Help with understing how this boolean expression was sinplified

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?

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

:-)

| 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

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"

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

"Maxim Vexler >"

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

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.

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.

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

This is rewritten as:

1. ab + ab' + ab + a'b

and then

1. a(b+b') + (a+a')b

and then

1. a + b

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