- Given the two formulas:
Identify which one is always correct and which one is sometimes wrong, and prove the result.

- Give an additional condition for the incorrect formula to be always true.

- The formula is false in general.
- To make the first formula in part (a) always correct, we add the condition that . The claim is then:

If , then .

*Proof.*Let be any element in , then and . But, means that or (since this is the negation of which means and , so its negation is or ). So, if , then and hence . On the other hand, if , then as well. Hence, .

For the reverse inclusion, let . Then or . If , then and . Since, , then we know (since every must be in ). Therefore, . On the other hand, if , then since (by our additional hypothesis) we know . Further, since , we know . Therefore, . So, .

Therefore, indeed, .∎

* Proof. * Let

Then,

Thus, the formula does not hold in this counterexample.∎

The formula is correct.

* Proof. * Let be any element of . This means and , which in turn means and . Since and we have . Then, since , we have . Thus, .

For the reverse inclusion, let be any element of . This gives us and . The first part in turn gives us and . But then we have in neither nor ; hence, . Since , we then have . Therefore, .

Therefore, .∎

for x in A-(B-C), how come is that equivalent to x in (A-B)UC being that the first affirmation says that x is in A necessarily and the second one means that x is either in C or A but not in B?

for example if A={1,2,3,4} B={2,3,4} and C={3,4,5}

A-(B-C) = A-{2} = {1,3,4} (i.e. (A-B)U(A⋂C))

while when you say

(A-B)UC = {1}UC = {1,3,4,5} (but 5 does not belong to A)

which is the whole point of this affirmation being wrong in the first place