The dependency preservation decomposition is a property to decompose database…

Concept. Let a schema R carrying a set of functional dependencies F be decomposed into fragments R1, R2, …, Rn. Write Fi = πRi(F) for the projection of F onto Ri — every dependency of F+ whose attributes all lie inside Ri. The decomposition is dependency preserving exactly when the fragments together retain the full constraining power of F, that is when (F1 ∪ F2 ∪ … ∪ Fn)+ = F+.

Why the test is on the closure, not on a literal list. Because the condition is stated on closures, an individual X → Y of F need not be written down in any single Fi. It is enough that Armstrong's axioms — reflexivity, augmentation and transitivity — regenerate X → Y from the union of the Fi. So the definition admits two routes for each dependency, and the textbook statement (Elmasri & Navathe) joins them with "either … or".

Applying it to a dependency X → Y of F.

1. Route one — X and Y sit together inside some fragment Ri and X → Y belongs to that fragment's projected set Fi. The dependency then survives the decomposition verbatim and can be enforced by examining Ri alone, with no join.

2. Route two — no fragment lists X → Y, yet applying Armstrong's axioms to F1 ∪ … ∪ Fn derives it. Enforcing the retained dependencies then forces X → Y to hold anyway, so nothing is lost.

3. If every X → Y of F falls under route one or route two, the closures match and the decomposition is dependency preserving. If even one dependency falls under neither, then (F1 ∪ … ∪ Fn)+ is a proper subset of F+ and that constraint can only be re-checked by joining fragments back together.

Worked check, route two in action. Take R(A, B, C) with F = {A → B, B → C, A → C} decomposed into R1(A, B) and R2(B, C). Here F1 = {A → B} and F2 = {B → C}, so A → C is listed by neither fragment; transitivity applied to A → B and B → C recovers it, the closures agree, and the decomposition is dependency preserving.

Contrast — the criterion does have teeth. Take R(A, B, C) with F = {A → B, B → C} decomposed instead into R1(A, C) and R2(B, C). Now F1 = {A → C} and F2 = {B → C}, and no sequence of Armstrong's axioms yields A → B from those two, so the closure is strictly smaller and this decomposition is not dependency preserving.

Contrast with the narrower statements.

• Demanding literal retention alone would reject the first worked decomposition above, which is dependency preserving even though A → C appears in no fragment.

• Naming the derivation route alone is technically wide enough — a dependency present in some Fi is trivially derivable from the union — but it drops the direct case that the standard definition states in its own words.

Result. The completion that reproduces the definition is the one asserting that each X → Y of F either appeared directly in one of the relation schemas Ri in the decomposed D, or could be inferred from the dependencies that appear in some Ri.