I have to get this expression A’B’C + A’BC + A’BC’ + AB’C + ABC to A'B+C. I did this but I can't finish it, I don't know how to.
A’B’C + A’BC + A’BC’ + AB’C + ABC
A'B(C+C')+C(A'B'+AB'+AB)
A'B+C(A'B'+AB'+AB)
That's it, I don't know how to solve that.
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. – dantopa Jun 26 '19 at 22:07
4 Answers
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\begin{align} E&=A'B+C(A'B'+AB'+AB)\\&=A'B+C((A'+A)B'+AB)\\ &=A'B+C(B'+AB) \end{align}
For boolean expressions $+$ also distributes over product, by example: $$ B'+AB = (B'+A)(B'+B)=B'+A $$ thus
\begin{align} E &=A'B+C(B'+A)\\ &=A'B+CB'+CA \\ &=(A'B+C)(A'B+B')+CA \\ &=(A'B+C)(A'+B')+CA\\ &=A'B+CA'+CB'+CA \\ &=A'B+C+CB' \\ &=A'B+C \end{align}
Picaud Vincent
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Thanks, im so lost though, what happened from here
(A′B+C)(A′B+B′)+CA=(A′B+C)(A′+B′)+CA=A′B+CA′+CB′+CA=A′B+C+CB′=A′B+C – Mica B Jun 26 '19 at 22:42
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Terms can be duplicated in boolean algebra, because or and and are idempotent: $X = X+X$ and $X = XX$.
$$\begin{align*} p &= A’B’C + \underline {A’BC} + A’BC’ + AB’C + ABC\\ &= (\underline {A’BC} + A'BC') + (A'B'C + \underline {A’BC} + AB'C + ABC)\\ &= A'B(C+C') + (A'+A)(B'+B)C\\ &= A'B + C \end{align*}$$
(Underlined terms were duplicated, no special meaning otherwise)
Also try using Karnaugh map which helps a lot.
peterwhy
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This way looks so much easier, thank you, i might solve this problem this way – Mica B Jun 27 '19 at 00:03
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There is this super-weird absorption laws in boolean algebra: $X+XY = X$ and $X(X+Y) = X$.
From your last line, as you "prematurely grouped" the $A'BC$ term into $A'B$, the absorption law can "recreate" an $A'BC$ out of $A'B$:
$$\begin{align*} p &= A'B+C(A'B'+AB'+AB)\\ &= (A'B + A'BC) + C(A'B ' + AB' + AB)\\ &= A'B + C(A'B + A'B' + AB' + AB)\\ &= A'B + C(A' + A)(B'+B)\\ &= A'B + C \end{align*}$$
Also try using Karnaugh map which helps a lot.
peterwhy
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i liked this way better than the other, yeah i know Karnaugh but i have to solve the expression both ways, thats how i knew i had to get A'B+C, thank you – Mica B Jun 27 '19 at 00:01
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@MicaB If you draw the Karnaugh map, you should see that the rectangles corresponding to $A'B$ and $C$ covers $A'BC$ twice. So $A'BC$ is a term that should appear twice. – peterwhy Jun 27 '19 at 22:51
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$$A'B'C'+A'BC'+A'BC+ABC=$$
$$=A'C(B'+B)+A'BC+ABC' , \{B'+B=1\}$$
$$=A'C'+B(A'C+AC')$$
$$=A'C' +B$$
vitamin d
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