0 votes 0 votes select all the right options $\forall_{t_1, \;t_2 \;\in \;r(R)}\big[(t_1(\alpha)=t_2(\alpha))\; \rightarrow (t_1(\beta)=t_2(\beta))\big]$ implies that there is $\alpha \rightarrow \beta$ functional dependency in relation $r$ with relation schema $R$, here $t_1$ and $t_2$ are the tuples in the instance $r$. If $K$ is a superkey of a relation instance $r$ of a relation schema $R$ then $\forall_{t_1, \;t_2 \;\in \;r(R)}\big[(t_1(K)=t_2(K))\; \rightarrow (t_1(R)=t_2(R))\big]$. Here $t_1$ and $t_2$ are the tuples in the instance $r$. A relational decomposition is said to be dependency preserving if enforcement of all functional dependencies should be possible on individual decomposed relations or when they joined. BCNF decomposition of a relation is default lossless decomposition. Databases databases database-normalization + – HM asked Sep 16, 2022 retagged Sep 16, 2022 by makhdoom ghaya HM 271 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes All Options A,B,C,D are correct. lalitver10 answered Sep 16, 2022 lalitver10 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes is wrong, there may not be real life dependency $\alpha \rightarrow \beta$, but an instance may coincidentally have the data that satisfy $\alpha \rightarrow \beta$. right, it is the definition of superkey. wrong. Not when they are joined. we must able to enforce functional dependencies on individual decomposed relations only. If we are able enforce the restrictions (functional dependencies) on merged relations then that merged relation preserves functional dependencies, not the decomposed relations. yes (I think, according general rule of BCNF decomposition, I’d like to know other views) HM answered Sep 16, 2022 HM comment Share Follow See all 0 reply Please log in or register to add a comment.