Question on Decomposition of Normal Forms

Question on Decomposition of Normal Forms –

Question 4 :
R(ABCDEFGH)
FD : {AB → C, AC → B, AD → E, B → D, BC → A, E → G}
Decompose the Relation R till BCNF.

Solution :
Step 1 : Find all the candidate keys of R.
         Candidate Key : {ABFH}, {BCFH} and {ACFH}

Step 2 : Checking For 2NF :
   (a) FD which violates 2NF :
       AB → C
       AC → B
       AD → E
       B → D
       BC → A
   (b)

(AB)+=
   {ABCDEG}

AB → C √
BC → A √ ,
  Because (BC)+ = {ABCDEG}
AC → B √ , 
  Because (AC)+ = {ABCDEG}
AD → E ×
B → D ×
E → G √

BC → A and AC → B are in 2NF also because as AB is the candidate key for
the relation "ABCDEG",
we can replace B by AC (Because AC → B) as AB = A(AC) = AC : CK
and we can replace A by BC (Because BC → A) as AB = (BC)B = BC : CK
Hence AB,BC,AC all three are the candidate keys for the relation "ABCDEG"

Check the CK of R is preserved in the decomposed relations- NO, 
So, make new relations for each CK which is not preserved.
Here, 3 Candidate Keys are there out of which 2 CK are not preserved-
{BCFH} and {ACFH}. So, making relations for each one as :

Hence the decomposition in 2NF :

Step 3 : Checking For 3NF :
         FD which violates 3NF :
         E → G

Hence the decomposition in 3NF :

Step 4 : Checking For BCNF :
         FD which violates BCNF : None
         Hence the decomposition is already in BCNF.