@Arjun Sir, Add Some spaces in option D, third term seems to be whole complement.

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16 votes

Consider the following Boolean expression.

$F=(X+Y+Z)(\overline X +Y)(\overline Y +Z)$

Which of the following Boolean expressions is/are equivalent to $\overline F$ (complement of $F$)?

- $(\overline X +\overline Y +\overline Z)(X+\overline Y)(Y+\overline Z)$
- $X\overline Y + \overline Z$
- $(X+\overline Z)(\overline Y +\overline Z)$
- $X\overline Y +Y\overline Z + \overline X\; \overline Y \;\overline Z$

6 votes

Best answer

$F=(X+Y+Z)(\overline X +Y)(\overline Y +Z)$

Taking complement of above expression;

$\overline {F}=\overline {(X+Y+Z)(\overline X+Y)(\overline Y+Z)}$

Applying De-Morgan’s law;

$\overline {F}=\overline{(X+Y+Z)}+\overline{(\overline X+Y)}+\overline{(\overline Y+Z)}$

$\overline{F}=(\bar X.\bar Y.\bar Z)+\overline{\overline{X}}.\overline Y+\overline{\overline{Y}}.\overline Z$

$\left [\because \overline{\overline{X}}=X, \text{Using double negation law} \right ]$

$\therefore \overline F=(\bar X.\bar Y.\bar Z)+(X.\overline Y)+(Y.\overline Z)$ $\quad \quad \to \text{Option (D)}$

Taking $\overline Y$ as common we get;

$\overline F=\overline Y \left[ (\bar X \bar Z)+X\right]+Y\overline Z$

$\left [ \because A+BC=(A+B)(A+C) \text{ Applying distributive law here}\right ]$

$\overline F=\overline Y\left[(X+\overline X)(X+\overline Z)\right]+Y\overline Z$

$\overline F=\overline Y\left[X+\overline Z\right]+Y\overline Z$

$\overline F=X \overline Y+\overline Y\overline Z+Y\overline Z$

Taking $\overline Z$ as common

$\overline F=X\overline Y+\overline Z(Y+ \overline Y)$

$\because \text{(Y+$\overline Y$=1) using complement law}$

$\therefore \overline F=X\overline Y+\overline Z$$\quad \quad \to \text{Option (B)}$

Applying distributive law here we get;

$\overline F=(X+\overline Z)(\overline Y+\overline Z)$$\quad \quad \to \text{Option (C)}$

So, correct options are $B,C,D.$

Option A is false and can be proved as follows:

Take $X = 0, Y = 1, Z = 0$

Now, $F = 0,$ since $(\bar Y + Z)$ term will be zero. So, $\bar F$ must be $1.$

But option A gives $0$ as the term $X + \bar Y$ evaluates to $0.$ So, option A is not equal to $\bar F.$

@Lakshman Patel RJIT sir kindly update option D, its not matching with arjun sir’s answer

$\overline{XYZ}$ should be $\overline{X}$.$\overline{Y}$.$\overline{Z}$

1

9 votes

$F = (X + Y + Z)(\overline{X} + Y) (\overline{Y} + Z)$

$ \overline{F} = \overline{(X + Y + Z)(\overline{X} + Y) (\overline{Y} + Z)}$

$ \overline{F} = \overline{X}\overline{Y }\overline{Z}+X\overline{Y}+Y\overline{Z} $ ----- **(D)**

$ \overline{F} = \overline{Y } (\overline{X}\overline{Z}+X)+Y\overline{Z} $

$ \overline{F} = \overline{Y } (\overline{X}+X)( \overline{Z}+X)+Y\overline{Z} $

$ \overline{F} = \overline{Y }( \overline{Z}+X)+Y\overline{Z} $

$ \overline{F} = \overline{Y }\overline{Z}+\overline{Y }X+Y\overline{Z} $

$ \overline{F} =\overline{Y }X+ \overline{Z}(\overline{Y }+Y) $

$ \overline{F} =\overline{Y }X+ \overline{Z} $ ----- **(B)**

$ \overline{F} = (\overline{Y }+\overline{Z}) (X+ \overline{Z}) $ ----– **(C)**

**Ans : B,C,D**

1 vote

The easiest way to find complement of a Boolean function is first find the dual of that function (replace + with • and 1 with 0) then interchange literals with their negations (i.e x with x')

f=(x+y+z).(x'+y).(y'+z)

Dual of f= (x.y.z)+(x'.y)+(y'.z)

f'=(x'.y'.z')+(x.y')+(y.z')

And using some Boolean laws u get 2,3 options also

f=(x+y+z).(x'+y).(y'+z)

Dual of f= (x.y.z)+(x'.y)+(y'.z)

f'=(x'.y'.z')+(x.y')+(y.z')

And using some Boolean laws u get 2,3 options also