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Consider the following logic circuit whose inputs are functions $f_1, f_2, f_3$ and output is $f$

Given that

• $f_1(x,y,z) = \Sigma (0,1,3,5)$
• $f_2(x,y,z) = \Sigma (6,7),$ and
• $f(x,y,z) = \Sigma (1,4,5).$

$f_3$ is

1. $\Sigma (1,4,5)$
2. $\Sigma (6,7)$
3. $\Sigma (0,1,3,5)$
4. None of the above

Here f2 should be : f2(x,y,z)=Σ(6,5) as given in original paper.

Can you show the original paper? But that doesn't any way change the answer here as min-term 5 is in f3.
Arjun sir great explanations
Thank you so much
Great explanation Arjun sir! Understood easily ! Thanks :)
what is the result of AND of Maxterm instead of minterms?

Maxterm Means$: A+B+C$

AND of Maxterm is equal to  $(A+B+C). (\bar{A}+B+C)$ Product of Sum(POS)

Option A is right choice for F3

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$f = ((f_1f_2)'f_3')' = f_1f_2 + f_3$

In minimum sum of products form, AND of two expressions will contain the common terms. Since $f_1$ and $f_2$ don't have any common term, $f_1f_2$ is $0$ and hence $f = f_3 =Σ(1, 4, 5).$

Correct Answer: $A$
by

here we dont have AND gate. We have NAND gate so f1 NAND(compl. of AND) f2 = {NULL}Compl. which mean f1 NAND f2 will have all the element. So, answer should be D
Here f1 and f2 have $\sum (5)$ in common then why its that nothing in common ?
if you read the question clearly  then your doubt is clear ....

Here we have NAND - NAND Circuit, we can convert it to following AND - OR circuit. (As NAND is bubbled OR). Now it is easy to solve this question. F1 AND F2 = 0. SO whatever f3 is directly passed to output. So answer is A.

sir how you convert NAND to AND and NOR to OR ..?

i got you answer but just only wants to know the hidden concept behind the your concept

and i think you should edit your answer "Here we have NAND - NAND Circuit to Here we have NAND - NOR Circuit"
Here we don't have any nor gate its nand - nand realization which is equivalent to and-or realization.
this should be the best answer

AND of two minterm expression gives common terms.

$f1\ AND\ f2\equiv f1\cap f2.$

So, $f1\ NAND\ f2\equiv$${(f1\cap f2)'}$

$f1\cap f2=\{0,1,3,5\}\cap\{6,7\}=\phi$

And, we have,

$f=[(f1\cap f2)' \cap (f3)']'=f1 \cap f2 \bigcup f3 = \{1,4,5\}$

But, $f1 \cap f2 = \phi$

$\Rightarrow \phi\ U f3 = \{1,4,5\}$

$\Rightarrow f3 =\{1,4,5\}$