Given that decomposition of r into r1 and r2 is lossy ===> common attribute(A) is neither key in r1 nor in r2
∴ A should be have duplicates, but on natural join we check the condition on A only ===> result of natural join ⊃ r
take the sample set:-
r:-
A B C D
1 a i x
1 b ii y
2 c iii y
r1 :-
A B C
1 a i
1 b ii
2 c iii
r2:-
A D
1 x
1 y
2 y
r1 * r2
A B C D
1 a i x
1 a i y --------> Spurious tuple
1 b ii x --------> Spurious tuple
1 b ii y
2 c iii y
∴ option C is right.
s ⊃ r ===> s ∪ r = s
s * r = s ∩ r , given that s * r = s ====> s ∩ r = s , but we know that s ⊃ r ===> s ∩ r = r