A relation R={A,B,C,D,E,F,G} is given with following set of functional…

Determine whether {A, B, D} is a candidate key by computing its closure and checking minimality.

Compute the closure of {A, B, D}:

• Start: {A, B, D}

• Use AD → E: add E → {A, B, D, E}

• Use B → C: add C → {A, B, C, D, E}

• Use BE → F (we have B and E): add F → {A, B, C, D, E, F}

• Use AF → G (we have A and F): add G → {A, B, C, D, E, F, G}

Since the closure equals all attributes {A,B,C,D,E,F,G}, {A,B,D} is a superkey.

Check minimality:

• Remove A: {B, D}+ gives {B, C, D} via B → C, but cannot derive E, F, or G.

• Remove B: {A, D}+ gives {A, D, E} via AD → E, but cannot derive B, C, F, or G.

• Remove D: {A, B}+ gives {A, B, C} via B → C, but cannot derive D, E, F, or G.

Because no proper subset of {A,B,D} is a superkey, {A,B,D} is minimal and therefore a candidate key.