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Simplify the Boolean expression to the minimum numbers of the literals.

 

  1. $ABC + A’B + ABC’$
  2. $X’YZ + XZ$
  3. $( X+ Y)’(X’ + Y’)$
  4. $XY + X(WZ + WZ’)$
  5. $(BC’ + A’D)(AB’ + CD’)$
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${\color{Red}{Literals:} }$Every occurrence of the variables in it's true(pure) and complemented(negation) forms.

 

$(5)$$(BC’ + A’D)(AB’ + CD’)$    Here number of literals$ = 8$

       we can write like this

      $\Rightarrow BC'AB'+BC'CD'+A'DAB'+A'DCD'$

     $\Rightarrow ABB'C'D+BC'CD'+A'AB'D+A'CDD'$

     $\Rightarrow 0+0+0+0=0$    $\left [ X.X'=0\right ]$     Here number of literals $ = 0$


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