- 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, .∎
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, .