Writing SQL query equivalent to relational division of relational algebra?

Question

EMPLOYEE (Ssn#, Fname, Lname, Bdate, Address, Sex, Salary, Super_ssn)
WORKS_ON (Essn#, Pno, Hours)
PROJECT (Pnumber#, Pname, Plocation)

Question 1: Write an SQL query (equivalent to relational division of relational algebra) to retrieve the project name which is contributed by all employees.

Question 2: Write an SQL query (equivalent to relational division of relational algebra) to retrieve the first name and last name of all employees who work on every project.

ITEM (Item#, Item_Name, Unit_Price)
ORDER (Order#, Ord_Date, Cust#, Ord_Amt)
ORDER_ITEM (Order#, Item#, Qty)
WAREHOUSE (Warehouse#, Warehouse_City)
SHIPMENT (Order#, Warehouse#, Ship_Date)
WAREHOUSE_ITEM (Item#, Warehouse#, Qty)

Question 3: Write an SQL query (equivalent to relational division of relational algebra) to list Item_Name and Quantity that are stored in all warehouses in Sydney.

My answer for Question 1:

SELECT Pname 
FROM PROJECT
WHERE EXISTS (
    SELECT Pname 
    FROM PROJECT, EMPLOYEE 
    WHERE PROJECT.Pnumber = WORKS_ON.Pno);

My answer for Question 2:

SELECT Fname, Lname 
FROM EMPLOYEE
WHERE EXISTS (
    SELECT Fname, Lname 
    FROM EMPLOYEE, WORKS_ON, PROJECT
    WHERE EMPLOYEE.Ssn = WORKS_ON.Essn 
    AND WORKS_ON.Pno = PROJECT.Pnumber);

My answer for Question 3:

SELECT Item_Name, Qty 
FROM ITEM, ORDER_ITEM
WHERE ITEM.Item# = ORDER_ITEM.Item#
AND EXISTS(
    SELECT * 
    FROM WAREHOUSE 
    WHERE Warehouse_City = 'Sydney');

(If these are correct, I will able to convert them in relational algebra with relational division.)

What are the mistakes in my SQL queries?

Answer 1

Here is a short and concise article describing the various techniques for relational division in SQL, with an example similar to yours using students and courses instead of employees and projects: http://www.inf.usi.ch/faculty/soule/teaching/2016-fall/db/division.pdf

It offers three equivalent solutions:

SELECT DISTINCT x.Student
FROM taken AS x
WHERE NOT EXISTS (
SELECT *
FROM required AS y
WHERE NOT EXISTS (
SELECT *
FROM taken AS z
WHERE (z.Student=x.Student)
AND (z.Course=y.Course)));
SELECT Student FROM Taken
EXCEPT
SELECT Student FROM (
SELECT Student,Course
FROM (select Student FROM Taken), Required
EXCEPT
SELECT Student,Course FROM Taken);
SELECT Student
From Taken
WHERE Course IN (SELECT Course FROM Required)
GROUP BY Student
HAVING COUNT() = (SELECT COUNT() FROM Required);

Answer 2

The third seems correct. The first and second strike me as wrong. You are looking for projects where any employee contributed, not where all contributed. In essence you are doing a division on an ANY rather than an ALL criteria.

In these cases you need to change your queries to use NOT EXISTS and look for mismatches.