We define $$\overline{\bigcup_{p\in P} Sp} =\bigcap_{p\in P} \overline{Sp}$$ and
$$\overline{\bigcap_{p\in P} Sp} =\bigcup_{p\in P} \overline{Sp}$$ Which are just another way to write de morgans laws. Prove these identities when P is a finite set. You can use the "regular" de morgans law to reason your transformations.
My proof by induction:
Basis: Prove the series for $k = 2$ where k is the number of sets , $k$ is an integer.
... I did this and it's not the focus of this post
Inductive Hypothesis: Suppose these laws work for any $k$ num of sets where $k\ge 2$, so $$\overline{\bigcup_{p\in \{p1,p2,...,pk\}} Sp} =\bigcap_{p\in \{p1,p2,..,pk\}} \overline{Sp}$$ as well as $$\overline{\bigcap_{p\in \{p1,p2,...,pk\}} Sp} =\bigcup_{p\in \{p1,p2,...,pk\}} \overline{Sp}$$
Inductive Step: I want to show that the laws work for $k+1$ so I want to show that firstly : $$\overline{\bigcup_{p\in \{p1,p2,...,pk,pk+1\}} Sp} =\bigcap_{p\in \{p1,p2,..,pk,pk+1\}} \overline{Sp}$$ and then same thing for the other law.
Let's focus on the first law only.
$\overline{\bigcup_{p\in \{p1,p2,...,pk,pk+1\}} Sp} = \overline{\bigcup_{p\in \{p1,p2,...,pk\}} Sp}$ $ \cap $ $\overline{Sp_{k+1}}$ //by DeMorgans
$= {\bigcap_{p\in \{p1,p2,...,pk\}}\overline Sp} \cap \overline{Sp_{k+1}}$ //by Inductive Hypothesis
. . .
Ok, now I am stuck, what else can I do so i deduce $=\bigcap_{p\in \{p1,p2,..,pk,pk+1\}} \overline{Sp}$ ?
Thanks anyone who helps.
Let K be a collection of subsets of some universal set thing S.
Then S - $\cup\ $K = $\cap\ ${ S - A : A in K }. Proof:
x in S - $\cup\ $K iff x in S and x not in $\cup\ $K
iff x in S and for all A in K, x not in A
iff for all A in K, x in S - K
iff x in $\cap\ ${ S - A : A in K }.
The other theorem is the dual of the first and has an immediate proof by taking compliments of both sides and replacing the subsets with their compliments.
As this theorem holds for any collection of subsets, it cannot be proved by induction.
I leave for the reader to transcribe this theorem and proof into that clumbsy indexed notation, if such be their want.
Recall that these are theorems; they are not definitions. The distinction is fundamentally important.