Candidate key = Key = Minimal Superkey

A superkey (and therefore a candidate key) that does not contain all attributes must contain at least an attribute set that is the left side of a functional dependency. Otherwise it would not be possible to generate an additional attribute from the attribute set of this superkey and this contradicts the definition of a superkey.

But a subset of ABCDE containing at least one of C,AB,AE or DE is already a superkey and to be minimal it is necessary that it contains only one of these subsets.

So the only subsets that must me investigated are that that are supersets of AC. But a superset of AC cannot be a minimal superkey because C already is a key.

Therefore the only candidate keys are C,AB,AE and DE

I would take following approach.

From left side of the relations we see 3 primary keys:

• Key AB with AB->CDE relation is represented in LAB table.
• Key C with C->D relation is represented in LC table.
• Key D with D->E and D->B relations are represented in LD table.

Does it make sense?

CREATE TABLE LD (
  D INT,
  B INT, /* B can be omitted here, and derived from AB */
  E INT,
  CONSTRAINT LD_PKEY PRIMARY KEY ( D )
);

CREATE TABLE LC (
  C INT,
  D INT,
  CONSTRAINT LC_PKEY PRIMARY KEY ( C ),
  CONSTRAINT LC_D_FKEY FOREIGN KEY ( D ) REFERENCES LD

);

CREATE TABLE LAB (
  /* Attributes D and E are dependent on C, so we don't store them here */
  A INT,
  B INT,
  C INT,
  CONSTRAINT LAB_PKEY PRIMARY KEY ( A, B ),
  CONSTRAINT LAB_C_FKEY FOREIGN KEY ( C ) REFERENCES LC
);