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De Morgan

$$y=\overline{a+\overline{b(\overline{c+d})}}+\bar{b}$$ $$y=\bar{a}(\overline{\overline{b(\overline{c+d})}})+\bar{b}$$ $$y=\bar{a}(b(\overline{c+d}))+\bar{b}$$ $$y=\bar{a}(b\overline{cd})+\bar{b}$$ $$y=\bar{a}b\overline{cd}+\bar{b}$$

This is as far as I've got on my simplification:

$$x+\bar{x}y=1+0\cdot1~or~0+1\cdot1$$

So it might be

$$1+0~or~0+1=1$$

But

$$\bar{a}b\overline{cd}+\bar{b}=\overline{acd}(b+\bar{b})=\overline{acd}\cdot 1=\overline{ac}d$$

I can't understand why it's will be \$y=\overline{acd}+b\$, so how I must to minimize this \$b\$?

Maybe I solved it! a'bc'd'+b'= (a+b'+c+d)'+b'= ((a+b'+c+d)b)'= (ab+b'b+cb+db)'= (ab+1+cb+db)'= ((a+c+d)b)'= (a+c+d)'+b'= =a'c'd'+b' Is it correct?

Ciao
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  • Ok thank you for editing your problem. Now I need a little clarification. What do you mean by your last sentence, "I can't [understand] why..." I don't understand what you're trying to say nor how you're relating this to your question. –  Jul 02 '18 at 15:36
  • I'm voting to close this question as off-topic because homework needs and attempt at a solution (even if you do have the answer) – Voltage Spike Jul 02 '18 at 15:41
  • @KingDuken the solution of the function should be this y=a'c'd'+b but I don't know how to get it.. I can't undersand how to semplify b+b' so why b it's desaper? Cos for the rule a+a'=1 so it's must be just a'c'd' but in the correct answer of this exercise is a'c'd'+b – Ciao Jul 02 '18 at 15:43
  • @laptop2d A small question. If one has done his homework but can't solve it. What should he do? Do not ask for the explanation of why it's wrong? – Ciao Jul 02 '18 at 15:46
  • You just need the intuitive connection...think about it this way. If b is false, the result will be unconditionally true due to the right term. Therefore, the only time the left term is needed is when b is true. So if the left term will affect the result in any way, it can be assumed that b is true--so it's optimized out with that assumption. – Cristobol Polychronopolis Jul 02 '18 at 15:49
  • @CristobolPolychronopolis Ok I don't understand well. For addition 0+1=1 For moltiplication 01=0. Semplifcation x+x'y 1+01 or 0+1*1 So it might be 1+0 or 0+1 (it's the same)=1. But a'bc'd'+b'=a'c'd'(b+b')=a'c'd'(1)=a'c'd' – Ciao Jul 02 '18 at 16:18
  • Get as far as you can but provide an answer, and tell us where your stuck. – Voltage Spike Jul 02 '18 at 16:18
  • Suppose I tell you that I'll be there if (a) my car starts, or if (b) my car doesn't start but I can get an Uber. Logically, I'm saying the same thing if I say "I'll be there if my car starts or I can get an Uber." I don't need the additional condition specifying that I'm getting an Uber only if my car doesn't start...if my car started, I'll be there Uber or not. The result is whether I show up, not how I got there. See the analogy? – Cristobol Polychronopolis Jul 02 '18 at 16:57
  • @CristobolPolychronopolis ok I understood your logic with the car, but not with the semplification. How can I say that I'll remove b because I've got a b '? – Ciao Jul 02 '18 at 17:37
  • @laptop2d My stuck is on this semplification. I can't understand how to do it well. So I don't know why my answer it's wrong. – Ciao Jul 02 '18 at 17:39
  • Maybe I solved it! a'bc'd'+b'=(a+b'+c+d)'+b'=((a+b'+c+d)b)'=(ab+b'b+cb+db)'=(ab+1+cb+db)'=((a+b+d)b)'=(a+b+d)'+b'=a'b'd'+b' Is it correct? – Ciao Jul 02 '18 at 18:36
  • No a'b'd'+b' is not correct. – StainlessSteelRat Jul 06 '18 at 21:45
  • @StainlessSteelRat a'c'd'+b' I wrote down wrongly the solution and didn't see it.. – Ciao Jul 06 '18 at 22:04
  • Voting to close, as this is not an electronics question. While boolean algebra is certainly necessary for designing digital logic electronics, pure math questions are better off at https://math.stackexchange.com/. – Lundin Nov 12 '18 at 10:51

1 Answers1

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De Morgan $$y=\overline{a+\overline{b(\overline{c+d})}}+\bar{b}$$ $$y=\bar{a}(b(\overline{c+d}))+\bar{b}$$ You are correct up to this point. Based on the comments, you are correct for the rest, but you fooled up the mathjax syntax as you deMorganed. Corrected we have: $$y=\bar{a}b\bar c \bar d+\bar{b}$$

You are also headed in the correct direction: $$X+\bar{X} Y = X + Y$$

Which is the redundancy law. The inclusion of \$X\$ means the \$\bar X\$ in \$\bar X Y\$ is redundant.

StainlessSteelRat
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