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# One correct and one incorrect formula for set complements

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

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

1. The formula is false in general.
2. 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, .∎

3. 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, .∎

### One comment

1. Anonymous says:

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