Boolean algebra simplification a'bc+ab'c+abc'+abc

Question

Can't figure out how to simplify $(^\neg a)bc+a(^\neg b)c+ab(^\neg c)+abc$, I'm really bad at this...

No clue, are just working with boolean algebra identities like de Morgan's law — Brandon Jerz, Sep 28 '14 at 21:19

score 0 · Answer 1 · answered Mar 14 '17 at 06:07

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A'B C + A B'C + A B C'+ A B C

Since We can use ABC = ABC + ABC So I use three of Copies:

A'B C + A B'C + A B C' + A B C + A B C + A B C

RE-ARRANGE: A'B C + A B C + A B'C + A B C + A B C'+A B C

GROUPING: BC(A'+A) + AC(B'+B) + AB(C'+C)

Since A'+A=1 & B'+B=1 & C'+C=1;

= BC+AC+AB

answered Mar 14 '17 at 06:07

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Welcome to math.stackexchange. You can type mathematical formulas in a beautiful way using MathJax. Here's a tutorial. – Scientifica Mar 14 '17 at 06:11

score 0 · Accepted Answer · answered Sep 28 '14 at 21:24

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$a'bc+ab'c+abc'+abc = a'bc+a(b'c+bc'+bc)$

Let's continue:

$a'bc+a*(b'c+bc'+bc)=a'bc+a(b'c+b(c'+c))=a'bc+a(b'c+b)=a'bc+ab'c+ab$

Why don't you try to see if you can simplify further (which I doubt is possible).

answered Sep 28 '14 at 21:24

Oria Gruber

I'm really confused as to how you completed the first step – Brandon Jerz Sep 28 '14 at 21:37
with numbers, $6+8+10=2(3+4+5)$. this is a similar concept. $ab'c+abc'+abc=a(b'c+bc'+bc)$. because $a$ appears in every term. – Oria Gruber Sep 29 '14 at 11:24
$(x+x'y) = (x + y)$ – sawan kumawat May 05 '19 at 17:32
Please read, Oria, Enforcement of Quality Standards. Consider yourself fully informed. You answered a very low quality answer, and fully. This violates quality standards on this site, so be aware of the consequences of repeatedly doing so. – amWhy Sep 20 '21 at 18:11

2 Answers2