Algebra Confusion: Proving (A+C)*(A'+B)*(B+C) = B*C

In summary: A'*B)=> B=A' (2)from (1) and (2) we get A'=BIn summary, the given statement (A*C)*(A'+B)*(B+C) = B*C is proven using the commutation and distribution properties of boolean algebra, along with the substitution A*B = 0. The final step involves showing that (A'+B) = B, which is done by considering the two possible values for A and B and using the Idempotence property. The proof also includes a demonstration of A' = B, given A*B = 0 and A+B = 1.
  • #1
zenity
13
0
Hi there,

I've never done this type of Algebra... and it's a bit confusing. I've completed this question, but I don't know if I took the right steps...

*=AND, +=OR, '=NOT

If A*B = 0, A+B=1

Prove: (A+C)*(A'+B)*(B+C) = B*C

I expanded it a bit and applied some of the rules. Can someone explain it step by step? I got the answer but I'm a bit confused at how I got the answer.

Thanks!
 
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  • #2
(A+C)*(A'+B)*(B+C)
= (A+C)*(B+C)*(A'+B) (commutation)
= ((A*B)+C)*(A'+B) (distribution)
= (0+C)*(A'+B) (substitution A*B = 0)
= C*(A'+B)

Now I don't know what "rules" you know or how formal you have to be, but you should be able to see that (A'+B) = B given A+B = 1. Clearly, if A'+B = 0, then B = 0. If A'+B = 1, we want B to also necessarily be 1. But suppose it's not, so suppose A'+B=1 but B=0. Then A' = 1, which means A=0. So both A and B would be 0, but we're given A+B = 1, which is a contradiction, so B does indeed have to be 1 if (A'+B) is 1.
 
  • #3
Mhm.. I understood all the steps until the last one...this is what I think...

Since AB = 0 and A+B=1

That leaves us with two choices for A or B. A can either be 0 or 1, and B must be the exact compliment.

Now we're stuck with C(A'+B). *Here's where I'm a bit confused*

If A is 0, B will be 1, then A' + B = 1
If A is 1, B will be 0, then A' + B = 0

From which of these two can we say C*(A'+B) = B*C?

in otherwords... how does (A'+B) = B?
 
Last edited:
  • #4
Well I've already told you how (A' + B) = B. However, if you didn't follow the explanation, then consider what you said yourself:

A can either be 0 or 1, and B must be the exact compliment.

So B = A', right? So (A' + B) = B + B = B (Idempotence).
 
  • #5
Ah... I think I understand what I just said. A'=B.. then you'd just have to sub it in.

Man, boolean algebra is mighty weird! I just started a week ago, still getting the hang of it. Thanks for your help!
 
  • #6
how about a proof of A'=B

given A*B=0
(or with A')
A'+A*B=A'
(and with B)
A'*B + A*B*B=A'*B
=> (A'+A)*B=A'*B
=> B=A'*B (1)

given A+B=1
(and with A') => A'*A + A'*B = A'
=> A'*B = A'

(substitute from (1) ) => B=A'
 

1. What is the purpose of proving the equation (A+C)*(A'+B)*(B+C) = B*C in algebra?

The purpose of proving this equation is to demonstrate the validity and accuracy of the algebraic expression. By proving it, we can show that the equation is true for all possible values of the variables A, B, and C, and it follows the rules and properties of algebra.

2. Can you explain the steps involved in proving (A+C)*(A'+B)*(B+C) = B*C in algebra?

In order to prove this equation, we need to use the distributive property and other algebraic rules and properties. We can expand the expressions using the distributive property and then simplify the terms by combining like terms. Eventually, we will reach a point where the left and right sides of the equation are equal, proving its validity.

3. What are the common mistakes to avoid while proving (A+C)*(A'+B)*(B+C) = B*C in algebra?

Some common mistakes to avoid include forgetting to use the distributive property, making errors while simplifying terms, and not considering all possible values of the variables. It is also important to double-check each step and ensure that the equation is being manipulated correctly.

4. How can I apply the concept of proving equations in my everyday life?

Proving equations in algebra is a fundamental skill that can be applied in various situations in everyday life. For example, it can be used to calculate and verify discounts, interest rates, and other financial transactions. It can also be applied in science and engineering to solve complex problems and validate equations and formulas.

5. Are there any tips or tricks to make proving (A+C)*(A'+B)*(B+C) = B*C easier?

One helpful tip is to work backwards from the desired outcome. Instead of trying to manipulate the left and right sides of the equation separately, start with the end goal and work backwards to see what steps are needed to reach it. It is also important to practice and familiarize oneself with the rules and properties of algebra to make the process easier.

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