Prove or disprove the following inference rules for functional dependencies. A proof can be made either by a proof argument or by using inference rules IR1 through IR3. A disproof should be performed by demonstrating a relation instance that satisfies the conditions and functional dependencies in the left-hand side of the inference rule but does not satisfy the dependencies in the right-hand side. a. \(\\{W \rightarrow Y, X \rightarrow Z\\} \vDash\\{W X \rightarrow Y\\}\) b. \(\\{X \rightarrow Y\\}\) and \(Y \supseteq Z \vDash\\{X \rightarrow Z\\}\) ?. \(\\{X \rightarrow Y, X \rightarrow W, W Y \rightarrow Z\\} \vDash\\{X \rightarrow Z\\}\) d. \(\\{X Y \rightarrow Z, Y \rightarrow W\\} \vDash\\{X W \rightarrow Z\\}\) e. \(\\{X \rightarrow Z, Y \rightarrow Z\\} \vDash\\{X \rightarrow Y\\}\) f. \(\quad\\{X \rightarrow Y, X Y \rightarrow Z\\} \vDash\\{X \rightarrow Z\\}\) \(\mathrm{g} .\\{X \rightarrow Y, Z \rightarrow W\\} \vDash\\{X Z \rightarrow Y W\\}\) h. \(\\{X Y \rightarrow Z, Z \rightarrow X\\} \vDash\\{Z \rightarrow Y\\}\) ¡. \(\\{X \rightarrow Y, Y \rightarrow Z\\} \vDash\\{X \rightarrow Y Z\\}\) j. \(\quad\\{X Y \rightarrow Z, Z \rightarrow W\\} \vDash\\{X \rightarrow W\\}\)

Step by step solution

01

Exercise a

From the dependency set, apply the augmentation rule on \(W \rightarrow Y\) and \(X \rightarrow Z\) to get \(WX \rightarrow YX\) and \(WX \rightarrow ZX\) respectively. By applying the union rule, you get \(WX \rightarrow YZX\) and thus \(WX \rightarrow Y\).

02

Exercise b

Since \(Y \supseteq Z\), you know that \(Y \rightarrow Z\). By applying the transitivity rule on \(X \rightarrow Y\) and \(Y \rightarrow Z\), it infers that \(X \rightarrow Z\).

03

Exercise c

By applying the augmentation rule on \(X \rightarrow Y\) and \(X \rightarrow W\), you get \(X \rightarrow YW\). Now apply the transitivity rule with \(XYW \rightarrow Z\), it infers that \(X \rightarrow Z\).

04

Exercise d

You can't infer \(XW \rightarrow Z\) as there is no relation between X and W.

05

Exercise e

You can't infer \(X \rightarrow Y\) because there isn't any dependency of X on Y.

06

Exercise f

By using the augmentation rule and making \(X \rightarrow YX\). Now applying the transitivity rule with \(XY \rightarrow Z\), we infer \(X \rightarrow Z\).

07

Exercise g

You can't infer \(XZ \rightarrow YW\) as W is completely independent of the other attributes.

08

Exercise h

You can't infer \(Z \rightarrow Y\) as there is no direct functional dependency of Z on Y.

09

Exercise i

By applying the transitivity rule on \(X \rightarrow Y\) and \(Y \rightarrow Z\), it infers that \(X \rightarrow Z\), and thus \(X \rightarrow YZ\).

10

Exercise j

You can't infer \(X \rightarrow W\) as there isn't any dependency between X and W.

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