If $P, Q, R$ are subsets of the universal set U, then $$(P\cap Q\cap R) \cup (P^c \cap Q \cap R) \cup Q^c \cup R^c$$ is
Answer D $\quad(P\cap Q\cap R)\cup (P^{c}\cap Q\cap R)\cup Q^{c}\cup R^{c}$
$=(P\cup P^{c})\cap (Q\cap R)\cup Q^{c}\cup R^{c}$ $=(Q\cap R)\cup Q^{c}\cup R^{c}$ $=(Q\cap R)\cup (Q\cap R)^{C}$ $= U.$
Can we treat these like Boolean expression and solve? Like PQR + P'QR + Q' + R'. and minimise this. Is this method always correct? @Praveen Sir? @Arjun Sir?
Praveen Saini if use Aspi R Osa 's method and found P.PQ then this equivalent to PQ or we take it as P.PQ ?
Praveen Saini sir
https://gateoverflow.in/3562/gate2006-it-23
i above link's Ques
in I, LHS=P+QR-PQR RHS=(P+Q-PQ).(P+R-PR) =P+PR-PR+PQ+QR-PQR-PQ-PQR+PQR =P+QR-PQR LHS=RHS So I is true but original ans is I is false
plz verify
whats wrong in my explanation
so option d
hope it might help....
Treating as a boolean expression like suggested in an answer here:
PQR + P'QR + Q' + R'
= (P+P') QR + Q' + R'
= QR + Q' + R'
= QR + Q'R' + Q'R + R'
= R(Q+Q') + R'(Q'+1)
= R + R'
= 1
Also R+R' means RUR' so its equal to U.