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Construct the required models and the actual customer–model purchases using relational projection and a natural join:The Cartesian product $I\times M$ lists every obligation: every customer paired with every model present in Car. The relational difference $(I\times M)-V$ consists of unfulfilled obligations. Remove the customer identifiers occurring in this database relation and only then recover names:This is an explicit relational division construction with all customers as the candidate universe. If there are no models, there are no missing obligations, so every customer qualifies. A version starting solely from customers occurring in $V$ would incorrectly omit customers with no purchases in that empty-model case.

The quantified object here is the model database attribute, so two makes sharing a model value contribute one required value. If a different application identified a model by the pair (make, model), both the required and purchased relational projections would use that pair. Also, Car contains unsold cars: a model available only in unsold inventory is still a required model, and then nobody qualifies. Testing identifiers before names prevents different customers with the same name from pooling their purchases.

The right-hand relational projection lists the models of cars for which at least one recorded sale was made by a salesperson earning strictly more than $30000$. The outer relational difference removes precisely those models from all models present in Car. Thus the answer isThis includes models with no sales at all, as well as models sold exclusively by salespeople whose salaries are at most $30000$. A model sold by both a low-paid salesperson and a high-paid salesperson is excluded. A salary exactly equal to $30000$ does not trigger exclusion. These are model-level statements: one unsold car of a model does not rescue that model if a different car of the same model had a qualifying sale.

Assume the two database queries have compatible output database attributes. ThenFor 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 $A\subseteq A^{\prime }$ be two ambient value domains containing its data and the query constants. Database query safety givesand thereforeAlthough 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.

Find the salesperson identifiers satisfying each price condition using relational selection and relational projection:Intersect these identifier sets, relational join the result to the salesperson records, and apply relational projection to names:If intersection is not a primitive of the chosen relational algebra, use $L\cap H=L-(L-H)$ with relational difference. There is no need to access Car: the foreign key already ensures that every sale refers to a car.

Intersecting names instead of identifiers is wrong. Two different salespeople called Alex, one making only $10000$ sales and the other only $30000$ sales, would make Alex appear in both name sets although neither person satisfies the condition. Nor can one sale satisfy both price tests: the two distinct prices require distinct sale records.

An algebra of operations on database relations. Classical relational algebra has relational selection, relational projection, relational join, Cartesian product, union and relational difference; database aggregation requires an extension.

A relational algebra operation expressing a universal matching requirement. An explicit relational difference construction can specify the intended candidate universe, including the case of an empty divisor.