# Boolean Algebra, possibly a typo in my textbook?

Discussion in 'Math' started by gragg, Feb 15, 2011.

1. ### gragg Thread Starter New Member

Feb 15, 2011
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0
Hi and sorry for my first post being a question. I'm a student of Computer Engineering and I've had to do a lot of these problems in previous classes. I'm a little rusty on them, and was never all that good to begin with..but I'm pretty sure there is a typo in this question and was hoping someone could confirm.

Prove the identity of this equation by using algebraic manipulation.
ABC' + BC'D' + BC + C'D = B + C'D

The reason for my belief is that...well, like I said I'm not very good at this but I'm pretty sure there's no conceivable way to actually cancel out that A in the first term. That and, I googled the exact question and searched for it here, both returning results of the exact same question, only with the third term containing an A'.

Just looking for some confirmation on this, but I do have another problem that's been driving me a little nuts too and is definitely solvable.

AD' + A'B + C'D + B'C = (A' + B' + C' + D') ( A + B + C + D)

I started to use DeMorgan's theorem but that didn't get me very far...this is starred as an extra hard problem in my text so either it is, or just the idea of it being one has got me discouraged

Thanks for looking =]. And for future reference, should I put this in homework help next time?

Last edited: Feb 15, 2011
2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
I can vouch that the equation is correct. I have confirmed it through the truth table. But after more than half an hour of meddling, I can't manage to prove it through Boolean operations.

I haven't met anything like this before! The trap is that you can't remove the extra C'Ds in the end of each expression, as it contributes differently in each expression.

P.S. Homework Help is the home of any school/college/uni related question, so if it fits the description, post it over there.

gragg likes this.
3. ### gragg Thread Starter New Member

Feb 15, 2011
3
0
How frustrating! I appreciate your trying though. I will post the answer if I figure it out or obtain it somehow before someone else does.

4. ### gragg Thread Starter New Member

Feb 15, 2011
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0
I have discovered the answer to this problem, but I only follow it so far. If anyone would be so kind as to guide me from the indicated point, I would appreciate it.

B(AC' + C'D' + C) + C'D
B(C'(A + D') + C) + C'D
B(A + D' + C) + C'D
B(A + D' + C)(C'D)' + C'D **************************
B(A + D' + C)(C + D') + C'D
B(C + D') + C'D
B(C'D)' + C'D
B + C'D

From the asterisked line, I do not recall the rule that allows you to add the (C'D)'. I understand that it is the same thing as the D' + C part of the term is follows. If anyone can explain to me the process from that line down, or even just name the boolean algebra rules that were used - I would appreciate it greatly.

5. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
What your book does is valid, but it seems like a cheap trick to me. Still, the process is valid.

Here is what it does:
Suppose you have an expression F and a variable A.
Form the expression G=F+A.
When A=1, you know that G is guaranteed to be 1.
When A=0, G=F. But since you know that A=0, you can equally write F=GA'.

Is that clear?