Assuming [tex]A'[/tex] denotes the complement of the set [tex]A[/tex], i.e. all elements in the universal set that do not belong to [tex]A[/tex], then we can prove this in general.
To establish equality between two sets, you need to show that they are both subsets of one another.
- Prove [tex](A\cup B)'\subseteq A'\cap B'[/tex]:
Let [tex]x\in(A\cup B)'[/tex].
By definition of set complement, this means [tex]x\not\in A\cup B[/tex].
By definition of set union, [tex]x\not\in A[/tex] and [tex]x\not\in B[/tex].
By definition of complement, [tex]x\in A'[/tex] and [tex]x\in B'[/tex].
By definition of set intersection, [tex]x\in A'\cap B'[/tex].
Therefore [tex](A\cup B)'[/tex] is a subset of [tex]A'\cap B'[/tex], because membership of some arbitrary element in the first set directly implies membership in the second set.
- Prove [tex]A'\cap B'\subseteq(A\cup B)'[/tex]:
Let [tex]x\in A'\cap B'[/tex]. The proof follows similarly as above.
By definition of intersection, [tex]x\in A'[/tex] and [tex]x\in B'[/tex].
By definition of complement, [tex]x\not\in A[/tex] and [tex]x\not\in B[/tex].
By definition of union, [tex]x\not\in(A\cup B)[/tex].
By definition of complement, [tex]x\in(A\cup B)'[/tex].
Therefore [tex]A'\cap B'[/tex] is a subset of [tex](A\cup B)'[/tex].
And hence both sets are equal, regardless of what the sets may be.
