If a relation schema \( R \) is in \(\mathbf{BCNF}\) with respect to a set of FDs \( F \), then every projection \( R_S \) (for any \( S \subseteq R \)) with projected dependencies \( \pi_S(F) \) is also in BCNF.
Because, If a nontrivial FD \( X \to A \) holds in \( R_S \), then \( X \to A \) holds in \( R \). Since \( R \) is BCNF, \( X \) is a superkey of \( R \)
Hence \( X \) determines every attribute of \( S \), so \( X \) is a superkey of \( R_S \). Thus BCNF is downward-closed under projection.
The analogous claim for \(\mathbf{3NF}\) is false. A relation can be in 3NF while some component of a lossless decomposition is not in 3NF (counterexample below).
Counter example (3NF not preserved under a lossless decomposition).
Let \( R(A,B,C,D) \) with
\[
F \;=\; \{\, AD \to B,\;\; AD \to C,\;\; B \to A \,\}.
\]
(i) \( R \) is in 3NF.
Compute keys: \( AD^+ = A B C D \) (since \( AD \to B \) and \( AD \to C \)), so \( AD \) is a key. Also \( BD \) is a key because \( B \to A \) and then with \( A D \to C \) we obtain \( C \)
Hence \( BD^+ = A B C D \). Prime attributes in \( R \) : \( A, B, D \) (and \( C \) is nonprime).
Check FDs: \( AD \to B \) and \( AD \to C \) have LHS as Superkey
\( B \to A \) has non-superkey LHS but RHS \( A \) is prime. Hence \( R \) satisfies 3NF (not BCNF due to \( B \to A \)).
(ii) Lossless decomposition \( R \to R_1(A,B,C) \) and \( R_2(B,C,D) \).
\( R_1 \cap R_2 = \{ B, C \} \). Since \( B \to A \), we have \( \{ B, C \} \to \{ A, B, C \} = R_1 \)
By the binary lossless-join test \((R_1 \cap R_2) \to R_1 \Rightarrow\) lossless.
(iii) \( R_1 \) is not in 3NF.
In \( R_1(A,B,C) \), the projected FDs include \( B \to A \). Keys of \( R_1 \) are \( \{ B, C \} \), so prime attributes in \( R_1 \) are \( B, C \) and \( A \) is nonprime. The FD \( B \to A \) has LHS not a superkey and RHS nonprime, so 3NF is violated in \( R_1 \).
Independence of lossless-join and dependency-preserving.
• Lossless but not dependency-preserving.
\( R(A,B,C) \) with \( F = \{ A \to B,\; B \to C \} \). Decompose into \( R_1(A,B) \) and \( R_2(A,C) \).
\( R_1 \cap R_2 = \{ A \} \) and \( A \to A B = R_1 \), so the decomposition is lossless. However, \( B \to C \) cannot be enforced within \( R_1 \) or \( R_2 \) alone, so it is not dependency-preserving.
• Dependency-preserving but lossy.
\( R(A,B,C) \) with \( F = \{ A \to B \} \). Decompose into \( R_1(A,B) \) and \( R_2(B,C) \).
The dependency \( A \to B \) is preserved (it remains in \( R_1 \)), but the decomposition is lossy since \( \{ B \} \) does not functionally determine all attributes of \( R_1 \) or of \( R_2 \).
Definitions :
• BCNF: For every nontrivial FD \( X \to A \) in \( F^+ \) on \( R \), \( X \) is a superkey of \( R \).
• 3NF: For every nontrivial FD \( X \to A \) in \( F^+ \) on \( R \), either \( X \) is a superkey of \( R \) \(\;\)or\(\;\) \( A \) is prime (belongs to some candidate key of \( R \)).
• Binary lossless-join test: For \( R \to R_1, R_2 \), if \( (R_1 \cap R_2) \to R_1 \) or \( (R_1 \cap R_2) \to R_2 \) holds in \( F^+ \), then the decomposition is lossless.
Conclusions :
• “If \( R \) is in 3NF, then every lossless decomposition is again 3NF?” — False.
• BCNF is preserved under projection
• Lossless-join and dependency-preserving are independent.
Reference :
Silberschatz–Korth–Sudarshan, Database System Concepts, Ch. 7 slides: BCNF/3NF definitions; lossless-join test; decomposition algorithms.
https://www.db-book.com/slides-dir/PDF-dir/ch7.pdf
