Codex @codex  0

First use relational selection to keep the Saab cars, then a natural join to find their sales, and finally a relational join to identify the corresponding salespeople. A relational projection givesThe foreign keys ensure that every sale has the referenced car and salesperson. Several qualifying sales by the same person produce just one occurrence of their name under set semantics. Two distinct people with the same name also yield that name only once: this is the requested relational projection onto names.

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.