Problem 13:
  A -> BC
  CD -> E
  B -> D
  E -> A


ADDITIONAL FD's in the closure of the FD's above (and Attribute closures/candidate keys based on this)
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A->B  decomposition
A->C  decomp
A->D  transitivity (A->B and B->D)
E->BC  trans   E->A and A->BCh
CD->BC trans
E->B  decomp
E->C  decomp
E->D  trans   (E->B and B->D)
E->E  refl
E->A  given
E->ABCDE union
E+ = {ABCDE} or E+ = R so E is a candidate key
A->A refl
A->CD  (union of above)
A->E  by transitivitiy CD->E
A->ABCDE  by union
A+ = {ABCDE}  or A+ = R so A is a candidate key
D is not a candidate key, because it doesn't determine anything by itself
B is not a candidate key, B->BD   nothing else
C is not a candidate key because it doesn't determine anything by itself
CD a candidate key?
CD -> E and E-> ABCDE,  then CD->ABCDE  CD+={ABCDE} CD+ = R so superkey
   CD is superkey easy to show above
   CD is minimal, because C and D are not candidate keys
AD is NOT a candidate key because A is a candidate key so it's not minimal.
Is BC also a candidate key? Yes:
BC -> B, C, D    and CD->E   so BC->E by trans,  BC->ABCDE or BC+={ABCDE} BC+=R
     so BC is a superkey
      So BC is a candidate key because mininmal (i.e. neither B nor C are candidate keys alone).

So while CD and BC might appear to be not minimal given that they have 2 attributes while A and E have only 1 attribute,
CD and BC are indeed minimal superkeys (and hence candidate keys) because neither have attributes which can be removed and still be superkeys, so they are minimal in the key sense of that word.