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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 $F$)?

1. $(\overline X +\overline Y +\overline Z)(X+\overline Y)(Y+\overline Z)$
2. $X\overline Y + \overline Z$
3. $(X+\overline Z)(\overline Y +\overline Z)$
4. $X\overline Y +Y\overline Z + \overline X\; \overline Y \;\overline Z$

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

Fixed 👍

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

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 $\bar F.$

Properties of Boolean Algebra

Actually Option D should contain

X' Y' Z'  and not (XYZ)' .

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

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

Done👍

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

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

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

why we are taking complement here
Read the question properly . It is asked in complement form thats why we are taking complement
after the statement ..  D .. instead of taking Y as common if we take Z as common we are not able to arrive at the answer.. is there any reasoning behind this can you explain if there is any think  wrong in my question..

We can solve this question using k-map.

Since it is a MSQ so we can easily explore all possibilities using k-map and answer correctly.

So, correct options are B,C,D.

by
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

### 1 comment

This is the way to find complement not dual in dual we don't do the negation of a variable like ( A < -> A')