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Given the function $F = P' +QR$, where $F$ is a function in three Boolean variables $P, Q$ and $R$ and $P'=!P$, consider the following statements.

$(S1) F = \sum(4, 5, 6)$

$(S2) F = \sum(0, 1, 2, 3, 7)$

$(S3) F = \Pi (4, 5, 6)$

$(S4) F = \Pi (0, 1, 2, 3, 7)$

Which of the following is true?

1. (S1)-False, (S2)-True, (S3)-True, (S4)-False
2. (S1)-True, (S2)-False, (S3)-False, (S4)-True
3. (S1)-False, (S2)-False, (S3)-True, (S4)-True
4. (S1)-True, (S2)-True, (S3)-False, (S4)-False

edited | 3k views
0
This question are concept building
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what does ' !P ' mean?
0

P ′ = ! P  means  ??

$F=P'+QR$,  draw the Kmap for this

We can find the minterm $\sum (0,1,2,3,7)$ and maxterm $\Pi (4,5,6)$

So, option A is correct: $(S1)$-False, $(S2)$-True, $(S3)$-True, $(S4)$-False

edited

$F = P{}' + QR$

for SOP we have :

$F = P{}.1.1 + 1.QR'=P{}'(Q+Q{}')(R+R{}') + (P+P{}')QR$

$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR+P{}'QR$

$P{}'QR + P{}'QR{}' +P{}'Q{}'R+P{}'Q{}'R{}'+PQR$

$F=\sum(0, 1, 2, 3, 7)$ (considering barred terms as 0 and unbarred as 1 and converting them to binary and then to decimal).

now for POS we have :

$F = P{}' + QR = (P{}'+Q)(P{}'+R) = (P{}'+Q+0)(P{}'+R+0)$

$(P{}'+Q+R.R{}')(P{}'+R+Q.Q{}')$

$(P{}'+Q+R)(P{}'+Q+R)(P{}'+Q+R{}')(P{}'+Q{}'+R)$

$(P{}'+Q+R)(P{}'+Q+R{}')(P{}'+Q{}'+R)$

$F=\prod (4, 5, 6)$(considering barred terms as 1 and unbarred as 0 and converting them to binary and then to decimal).

http://mcs.uwsuper.edu/sb/461/PDF/sop.html

edited
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But K-map is better time saving approach rt?
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may be but that also kind of varies from individual to individual don't you think ?
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Yes. Sure :)
But you needn't solve for both min terms and max terms rt? Solve one and take complement for other.
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i was thinking of doing that, but just did it to show the general method to solve it both ways( POS ans SOP). Feel free to edit. S2 and S3 are true Option A is correct

P'QR + P'QR' +P'Q'R + P'Q'R' + PQR = ⋿(0,1,2,3,7) = ∏(4,5,6)
option A is correct.

F=~P+QR

=011+000+010+001+011+111=∑(0,1,2,3,7)=Π(4,5,6)