B→D violates because B is not superkey and D is not a superkey → only 3NF, not BCNF .
A weak solution: "π_fname, lname (σ_dno=5 (EMPLOYEE ⨝ WORKS_ON ⨝ PROJECT))" A solution explains: elmasri navathe database system solution manual better
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If your answer differs, use the manual to find exactly where your logic diverged. B→D violates because B is not superkey and
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