Identify the correct operation which produces the below given output based on…

Concept: R1∩R2 (intersection) returns only the tuples common to both relations. Intersection and set difference are both defined only for union-compatible relations — the same number of attributes over matching domains, which R1 and R2 satisfy here. Set difference is the more fundamental of the two: for any two union-compatible relations A and B, the identity A − (A − B) = A ∩ B always holds — subtracting from A everything that is not in B leaves exactly what A shares with B. So an expression written using only "−" can still compute an intersection.

Application:

1. R1 holds the tuples (a1, 9), (b1, 5), (c1, 6), and (d1, 11).

2. R2 holds the tuples (b1, 5) and (f1, 8).

3. Computing R1∩R2 directly: the only tuple lying in both lists is (b1, 5), so R1∩R2 = {(b1, 5)}, which is exactly the output shown in the question.

4. Evaluating R1−(R1−R2) step by step: the inner R1−R2 removes every R1 tuple that also lies in R2, leaving the unshared tuples (a1, 9), (c1, 6) and (d1, 11), so R1−R2 = {(a1, 9), (c1, 6), (d1, 11)}.

5. Subtracting that three-tuple result from R1 again removes exactly those three tuples, so R1−(R1−R2) = {(b1, 5)} — the same relation the intersection produced, exactly as the identity A − (A − B) = A ∩ B predicts.

Cross-check — weighing the other choices:

• R1−R2 = {(a1, 9), (c1, 6), (d1, 11)} — three tuples rather than the single-tuple output, so that expression alone cannot produce what the question shows.

• Choosing R1∩R2 by itself, or R1−(R1−R2) by itself, leaves the answer half-stated: each of them independently evaluates to {(b1, 5)}, so neither expression is the one and only way to produce the output.

Result: R1∩R2 and R1−(R1−R2) are two different-looking ways of writing the very same operation, and each independently evaluates to {(b1, 5)} — precisely the output shown. Because the output can be produced by either expression, the answer that recognises both of them together is the complete one.