Suppose that two relations R(A, B) and S(A, B) have exactly the same schema.…

Answer: The following equalities are true: R ∩ S = R - (R - S); R ∩ S = S - (S - R); R ∩ S = R NATURAL JOIN S. The equality with the Cartesian product is false.

• Proof that R ∩ S = R - (R - S): If x ∈ R ∩ S then x ∈ R and x ∈ S, so x ∉ (R - S), hence x ∈ R - (R - S). Conversely, if x ∈ R - (R - S) then x ∈ R and not(x ∈ R and x ∉ S). Since x ∈ R, we must have x ∈ S, so x ∈ R ∩ S. Therefore the sets are equal.

• Proof that R ∩ S = S - (S - R): This is the symmetric argument of the previous item: membership in S but not in S - R forces membership in R, so the result is the intersection.

• Reason that R ∩ S = R NATURAL JOIN S when schemas are identical: A natural join matches tuples on all common attributes. If the two relations have exactly the same attributes, a tuple appears in the natural join precisely when that same tuple is present in both relations. Thus the natural join returns exactly the intersection.

• Why R ∩ S ≠ R × S in general: The Cartesian product forms combined tuples from R and S, not the tuples that appear in both relations. For example, let R = {(1,2)} and S = {(1,2)}. Then R ∩ S = {(1,2)} but R × S = {((1,2),(1,2))}, which has a different structure and meaning, so the equality fails.