 Solution

ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/databases/1/c/i/solution

Assume the two database queries have compatible output database attributes. Then

True: Q_{1} - Q_{2} is safe.

For a finite input database, the result of the relational difference is a subset of the finite result of

Q_{1}

, so it is finite and introduces no new values. For the usual domain independence of a database query, let

D

be the input database and let

A \subseteq A^{'}

be two ambient value domains containing its data and the query constants. Database query safety gives

Q_{i}^{A} (D) = Q_{i}^{A^{'}} (D) (i = 1, 2),

and therefore

Q_{1}^{A} (D) ∖ Q_{2}^{A} (D) = Q_{1}^{A^{'}} (D) ∖ Q_{2}^{A^{'}} (D) .

Although a domain relational calculus expression for the relational difference uses negation, its positive membership requirement in

Q_{1}

supplies the necessary range restriction. Negation is not by itself a reason for failure of database query safety.

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