Prove or disprove that `T=R÷S` implies `S=R÷T`?
First, your answer is "Yes" but without any proof, only a single example. You have proven nothing about the general case. If I were to mark your answer, you'd get a straight 0 (OK, maybe a 1 because you understand what relational division is and how to perform it) and a note to read up on mathematical proofs.

For the actual question, the answer is: No, it is not correct. Now to prove that, we do only need a single counter-example. Here it is:

 R                        S          ==&gt;    T=R÷S
 ------------         ---------             -----
| A | B | C |         | A | B |             | C |
| 1 | 2 | 4 |         | 1 | 2 |             | 4 |
| 1 | 2 | 9 |                               | 9 |
| 3 | 3 | 4 |
| 3 | 3 | 8 |
| 3 | 3 | 9 | 
| 4 | 1 | 2 |


But:

 R                      T      ==&gt;       R÷T != S
 ------------         -----           ---------
| A | B | C |         | C |           | A | B |
| 1 | 2 | 4 |         | 4 |           | 1 | 2 |
| 1 | 2 | 9 |         | 9 |           | 3 | 3 |         
| 3 | 3 | 4 |
| 3 | 3 | 8 |
| 3 | 3 | 9 | 
| 4 | 1 | 2 |




This is actually similar to integer division (when we discard the remainder). Sometimes it's Yes:

37 / 7 = 5 (the remainder 2 is discarded) and
37 / 5 = 7 (the remainder 2 is discarded)


and sometimes is No:

37 / 10 = 3 (the remainder 7 is discarded) but
37 / 3 = 12 (the remainder 1 is discarded)


and of course 12 != 10, and the single counter-example disproves the general case. 



What can be proven is this:


If R÷S = T, then S is a (not necessarily strict) subset of R÷T.