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...

Answer 1

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

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).

Answer 2

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