This is the division operator. When we say (r / s)- first of all, the attribute set of r must be a super set of that of s. Now, (r/s) will select all tuples from r, which doesn't have any missing corresponding tuple in s- i.e., it works like a quotient.
Example:
R(stud_no,name,sex,course) and S(stud_no,course)
Let $r = \{<10, \text{Rahul}, M, CS>, <10, \text{Rahul}, M, Che>, <11, \text{Ben}, M, CS>\},\\ s = \{<CS>, <Che>\}$
$$∏_{R−S}(r) - \left(∏_{R−S} \left(∏_{R−S}(r) \times (s) \right) -∏_{R−S,S}(r)\right)$$
- $∏_{R−S}(r)$ - Select all attributes of R but not S from r.
$\{<10, \text{Rahul},M>, <11, \text{Ben},M>\}$
- $∏_{R−S,S}(r)$ - Select all attributes of R but not S union all attributes of S from r, which is same as selecting all attributes from r, but with attributes of S at end so that we can subtract correctly.
$\{<10, \text{Rahul}, M, CS>, <10, \text{Rahul}, M, Che>, <11, \text{Ben}, M, CS>\}$
- $\left(∏_{R−S} \left(∏_{R−S}(r) \times (s) \right) -∏_{R−S,S}(r)\right)$ - Take the cross product of r (after removing attributes of s) and s and subtract all tuples in r. So, this returns the tuples in $r\times s$ for which $r$ doesn't have a tuple.
$∏_{R−S}\{<\text{Rahul}, M, 10, CS>, <\text{Rahul}, M, 10, Che>, <\text{Ben}, M, 11, CS>, <\text{Ben}, M, 11, Che>\}\\ - \{<10, \text{Rahul}, M, CS>, <10, \text{Rahul}, M, Che>, <11, \text{Ben}, M, CS>\} $
$=∏_{R−S} \{ <\text{Ben}, M, 11, Che>\} $
$ = \{<11, \text{Ben}, M>\}$
So, $∏_{R−S}(r) - \left(∏_{R−S} \left(∏_{R−S}(r) \times (s) \right) -∏_{R−S,S}(r)\right) \\ = \{<10, \text{Rahul}, M>, <11, \text{Ben}, M>\} - \{<11, \text{Ben}, M>\} \\= \{<10, \text{Rahul}, M>\}$
That is using division operator we are selecting the students who have selected ALL the courses.
