1. ER Data Model (Ch2 from H.G-M)
2. Relational Model (Ch3 from H.G-M, you may skip Section 3.7)
3. Relational Algebra (Ch5 from H.G-M, you may skip Section 5.3)
4. SQL (Ch6 from H.G-M)

F = { A -> BC, ACD -> E, B -> D, C -> D, AB -> E, E -> BC}

1. ACD -> E can be replaced by AC -> E since C -> D is given.
2. AB -> E can be replaced by A -> E since A -> B is given.
3. AC -> E can be replaced by A -> E since A -> C is given. (in class I combined it as ABC -> E to be replaced
with A -> E because A -> BC was given.)

Modified F up to this point:
= {A -> BC, A -> E, B -> D, C -> D, E -> BC}

Note: do not combine 1st and the 5th FD and reduce it to A -> BC. Instead,
split them into FD's with a single attribute on the right.
So, set of FDs now is:
= { A -> B, A -> C, A -> E, B -> D, C -> D, E -> B, E -> C}
You can now eliminate A -> B because of A -> E, and E -> B;
eliminate A -> C because of A -> E and E -> C.
Note: you could have eliminated A -> BC because of A -> E and E -> BC.

Now the minimal cover is:

Fc = {A -> E, E -> B, E -> C, B -> D, C -> D}