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Asked: 2024-09-23T15:33:29+05:30 In:

How can I simplify the boolean expression A + B(A’B + C’B) + C? I’m looking for a step-by-step breakdown of the simplification process.

anonymous user

I’m working on simplifying a Boolean expression, and I could really use some help. The expression I’m dealing with is A + B(A’B + C’B) + C. It feels a bit overwhelming, and I want to make sure I simplify it correctly.

First off, I thought I could distribute B over the terms in the parentheses. If I do that, I’d end up with B*A’B + B*C’B. When I worked that out, I noticed that the B in B*A’B and B*C’B both have a common B, and that got me thinking about how I can bring in some other Boolean laws to make things easier.

Then I was looking into the term A’B and C’B. It struck me that since I have A + C in the original expression, I might be able to simplify that part where B is tied with those two terms. So, I wonder if A + A’B will simplify further. Maybe there’s a law I’m overlooking?

After breaking it down, I kept hearing that the consensus theorem might help here, which suggests that you can eliminate A’B in the presence of A. This could potentially clean up the expression a bit. But then, there’s that C hanging around. Should I incorporate that into my simplification somehow?

Once I simplify everything, I want to make it as clean and straightforward as possible. Ideally, I’d love it if I could spot any redundancies along the way, especially if I can eliminate any unnecessary terms.

So what do you all think? How would you approach simplifying this expression? I’d really appreciate a step-by-step walkthrough or any tips on the laws and rules that could apply here. I’m open to any insights or methods you might have in tackling expressions like this! Let’s see if we can get this all sorted out together.

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    2 Answers

    1. anonymous user



      Simplifying Boolean Expression

      Simplifying the Boolean Expression: A + B(A’B + C’B) + C

      It looks like you’re on the right track with this expression! Let’s break it down step by step:

      Step 1: Distribute B

      You mentioned distributing B over the parentheses:

          A + B(A'B + C'B) + C
    = A + BA'B + BC'B

      Now, since B is common in both BA’B and BC’B, that’s a nice observation!

      Step 2: Simplifying BA’B

      Notice here that BA’B simplifies to just AB because:

          BA'B = B(A'B) = AB

      That’s because anything AND-ed with itself simplifies to itself in Boolean algebra.

      Step 3: Combine Like Terms

      Your expression now looks like this:

          A + AB + BC'B + C

      Now focusing on BC’B, that’s still just BC’ because of the similar reasoning as before.

      Step 4: New Expression

      Now we have:

          A + AB + BC' + C

      Here’s where the consensus theorem you mentioned comes into play!

      Step 5: Apply Consensus Theorem

      The consensus theorem states that:

          A + A'B = A + B

      So we can apply it here:

          A + AB = A

      Now our expression simplifies further:

          A + BC' + C

      Step 6: Final Expression

      You could even notice that A + C is present in the original, so now the final simplified expression remains:

          A + C + BC'

      No further simplifications seem possible without losing valuable information from the expression. So you’ve done a great job simplifying it here!

      Tips on Future Simplifications:

      • Always look for common factors.
      • Using laws like Distribution, Absorption, and Consensus can save time.
      • Check for redundancy – if you have terms that don’t change the output, they can often be eliminated!

      Hope this helps you feel less overwhelmed! Boolean expressions can get tricky, but breaking them down step-by-step really helps!


    2. anonymous user


      To simplify the given Boolean expression A + B(A’B + C’B) + C, we start by applying distribution. The expression inside the parentheses becomes B(A’B) + B(C’B), which simplifies to BA’B + BC’B. Note that B*B = B, so we can rewrite it as BA’ + BC’. Now, incorporating this back into the original expression yields A + BA’ + BC’ + C. Next, we can use the consensus theorem, which states that A + A’B = A + B, allowing us to reduce A + BA’ to A + B.

      Now, the expression is simplified to A + B + BC’ + C. We can observe that C can absorb BC’ since if C = 1, then BC’ = 1 has no effect, and if C = 0, we only care about the first two terms, making B valid in that case as well. Thus, this simplifies further to A + B + C. Therefore, the final simplified expression is A + B + C, which is both clean and straightforward with no unnecessary terms.


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