# Boolean algebra expression Floyd Digital Logic

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Simplify the following expression

AB’C + A’BC + A’B’C

Solution given is A’C + B’C can someone show me how?
0
Draw a K map and simplify.

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

=AB'C+A'C(B+B')

=AB'C+A'C x 1

=C(AB'+A')

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

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

=A'C+B'C

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

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

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

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

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

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

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

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

edited

$A\bar{B}C + \bar{A}BC + \bar{A}\bar{B}C$

Taking C common,

=$(A\bar{B} + \bar{A}B + \bar{A}\bar{B} ) . C$

=$((A \bigoplus B) + \bar{A}\bar{B} ) . C$

Remember this shortcut, [ $(P \bigoplus Q) + PQ => P + Q$], also we know that $A \bigoplus B = \bar{A} \bigoplus \bar{B}$, negating both inputs have no effect on output of XOR and XNOR. Hence we can write

$(A \bigoplus B) + \bar{A}\bar{B} = \bar{A}+\bar{B}$

Hence it directly becomes,

=$(\bar{A}+\bar{B}).C$

=$\bar{A}.C+\bar{B}.C$

Another method is to draw K-MAP from given min terms and then minimize :

$f(A,B,C) = A\bar{B}C + \bar{A}BC + \bar{A}\bar{B}C$

101          011           001

$f(A,B,C) = \sum \small m (1,3,5)$

AB’C + A’BC + A’B’C

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

(A+A')B'C+A'BC                   //A+A' = 1

B'C+A'BC

(B'+A'B)C                      //(A+A'B) = A+B

(B'+A')C

A'C+B'C

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