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Find the FDs that hold for a subrelation.

Given $R(ABCDE)$ with FD set $ \{ A \rightarrow B, B \rightarrow E, C \rightarrow D \} $

Decomposed into :

$R_1 (ABE) :$ (Non-trivial)FDs that hold in $R_1$ are $\{ A \rightarrow B, B \rightarrow E \}$

$R_2 (CD) :$ (Non-trivial)FDs that hold in $R_2$ are $\{ C \rightarrow D \}$

$R_3 (AC) :$ (Non-trivial)FDs that hold in $R_3$ are $\{ \}$

So, clearly, ALL FDs of $R$ are preserved by this Decomposition i.e. Union of FD set of $R_1,R_2,R_3$ is FD set of $R.$

The Decomposition is Lossless as well because $R_2,R_3$ have common set of attributes $\{ C \} $ which is super key in $R_2$, So, we can merge them in Lossless manner and then we can merge with $R_1$ in lossless manner.

$R_1,R_2,R_3$ ALL are in 2NF.

$R_2,R_3$ are in BCNF.

$R_1$ is in 2NF but NOT in 3NF because there is Transitive dependency $B → E$

So, Option A is correct answer.

Given $R(ABCDE)$ with FD set $ \{ A \rightarrow B, B \rightarrow E, C \rightarrow D \} $

Decomposed into :

$R_1 (ABE) :$ (Non-trivial)FDs that hold in $R_1$ are $\{ A \rightarrow B, B \rightarrow E \}$

$R_2 (CD) :$ (Non-trivial)FDs that hold in $R_2$ are $\{ C \rightarrow D \}$

$R_3 (AC) :$ (Non-trivial)FDs that hold in $R_3$ are $\{ \}$

So, clearly, ALL FDs of $R$ are preserved by this Decomposition i.e. Union of FD set of $R_1,R_2,R_3$ is FD set of $R.$

The Decomposition is Lossless as well because $R_2,R_3$ have common set of attributes $\{ C \} $ which is super key in $R_2$, So, we can merge them in Lossless manner and then we can merge with $R_1$ in lossless manner.

$R_1,R_2,R_3$ ALL are in 2NF.

$R_2,R_3$ are in BCNF.

$R_1$ is in 2NF but NOT in 3NF because there is Transitive dependency $B → E$

So, Option A is correct answer.