First part :
I want to prove the following De Morgan's law : ref.(dfeuer)
$A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $
Second part:
Prove that $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C) $
Proof:
Let $y\in (A\setminus B) \cup (A\setminus C)$
$(A\setminus B) \cup (A\setminus C) = (y \in A\; \land y \not\in
B\;) \vee (y \in A\; \land y \not\in C\;)$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A\ \land (
y \not\in B\; \vee \; y \not\in C ) $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= y \in A \land
(\lnot ( y\in B) \lor \lnot( y\in C) )$
According to De-Morgan's theorem :
$\lnot( B \land C) \Longleftrightarrow (\lnot B \lor \lnot C)$ thus
$y \in A \land (\lnot ( y\in B) \lor \lnot( y\in C) ) = y \in A \land y \not\in (B \land C)$
We can conclude that $(A\setminus B) \cup (A\setminus C) = A\setminus (B \cap C)$
Best Answer
I'll guide you through one direction. You will need to figure the other out yourself.
Let $x\in A \setminus (B\cap C)$.
Then by the definition of $\setminus$, $x\in A$ and $x\notin B\cap C$. The latter statement can be translated into $\lnot (x \in B \cap C)$.
By the definition of $B\cap C$, $x\in B\cap C \iff (x \in B) \land (x\in C)$.
Thus we conclude that $\lnot ((x \in B) \land (x\in C))$.
By De Morgan's laws for logic, we can rewrite this as $\lnot(x\in B)\lor\lnot(x\in C)$.
Suppose that $\lnot (x\in B)$. Then since $x\in A$, $x\in A\setminus B$.
Suppose instead that $\lnot (x \in C)$. Then since $x\in A$, $x\in A \setminus C$.
As one of these must be true, $(x\in A\setminus B)\lor (x\in A \setminus C)$.
By the definition of union, $x\in (A\setminus B)\cup(A \setminus C)$.