First determine the key.
Starting with $BC:$
From $C\to D$, obtain $D$.
Now $B,C,D$ are available, so $BCD\to A$ gives $A$.
Then $A\to F$ gives $F$, and $F\to G$ gives $G$.
Also, $BC\to E$.
Thus, $(BC)^+=\{A,B,C,D,E,F,G\}$.
Hence, $BC$ is a candidate key.
So A is correct.
Now simplify the FD set.
Because $C\to D$, the $D$ in $BCD\to A$ is extraneous.
Therefore, $BC\to A$.
Also, $A\to G$ is redundant because $A\to F$ and $F\to G$ already imply it.
Thus a minimal cover is
$\{BC\to A,\ BC\to E,\ A\to F,\ F\to G,\ C\to D\}$.
So B is correct.
Now group dependencies with the same determinant.
From $BC\to A$ and $BC\to E$, create $R_1(A,B,C,E)$.
The remaining relations are $R_2(A,F),$ $R_3(F,G),$ $R_4(C,D)$.
$\therefore \{ABCE,AF,FG,CD\}$ is the synthesized decomposition.
So C is correct.
The relation $ABCE$ already contains the original candidate key $BC$.
Therefore, an additional key relation is unnecessary.
So D is incorrect.
Finally $:$
- in $ABCE,$ $BC$ is a key
- in $AF,$ $A$ is a key
- in $FG,$ $F$ is a key
- in $CD,$ $C$ is a key
Hence each component is actually in BCNF.
So E is correct.
