4 Consider the following relational schema, DDL statements and tables:

EMPLOYEE( EmployeeID, EmployeeName, SupervisorID, DepartmentID)

PROJECT (ProjectID, EmployeeID)

DEPARTMENT( Department ID, DepartmentName)

CREATE TABLE EMPLOYEE

( EmployeeID INT PRIMARY KEY,

EmployeeName VARCHAR(50) NOT NULL,

SupervisorID INT DEFAULT 9,

DepartmentID INT,

FOREIGN KEY (SupervisorID) REFERENCES EMPLOYEE (EmployeeID)

ON DELETE SET DEFAULT ),

FOREIGN KEY (DepartmentID) REFERENCES DEPARTMENT(DepartmentID)

ON UPDATE SET NULL);

CREATE TABLE PROJECT (

ProjectID INT PRIMARY KEY,

EmployeeID INT DEFAULT 9,

FOREIGN KEY (EmployeeID) REFERENCES EMPLOYEE (EmployeeID)

ON UPDATE CASCADE);

CREATE TABLE DEPARTMENT(

DepartmentID INT PRIMARY KEY,

DepartmentName VARCHAR(50),

);

EMPLOYEE

EmployeeID	EmployeeName	SupervisorID	DepartmentID
6	A	7	6
7	B	7	6
1	C	8	7
8	D	9	3
9	E	NULL	6

PROJECT

ProjectID	EmployeeID
1	6
2	7
3	1
4	8
5	7
6	1

DEPARTMENT

DepartmentID	DepartmentName
3	X
6	Y
7	Z

Show the status of the above tables after each of the following operations:

a. Inserting a new employee whose EmployeeID, EmployeeName, SupervisorID, DepartmentID are 12, F, NULL,7 If it is rejected, explain.

EMPLOYEE

EmployeeID	EmployeeName	SupervisorID	DepartmentID
6	A	7	6
7	B	7	6
1	C	8	7
8	D	9	3
9	E	NULL	6

PROJECT

ProjectID	EmployeeID
1	6
2	7
3	1
4	8
5	7
6	1

DEPARTMENT

DepartmentID	DepartmentName
3	X
6	Y
7	Z

b. Inserting a new employee whose EmployeeID, EmployeeName, SupervisorID, DepartmentID are 12, F, 10,3. If it is rejected, explain.

EMPLOYEE

EmployeeID	EmployeeName	SupervisorID	DepartmentID
6	A	7	6
7	B	7	6
1	C	8	7
8	D	9	3
9	E	NULL	6

PROJECT

ProjectID	EmployeeID
1	6
2	7
3	1
4	8
5	7
6	1

DEPARTMENT

DepartmentID	DepartmentName
3	X
6	Y
7	Z

c. Deleting the Department ID = 6. If it is rejected, explain.