Chapter 16, Problem 2PE

a.

Explanation of Solution

To prove:

$E_1{\left|><\right|}_{\theta }\left(E_2-E_3\right)\equiv E_1{\left|><\right|}_{\theta }{E}_2-E_1{\left|><\right|}_{\theta }{E}_3$

Proof:

• To enhance the efficiency of ${\text{E}}_{\text{1}}|><|$ the queries, equivalences such as commutative, distributive, and associative properties can be applied.
• In the given equivalence, distributive property is applied, which means the natural join is distributive over difference operation.
• Let $E_1{\left|><\right|}_{\theta }{E}_2$ as “R1” and $E_1{\left|><\right|}_{\theta }{E}_3$ as “R2”.
• If a record “r” belongs to left hand side (LHS) of the given equivalence, it will also be the member of “R1” part of right hand side (RHS) of the above equation

b.

Explanation of Solution

To prove:

${\sigma }_{\theta }\left({}_A{\gamma }_F\left(E\right)\right)\equiv \left({}_A{\gamma }_F\left({\sigma }_{\theta }\left(E\right)\right)\right)$ where “θ” uses only attributes from “A”.

Proof:

• It is given that, “θ” uses only attributes from “A”.
• In the LHS of the equation, if  a record “r” in the result of ${}_A{\gamma }_F\left(E\right)$ is selected by the selection operation, then on the right hand side, all the records in relation “E” having value in “A” is equal to r[A] are selected by the selection operation and it is represented as,

  

c.

Explanation of Solution

To prove:

${\text{σ}}_{\text{θ}}$ (E₁⟕E₂) ≡ ${\text{σ}}_{\text{θ}}$ (E₁)⟕E₂, where “θ” uses only attributes from “E₁”.

Proof:

• It is given that, “$\theta$” uses only attributes from “A”.
• In the LHS of the equation, if  a record “r” in the result of “E₁⟕E₂” is selected by the selection operation, then on the right hand side, all the records in relation “E₁” having value is equal to r[E₁] are selected by the selection operation and it is represented as,

$\forall \text{r,r}\notin {\text{σ}}_{\text{θ}}$(E₁⟕E₂)  $\Rightarrow \text{r}\notin {\text{σ}}_{\text{θ}}\text{\hspace{0.17em}<}$