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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 | 2.1k views
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This question are concept building
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what does ' !P ' mean?
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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.

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)

S2 and S3 are true

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