Final answer:
The question requires showing the equivalence of (S∪T)' and S' ∩ T' according to DeMorgan's Law in set theory. By analyzing the membership of an element x in both sets, we can confirm that these sets are equivalent since they contain exactly the same elements, which are not in S or T.
Step-by-step explanation:
The question asks to show how (S∪T)' is equivalent to S' ∩ T' using DeMorgan's Law. In set theory, the prime (') denotes the complement of a set, the union (∪) represents the combination of two sets, and the intersection (∩) represents the common elements in both sets. According to DeMorgan's Law, the complement of the union of two sets is the intersection of their complements. To prove this, let's assume an element x.
- If x is in (S∪T)', then x is not in S∪T. This means x is neither in set S nor in set T.
- Since x is not in S, x must be in set S', and since x is not in T, x must also be in set T'. Thus, x is in S' ∩ T'.
- Conversely, if x is in S' ∩ T', then x is both in S' and in T'. Therefore, x cannot be in either S or T, which means x must be in (S∪T)'.
This proves that (S∪T)' and S' ∩ T' are indeed equivalent, as per DeMorgan's Law.