Consider the relations \(r(A, B)\) and \(s(B, C)\), where \(s.B\) is a primary…

Answer: r LOJ (σ_{B<5}(s)) is NOT equivalent to the original query.

Key facts:

• s.B is a primary key and r.B is a foreign key referencing s.B; there are no nulls. Therefore every tuple of r has a matching tuple in s.

• The original query r ⋈ (σ_{B<5}(s)) returns exactly those r tuples whose matching s has B<5, joined with their s attributes.

Why the three equivalent expressions match the original query:

• σ_{B<5}(r ⋈ s): Since every r tuple has a matching s tuple, performing the join first and then filtering B<5 keeps exactly those joined pairs where the matched s has B<5, same as the original.

• σ_{B<5}(r LOJ s): Because of referential integrity (every r matches an s), the left outer join behaves like an inner join here. Filtering B<5 afterward therefore yields the same result as the original query.

• σ_{B<5}(r) LOJ s: Selecting r tuples with B<5 first restricts to r rows whose matching s tuples have B<5; left outer-joining s then matches those s tuples, giving the same pairs as the original.

Why r LOJ (σ_{B<5}(s)) is not equivalent:

• Filtering s before a left outer join removes s tuples with B>=5. For r tuples whose matching s had B>=5, the left outer join now produces those r tuples with NULLs for s attributes instead of excluding them. The original query excludes those r tuples because their matching s was filtered out of σ_{B<5}(s).

• Counterexample: s = {(B=4,...), (B=6,...)} and r = {(B=4,...), (B=6,...)}. The original query r ⋈ (σ_{B<5}(s)) returns only the row with B=4. The expression r LOJ (σ_{B<5}(s)) returns both r rows, with the B=6 row having NULL s attributes, so the results differ.

Conclusion: r LOJ (σ_{B<5}(s)) returns extra r tuples when their matching s tuples were removed by the selection, so it is not equivalent to the original query; the other listed expressions are equivalent under the given constraints.