1 vote

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

**$ABC + A’B + ABC’$****$X’YZ + XZ$****$( X+ Y)’(X’ + Y’)$****$XY + X(WZ + WZ’)$****$(BC’ + A’D)(AB’ + CD’)$**

1 vote

Best answer

${\color{Red}{Literals:} }$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$**