### Database Schema:
CREATE TABLE taughtBy (
	course_id	INTEGER	NULL	PRIMARY KEY,
	p_id	INTEGER	NULL	PRIMARY KEY,
	foreign key (course_id) references course (course_id),
	foreign key (p_id) references person (p_id)
);

CREATE TABLE course (
	course_id	INTEGER	NULL	PRIMARY KEY,
	courseLevel	TEXT	NULL
);

CREATE TABLE person (
	p_id	INTEGER	NULL	PRIMARY KEY,
	professor	INTEGER	NULL,
	student	INTEGER	NULL,
	hasPosition	TEXT	NULL,
	inPhase	TEXT	NULL,
	yearsInProgram	TEXT	NULL
);

CREATE TABLE advisedBy (
	p_id	INTEGER	NULL	PRIMARY KEY,
	p_id_dummy	INTEGER	NULL	PRIMARY KEY,
	foreign key (p_id) references person (p_id),
	foreign key (p_id_dummy) references person (p_id)
);

### User's Query:
For the professor who advised student no.6, please list the IDs of the courses he or she teaches.

### Extra Knowledge:
professor refers to p_id_dummy and professor = 1; student no.6 refers to advisedBy.p_id = 6; IDs of the courses refers to taughtBy.course_id
SELECT T2.course_id FROM taughtBy AS T1 INNER JOIN course AS T2 ON T1.course_id = T2.course_id INNER JOIN advisedBy AS T3 ON T3.p_id = T1.p_id WHERE T1.p_id = 9