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The online calculator(not posting the name because I don't know if its's allowed) is giving me a different result and I can't find what I'm doing wrong:

Mine:

$A'BC' + A'BC + AB'C + ABC' + ABC= BC'(A'+A) + BC(A' + A) + AB'C= BC' + BC + AB'C= B(C' + C) + AB'C= B + AB'C$

The calculator's:

$=AC + BC' + BA'$

EL02
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  • Could you explain what $A$, $A'$, etc. are? Posting the name of your calculator is fine. – preferred_anon Dec 23 '20 at 21:51
  • @preferred_anon It's boolean algebra, that's why i tagged it. A is just a variable for 0/1 and A' is basically not(A). The website i used is: boolean-algebra.com – EL02 Dec 23 '20 at 21:55
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    You are both right. $BC' + BA' + AC = BC' + BA' + ABC + AB'C = B + AB'C$. – martini Dec 23 '20 at 21:57

1 Answers1

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The two answers are equivalent:

$$\begin{align*} AC+BC'+BA'&=AC+B(A'+C')\\ &=AC+B(AC)'\\ &=AC(B+B')+B(AC)'\\ &=ACB+ACB'+B(AC)'\\ &=B\big(AC+(AC)'\big)+AB'C\\ &=B+AB'C \end{align*}$$

Brian M. Scott
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