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+1 vote
Simplify the following expression

AB’C + A’BC + A’B’C

Solution given is A’C + B’C can someone show me how?
asked in Digital Logic by Active (1.2k points) | 90 views

2 Answers

+2 votes

:AB'C + A'BC +A'B'C


=AB'C+A'C x 1


=C[(A+A')(B'+A')]                     [Distributive Law XY+Z==(X+Z)(Y+Z)]

=C x 1 x (B'+A')


answered by Active (4.5k points)
+1 vote

Let $f(A,B,C)=A\cdot\overline{B}\cdot C+\overline{A}\cdot B\cdot C+\overline{A}\cdot\overline{B}\cdot C$

Take common from first and last term

$f(A,B,C)=\overline{B}\cdot C\cdot(A+\overline{A})+\overline{A}\cdot B\cdot C$

$f(A,B,C)=\overline{B}\cdot C\cdot1+\overline{A}\cdot B\cdot C$                      ${\color{Red}{[X+\overline{X}=1]} }$

$f(A,B,C)=\overline{B}\cdot C+\overline{A}\cdot B\cdot C$

$f(A,B,C)=C(\overline{B}+\overline{A}\cdot B)$

$f(A,B,C)=C(\overline{B}+\overline{A})\cdot(\overline{B}+B)$                   ${\color{Magenta} {[X+Y\cdot Z=(X+Y)\cdot(X+Z)]}}$

$f(A,B,C)=C(\overline{A}+\overline{B})\cdot 1$

$f(A,B,C)=\overline{A}\cdot C+\overline{B}\cdot C$

answered by Boss (39.8k points)
edited by

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