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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. $X\overline Y +Y\overline Z + \overline X\; \overline Y \;\overline Z$
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6 Answers

Best answer
15 votes
15 votes

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

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

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13 votes
13 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

5 votes
5 votes

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.

4 votes
4 votes

The simplest way to answer this question is finding the minterms and maxterms in $\bar{F}$ then comparing with each option,

if the given expression is equivalent then its maxterms/minterms as same as $\bar{F}$

 

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

Lets try to convert this pos form into canonical pos form so that we can find maxterms of function F.

Canonical POS = $(X + Y + Z)(\bar{X} + Y + (Z.\bar{Z}))((X\bar{X}) + \bar{Y} + Z)$

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



We know that, when it comes it maxterms, uncomplemented variable(X) indicates value 0 , complemented variable indicates ($\bar{X} $) indicates value 1.

NOTE: – Minterms of F = Maxterms of $\bar{F}$     and maxterms of F = minterms of $\bar{F}$


So, Maxterms in F = $\prod (0 , 2, 4, 5, 6)$   ,  Minterms of F = $\sum (1, 3, 7)$

From the above note Maxterms of $\bar{F}$ = $\prod (1, 3, 7)$   , Minterms of $\bar{F}$  = $\sum (0 , 2, 4, 5, 6)$
 

Now lets verify each option,

 



A)$(\bar{X} + \bar{Y} + \bar{Z})(X + \bar{Y})(Y + \bar{Z})$

So Canonical pos of this expresion is = $(\bar{X} + \bar{Y} + \bar{Z}) (X + \bar{Y} + (Z.\bar{Z})) ((X.\bar{X}) +Y + \bar{Z})$

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

                                                            =$\prod (7, 2, 3, 1, 5)$

As the maxterms of this expression is not same as $\bar{F}$ , both are not equivalent.

 

B)$X\bar{Y} + \bar{Z}$

Minterms of this Expression = $X\bar{Y}\bar{Z} + X\bar{Y}Z  + \bar{X}\bar{Y}\bar{Z}  + \bar{X}Y\bar{Z}+ X\bar{Y}\bar{Z} + XY\bar{Z} $

                                             =$\sum(0, 2 ,4 , 5, 6)$

which is same as $\bar{F}$, so these two exp are equivalent.



C) Just take $\bar{Z}$ common, then we will get  $X\bar{Y} + \bar{Z}$ which is option B so , These Two are also equivalent.



D)Just find out the minterms, the result is , minterms = $\sum(0, 2 ,4 , 5, 6)$, same as $\bar(F)$ so this is also equivalent.

 

Option B , C , D are correct.

Answer:

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