• ❑ b
▼ ❑ Section 2
• ❑ 1 σc(R∩S) and there is an index on S. Assuming that there C attributes in
both R and S, the options are:
▼ ❑ 2
• ❑ a πL(R∪S) ≠ πL(R)∪πL(S)
R(C1, C2) = {(1,1) (1,2) (1,2)}
S(C1, C2) = {(1,3) (1,4) (1,5)}
L = C1
• ❑ b πL(R–S) ≠ πL(R)-πL(S)
S(C1, C2) = {(1,1) (1,3)}
• ❑ 3 πL(R∪BS) = πL(R)∪BπL(S)
Attributes eliminated do not appear in results. Attributes eliminated are not
▼ ❑ 4