ISI 2019 | PCB CS | Question: 10

Question

Let $R$ be a relation with functional dependencies $\mathcal{F}$. For any subset of attributes $X \subseteq R$, the closure of $X$ is defined as the set

$$ X^{+}=\{A \in R \mid X \rightarrow A \text { holds with respect to } \mathcal{F}\} . $$

For two non-empty attribute sets $Y$ and $Z$ in $R$, prove or disprove each of the following statements:

1. $\left(Y^{+} Z\right)^{+}=(Y Z)^{+}$
2. $(Y Z)^{+}=Y^{+} Z^{+}$

Answer 1

Ans -(a)

It is correct in the worst case Y+ will drive anything(except itself) so (Y+Z)+ will become (YZ)+  in best case Y+ can drive some more attributes in this case at RHS (YZ)+ here will also Y drive those same attribute so LHS and RHS are equivalent

Ex - Let R (A,B,C,D,E,F) and F={A--->C, BC---->DE} Y=A, Z= EF

(Y+z)+ ={ACEF} (YZ)+={AEFC} Hence Proved

\

Ans -(b)

Ex - Let R (A,B,C,D,E,F) and F={A--->C, BC---->DE, C---->D}   Y=A,Z=B

(YZ)+ ={ABCDE}, Y+Z+={ACDB}

so LHS and RHS are not equivalent