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Answer:
relation of FDs are {a, b, d, h}. From these all but {h}
seem plausibly part of keys. Calculating the closures of likely candidates gives:
{a, b} + = {a b, c, i}
{b, d} + = {b, d, e, f}
{a, d} + = {a, d, g, h, i, j}
{a, b, d} + = {a, b, c, d, e, f, g, h, i, j}
decompose r into 2nf and then 3nf relations shown in attachment.
Step- by-Step Explanation
Explanation is given in the attachment.
