Next step in progression from 3NF to BCNF I am working with relation H that is detailed below:

H = {A,B,C,D}


It shows the following functional dependencies (FD[s]):

F = { A -&gt; B, B -&gt; C, {C,D} -&gt; A}


The candidate keys are

{A,D}, {B,D}, {C,D}


I belive relation H is in 3NF.

I also believe there are two FD that prevent this relation from satisfying BCNF: 

A -&gt; B and B -&gt; C

Then I have:

A -&gt; B 
H1 = {A,B}  F1 = {A -&gt; B} &lt;- this is BCNF
H2 = {A,C,D} F2 = {A -&gt; C, {C,D} -&gt; A} &lt;- this isnt' BCNF

B -&gt; C
H11 = {B,C} F11 = {B -&gt; C} &lt;- this is BCNF
H12 = {A,B,D} F12 = {A -&gt; B} &lt;- this is in BCNF


The questions

I have partial information about this algorithm, so I am not sure about certain things:


What should I do with H2? Just throw it away and say that H1, H11, H12 are decompositions of H to meet BCFN?
Can this relation be converted to BCNF with functional dependency preservation? (I would say no! because it is impossible to derive the FD {C,D} -&gt; A from F1 sum F11 sum F12. Can you confirm that?
First, you assumption about the 3NF is correct.

Then, in the analysis algorithm to find the BNCF, when you start to remove a dependency X → Y since it violates the BCNF, you should put in the first relation H1 not only XY, but X+, while in the second relation you should have H2 = H - X+ + X.

So, in the first step, the two resulting relations are:

H1 = {A,B,C) with dependencies A → B and B → C and key `A`,
H2 = {A, D} (with no dependencies)


In H1 the dependency B → C violates the BCNF since B is not a superkey, and you can decompose it in:

H11 = {A, B} with the dependency A → B and key A
H12 = {B, C} with the dependency B → C and key B


So, at the end, the final decomposition is H11, H12 and H2. Note that this decomposition determines the loss of the dependency {C, D} → A, and this happens also if we start by considering first the dependency B → C (obtaining the same decomposition).