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+25 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 $f$

in Digital Logic by Veteran (52.2k points)
retagged by | 2.9k views
Ans is 3 or 4 ?
This question needs a better Explanation

8 Answers

+21 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). 

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 $y$. 

by Veteran (424k 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.

@Arjun sir

In above K-map,number of prime implicant $(PI) = 3$

and number of essential prime implicant$(EPI) = 3$


@Lakshman Patel RJIT


Just wanted to confirm, Is Implicants is equal to Number of minterms or is it product terms in SOP?


A normal product term that implies $Y.$

Example: For the function $Y = AB + ABC + BC,$ the implicants are $AB, ABC,$ and $BC$ because if any one of those terms are true, then $Y$ is true.



@Lakshman Patel RJIT

What if we minimize the function? You can take this example where f is reduced to $\bar{x}+\bar{y}+z$. Can we say implicants are $\bar{x}, \bar{y}$ and z?

Edit:- Got it. Thanks anyway.

+29 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

by Veteran (56.8k 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. 


+12 votes


by Active (3.3k points)
Why not group last row instead of last two as xz?
+6 votes

$xz$ is an implicant and $\neg y$ is both prime and essential prime implicant. The sop would be $z+\neg x+\neg y$. 

Implicant: Something that implies a function is its implicant
Prime implicant: The most reduced (minimal) implicant
Essential prime implicant: The prime implicant which cannot be avoided in any SOP


by Loyal (6.9k points)
edited by
thanks @Marv
need better explanation
+4 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.

by Boss (14.4k points)
0 votes

 prime implicant isthe biggest subcube possible . if you minimize Z+~X+~Y ,~Y is one of the prime implicant of this not Y .

if you see z is a prime implicant so xz is an implicant of this function .

so option 3 . 

by Boss (10.4k points)
0 votes
Implicant- A product term that covers one or more minterms.

Prime Implicant- Not subset of higher implicant

EPI- There are many PIs but an EPI is one which has atleast one cell that is not covered by any other PI


Although C is the answer but I have a doubt that if we expand xz as x(y+y')z         xyz + xy'z are minterms
by Junior (577 points)
–2 votes
annswer - C
by Loyal (8.6k points)
y'x is not a prime implicant as it is included in implicant xz
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