7.13

Show that the decomposition in Exercise 7.1 is not a dependency-preserving decomposition.

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There are several functional dependencies that are not preserved. We discuss one example here. The dependency $B\to D$ is not preserved. $F_1$, the restriction of $F$ to $(A,B,C)$ is $A\to ABC,A\to AB,$ $A\to AC,A\to BC,A\to B,A\to C,A\to A,B\to B,C\to C,AB\to AC,AB\to ABC,AB\to BC,AB\to AB,AB\to A,AB\to B,AB\to C,AC(\text{same as }AB),BC(\text{same as }AB),ABC(\text{same as }AB)$. $F_2$, the restriction of $F$ to $(A,D,E)$ is $A\to ADE,A\to AD,A\to AE,A\to DE,A\to A,A\to D,A\to E,D\to D,E(\text{same as }A),AD,AE,DE,ADE(\text{same as }A).$

$(F_1\cup F_2{)}^{+}$ is easily seen not to contain $B\to D$ since the only FD in $F_1\cup F_2$ with $B$ as the left side is $B\to B$, a trivial FD. Thus $B\to D$ is not preserved.

A simpler argument is as follows: $F_1$ contains no dependencies with $D$ on the right side of the arrow. $F_2$ contains no dependencies with $B$ on the left side of the arrow. Therefore for $B\to D$ to be preserved there must be a functional dependency $B\to \alpha$ in $F_1^{+}$ and $\alpha \to D$ in $F_2^{+}$ (so $B\to D$ would follow by transitivity). Since the intersection of the two schemas is $A$, $\alpha =A$. Observe that $B\to A$ is not in $F_1^{+}$ since $B^{+}=BD$.