# 2NF decompositions are dependency preserving?

I have read that 1NF, 2NF and 3NF decompositions are lossless and dependency-preserving.

Consider this example on a relation $$R(A,B,C,D)$$ with functional dependencies set as $$FD =$${ $$AB \rightarrow CD, A \rightarrow C, BC \rightarrow D$$ }

Here when we do 2NF decomposition we get $$R_{1}(A,C)$$ with $$FD =$${$$A \rightarrow C$$} and $$R_{2}(A,B,D)$$ with $$FD =$${$$AB \rightarrow D$$}

The functional dependency $$BC \rightarrow D$$ is lost when we join but we know that 2NF is dependency preserving so why is it that we are unable to preserve the original FD?

• "I have read that 1NF, 2NF and 3NF decompositions are lossless and dependency-preserving." No, you haven't. Please quote an actual claim from a published competent source. Commented Apr 3, 2023 at 2:30
• Your "I have these FDs" doesn't make sense. "These are all the FDs that hold"?--Not possible. "These are all the non-trivial FDs that hold"?--Not possible. "These are some FDs that hold"?--Question can't be answered. Find out what a cover is & what the exact conditions are to apply a particular definition/rule/algorithm. To determine CKs & NFs we must be given FDs that form a cover. Sometimes a minimal/irreducible cover. And the set of all attributes must be given. See this answer. Commented Apr 3, 2023 at 2:30

Although you decomposed the table using the dependency that violates 2NF $$(A\rightarrow C)$$, the new tables are in BCNF, and decomposition to BCNF is not always dependency preserving.