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+21 votes

Let $f(x, y, z)=\bar{x} + \bar{y}x + xz$ be a switching function. Which one of the following is valid?

  1. $\bar{y} x$ is a prime implicant of $f$

  2. $xz$ is a minterm of $f$

  3. $xz$ is an implicant of $f$

  4. $y$ is a prime implicant of $f$

asked in Digital Logic by Veteran (59.8k points)
retagged by | 2.5k views

This one ..

5 Answers

+18 votes
Best answer

In sum of terms, any term is an implicant because it implies the function. So, $xz$ is an implicant and hence C is the answer. Still, lets see the other options. 

If no minimization is possible for an implicant (by removing any variable) it becomes a prime implicant. 

If a prime implicant is present in any possible expression for a function, it is called an essential prime implicant. (For example in K-map we might be able to choose among several prime implicants but for essential prime implicants there won't be a choice). 

  x'y' x'y xy xy'
z' $1$ $1$   $1$
z $1$ $1$ $1$ $1$


&    \textbf{x'y'}&  \textbf{x'y}&\bf{ xy}& \textbf{xy'} \\\hline
\textbf{z'}&1&1&&1 \\\hline \bf{z}&1&1&1&1\\ \hline   

So, $f = x' + y'x + xz$
$= y' + x' + z$ (could be also derived using algebraic rules as in )

So, the prime implicants are $x', y'$ and $z$. Being single variable ones and with no common variables, all must be essential also.  

Now, a choice is false, as $y'$ is a prime implicant and hence, $y'x$ is just an implicant but not prime. 
b choice - $xz$ is not a minterm. A minterm must include all variables. So, $xyz$ is a minterm so, is $xy'z$, but not $xz$. 
d choice - $y'$ is a prime implicant not $y$. 

answered by Veteran (396k points)
edited by

any term is an implicant because it implies the function.Can you explain it a bit.Not able to understand
Whenever a term in SOP form is 1, the function value becomes 1. i.e., any term implies the function.
+27 votes

Answer: C

f(x,y,z) = x' + y'x + xz

An implicant of a function is a product term that is included in the function.

so x', y'x and xz ,all are implicants of given function.

A prime implicant of a function is an implicant that is not included in any other implicant of the function. 

option a)   y'x is not a prime implicant as it is included in xz [ xy'z+ xyz]

option d) y is not a prime implicant as it include in both x' and xz.

a product term in which all the variables appear is called a minterm of the function

option b) xz is not a minterm

answered by Veteran (55.9k points)
so which is the prime implicant?
Draw kmap then derive .... Any subcube which is not fully part of other sub cube in Kmap is Prime implicant ....
@Praveen sir.
How $xy'$ is included in $xz?$
$xy' \leftrightarrow \color{RED}{xy'z} + xy'z'$
$xz \leftrightarrow xyz + \color{RED}{xy'z}$
$xy'z'$ is not included in $xz.$

Similarly, how $y$ is included in $x' \ and \ xz?$
$y$ is not even implicant.

   y'x is not a prime implicant as it is included in xz [ xy'z+ xyz]

Indeed, y'x is not a prime implicant But the given reason(explanation) for it is Wrong.

 y is not a prime implicant as it include in both x' and xz.

Again, Wrong reason.  


xy′z′ is not included in xz.

Similarly, how y is included in x′ and xz?
y is not even implicant.

You are absolutely right. 


+11 votes


answered by Active (3.2k points)
Why not group last row instead of last two as xz?
+3 votes

let suppose Y=f(A,B,C)

the expression for y=AB+ABC+BC

1)Implicant : for y the implicants are AB,ABC,BC basically in simple words implicant is 'for any SOP , the product term that implies our y is basically implicant ' i mean when your implicant would be 1 then y would be 1.

for given sop set any product term (AB or ABC or BC) 1 and it will make y=1.

2)Prime implicant : Implicants whose removal don't now imply y (mean their presence is prime)are called PI.

for y PI are AB,BC as if they are removed and here we lost y. ABC is not PI as if you remove it from y then still our y is not lost.

3)Essential PI : An EPI of a function y is one that cover a minterm of F , not covered by any other PI of y.

Minterm for given function are x'(y+y')(z+z') and y'z(x+x') and xz(y+y')

When you would solve given function using K-Map you will get f=x'+y'+z

For f if any one of (x' , y'z , xz) is 1 then it would make f=1 so these all are implicants.

as f= x'+y'+z and if we remove anyone of x' or y' or z then here we lost f so all three are PI.

answered by Boss (13k points)
–2 votes
annswer - C
answered by Loyal (8.8k points)
y'x is not a prime implicant as it is included in implicant xz

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