I have noticed some of the newer textbooks include this statement in their list of identities. Yes, it's possible to get it through different methods, but I think since it's used so frequently, they've started teaching it as being an identity.
Boolean algebra
- Thread starter lusito92
- Start date
Rich (BB code):
A + AB = A(1) + A(B)
= A(1+B)
= A(1)
= A
A + A'B = A(1) + A'B
= A(1+B) + A'B
= A(1) + AB + A'B
= A + AB + A'B
= A + (A+A')B
= A + (1)B
= A + BA' + AB = A' + B
You can always imagine replacing A' with C so that you have
C + C'B = C + B = A' + B
Note that A and B can be anything, so
wx'yz + w(x'y)'z can be written as
(x'y)(wz) + (x'y)'(wz)
Now A = x'y and B = wz, so the result is
wx'yz + wz
You can find this helpful in problems like
x'+y' + xyz'= x'+y'+ z'
If you apply DeMorgan's to the first two terms.
Hmmm... interesting... I guess I just need to buy a new textbook!
I don't have a problem with that, per se, as long as the students have to prove it before they can then use it as an identity. The same is true for all of the identities. Unlike in arithmetic where you start of with some axioms you just have to accept (and the deeper you go in mathematics the fewer axioms you have because you gain the ability to prove some of the things that us mere mortals have to merely accept), in Boolean Algebra you have three definitions, namely what the operators NOT, AND, and OR do, and from there everything can (and should) be proven explicitly using only the definitions and whatever ever identities have been proven previously.
Yes, I agree. The people teaching the students, and/or the books themselves, should definitely show the student how they arrived to that "identity". It must be derived first, and once the student understands where it came from, they can use it freely.
I'm with you 100%.
Matt
Do what WBahn suggested in post # 22
We aren't just going to give you the answers. You have been given enough information to solve this on your own. Loook at WBahn's post, read it over until you understand it. I know you can do it.
Matt
The term XY'' is probably not what you mean. This, as written, means
XY'' = X((Y')') = XY
Like exponentiation, the ' operator binds to the single factor immediately to its left.
I think you mean (XY)', in which case you are correct.
when I do '' I mean NOT XY that is (XY)' right? the whole term I want to say NOT, but Im doing in the paper with the bars but I cant do it here
yes, use parenthesis for grouping.
A' + B' = (AB)' : the whole thing, A ANDed with B, is NOTed
Correct.
When you write XY" (double quotes), that is pretty much indistinguishable from XY'' (two single quotes). So don't use doublle quotes for anything (in this context) and use parens to do the grouping and to make the grouping unambiguous. So even if you mean XY' to only have Y inverted, write it as X(Y').
When you use an overbar, the overbar can serve to do the grouping as well, though you have to be careful not to accidentally join two bars together -- and parens are very useful to prevent that from happening, as well.
Imagine (X')(Y') when draw with overbars. It is very, very easy to mess up and see a single overbar and take the term to be (XY)', which is NOT the same.
Ok thanks just the last question,
(C'AB)' + (A'+B)' = 1
(C')' + A' + B' + (A')' + B' = 1 <-- I separate every single term as (AB') = A' + B'
C + A' + B' + A + B' = 1 <-- I used the identity (X')' = X
C + 1 + B' = 1 <-- I used the identities A' + A = 1 and, B' + B' = B'
(C + B') + 1 = 1 <-- the same as x + 1 = 1 is this correct?
http://forum.allaboutcircuits.com/editpost.php?do=editpost&p=572420
(C'AB)' + (A'+B)' = 1
(C')' + A' + B' + (A')' + B' = 1 <-- I separate every single term as (AB') = A' + B'
C + A' + B' + A + B' = 1 <-- I used the identity (X')' = X
C + 1 + B' = 1 <-- I used the identities A' + A = 1 and, B' + B' = B'
(C + B') + 1 = 1 <-- the same as x + 1 = 1 is this correct?
http://forum.allaboutcircuits.com/editpost.php?do=editpost&p=572420
Yes. And you didn't have to show that B' + B' goes to B' since once you had shown that you had 1 + (a bunch of stuff) you are done.
You seem to be getting it down pretty good.
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