Help with understing how this boolean expression was sinplified

[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"

 

[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 ?

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 don't know based on what/how he did it, but check it yourself: both
functions produce the same result (IE they are the same function).

What I'm trying to figure out based on what boolean algebra theorem the
expression was simplified?

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

 

[Active8]

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.

 

[Active8]

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

 

[Rather Play Pinball]

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

 

[mark thomas]

The artist formerly known as Maxim Vexler <hq4ever (at) 012 (dot) net (dot)
il> >" < 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 ?
|
| Appreciate the help,
| Maxim.
|
| Example was taken from "Art Of Assembly (DOS-16bit), Chapter 02, Page 48"

I'll give you a hint:

x' + xy
= x'(1 + y) + xy

 

[Rich Grise]

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.

My point exactly!

I don't know based on what/how he did it, but check it yourself: both
functions produce the same result (IE they are the same function).

What I'm trying to figure out based on what boolean algebra theorem the
expression was simplified?

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'

How does it work if you go,
a + a'b + bb' = ?

[by wtf?] a + b(a + a')
[by P5] a + b*1 =
[by Th4] = a + b

Cheers!
Rich

 

[Ratch]

"Maxim Vexler <hq4ever (at) 012 (dot) net (dot) il> >" <"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 =

There is mistake in step 4 and step 6. Using the absorption theorem:
x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
Ratch

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"

 

[Ratch]

hq4ever (at) 012 (dot) net (dot) il> wrote in message

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 =

There is mistake in step 4 and step 6. Using the absorption theorem:
x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
Ratch

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"

 

[Thaas]

"Maxim Vexler <hq4ever (at) 012 (dot) net (dot) il> >" <"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 =

There is mistake in step 4 and step 6. Using the absorption theorem:
x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
Ratch

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"

Step 6 should have been:

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

Since x + x'y = x + y:

7. a + (a + b) =

Since x + x = x:

8. a + b

In the quoted text the author would have us believe between steps 5
and 6 that:

a'b + aa' = b(a + a')

Which simplifies to:

a'b = b

A fallacy.

 

[Ratch]

"Maxim Vexler <hq4ever (at) 012 (dot) net (dot) il> >" <"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 =

There is mistake in step 4 and step 6. Using the absorption theorem:
x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
Ratch

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"

Step 6 should have been:

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

Since x + x'y = x + y:

If one knows that the identity x + x'y = x + y is true, then a + a'b =
a + b, and one does not need to go further. Ratch

 

[Thaas]

"Maxim Vexler <hq4ever (at) 012 (dot) net (dot) il> >" <"Maxim Vexler
<hq4ever (at) 012 (dot) net (dot) il> wrote in message
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 =

There is mistake in step 4 and step 6. Using the absorption theorem:
x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
Ratch

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"

Step 6 should have been:

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

Since x + x'y = x + y:

If one knows that the identity x + x'y = x + y is true, then a + a'b =
a + b, and one does not need to go further. Ratch

Yeah, I just crawled back out of bed on that realization. Still, the
fallacy remains. The author mistyped something between 5 and 6.

 

[Thaas]

"Maxim Vexler <hq4ever (at) 012 (dot) net (dot) il> >" <"Maxim Vexler
<hq4ever (at) 012 (dot) net (dot) il> wrote in message
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 =

There is mistake in step 4 and step 6. Using the absorption theorem:
x = x+xy=x+xy+xz+x(w+v+...) , we continue the correct way.

a+a'b= a+a'b+ab=a+b(a'+a)=a+b*1=a+b, which is what you are looking for.
Ratch

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"

Step 6 should have been:

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

Since x + x'y = x + y:

If one knows that the identity x + x'y = x + y is true, then a + a'b =
a + b, and one does not need to go further. Ratch

Yeah, I just crawled back out of bed on that realization. Still, the
fallacy remains. The author mistyped something between 5 and 6.

arrrgh! You're right, it was the use of aa' for 0 from step 4 to 5
that threw the proof offcourse. Should've stayed in bed.

 

[Fred Bloggs]

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]

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.