The relation schemas 𝑅1 and 𝑅2 form a Lossless join decomposition of 𝑅 if…

Lossless-join criterion: For a decomposition of relation R into R1 and R2, let X = R1 ∩ R2. The decomposition is lossless if and only if X functionally determines the attributes of at least one of the two relations (that is, X -> (R1 - R2) or X -> (R2 - R1)).

Why this works (sketch):

• If X -> (R1 - R2), then every tuple in R1 is determined by its values on X. When you join R1 and R2 on X you cannot create spurious tuples because matching on X recovers the original R1 tuples; hence the natural join reconstructs R exactly.

• Symmetrically, if X -> (R2 - R1) then joining also recovers R without loss.

Remarks: The condition requires at least one of these functional dependencies. If both hold (i.e., X determines both R1 - R2 and R2 - R1), that is a stronger statement implying X determines all attributes of R and thus is also lossless, but both are not required.

Mapping this to the provided answer choices: any choice that includes the dependency that the common attributes determine the unique attributes of at least one relation satisfies the lossless-join condition. The option that asserts both such dependencies holds is sufficient but stronger than necessary.