Using Relational Algebra, print all tuples where the project city and parts city are the same when the parts and projects tables are joined together.

Relational Algebra that is used in my class:

() Select [] Project * Natural join - Difference & Intersection | Union

parts('pno', 'pname', 'color', 'weight', 'pcity') ===== ('p1', 'Nut', 'Red', '12', 'London') ('p2', 'Bolt', 'Green', '17', 'Paris') ('p3', 'Screw', 'Blue', '17', 'Rome') ('p4', 'Screw', 'Red', '14', 'London') ('p5', 'Cam', 'Blue', '12', 'Paris') ('p6', 'Cog', 'Red', '19', 'London') ('p7', 'Bolt', 'Black', '7', 'New York') ('p8', 'Cam', 'Yellow', '21', 'London')

projects('jno', 'jname', 'jcity') ======== ('j1', 'Sorter', 'Paris') ('j2', 'Punch', 'Rome') ('j3', 'Reader', 'Athens') ('j4', 'Console', 'Athens') ('j5', 'Collator', 'London') ('j6', 'Terminal', 'Oslo') ('j7', 'Tape', 'London') ('j8', 'Ruler', 'New York')

suppliers('sno', 'sname', 'status', 'scity') ========= ('s1', 'Smith', '20', 'London') ('s2', 'Jones', '10', 'Paris') ('s3', 'Blake', '30', 'Paris') ('s4', 'Clark', '20', 'London') ('s5', 'Adams', '30', 'Athens')