+b') + (a+a')b

and then

4. a + b

Example was taken from "Art Of Assembly (DOS-16bit), Chapter 02, Page 48"

That book was written by a moron, throw it in the trash.