How to find all pizzerias that serve every pizza eaten by people over 30?

Try doing a join using conditions rather than a cross. The conditions would be sure that you match up the records correctly (you only include them if they are in both relations) rather than matching every record in the first relation to every record in the second relation.

Definitely this is the concept of division operator in relational algebra.

But I tried on that course. The RA Relational Algebra Syntax doesn't support dev operator. So I used diff and cross instead. Here is my solution:

\project_{pizzeria}(Serves)
\diff
\project_{pizzeria}(
    (\project_{pizzeria}(Serves) 
    \cross 
    \project_{pizza}(\project_{name}(\select_{age>30}(Person))\join Eats))
    \diff
    \project_{pizzeria,pizza}(Serves)
)

The solution is the join div operator http://en.wikipedia.org/wiki/Relational_algebra#Division_.28.C3.B7.29

See http://oracletoday.blogspot.com/2008/04/relational-algebra-division-in-sql.html

1. On slide 6, note that n is (3 1 7).

2. On the next slide, o / n results in (4 8).

3. If o would also have (12 3) and (12 1) but not (12 7), 12 would not be part of o / n.

You should be able to fill in an example in the formula on Slide 16 and work it out.

1. In your case, we take ɑ to be:

   Chicago Pizza   cheese  cheese
   Chicago Pizza   cheese  supreme
   Chicago Pizza   supreme cheese
   Chicago Pizza   supreme supreme
   Dominos         cheese  cheese
   Dominos         cheese  supreme
   

2. Then we take β to be:

   cheese cheese
   cheese supreme
   supreme cheese
   supreme supreme
   

3. The result of ɑ / β would then be:

   Chicago Pizza
   

Dominos is not part of this because it misses (supreme cheese) and (supreme supreme).