If all rows are different then you get all options provided are correct.
Therefore assume there are duplicate values in P and Q
option A :- P natural_join ∏B(Q)
no.of rows in the result = |P|
option B :- ∏A (P) natural_join Q
no.of rows in the result = |Q|
option C :- P natural_join Q
Here A and B both attributes are common therefore The property of natural join is applied on both columns at a time
∴ P.A = Q.A and P.B = Q.B ===> if you have duplicates this results should be less than Option A and Option B
option D :- P ∩ Q
Here A and B both attributes are common therefore The property of intersection is applied on both columns at a time
∴ P.A = Q.A and P.B = Q.B ===> Option C and Option D are equal.
let P have additionally one more attribute C, and Q have additionally one more attribute D
then
Option C :- P.A = Q.A and P.B = Q.B
Option D :- P.A = Q.A and P.B = Q.B and P.C = Q.D
in this case Option D is right but not Option C
