de morgan law $A\setminus (B \cap C) = (A\setminus B) \cup (A\setminus C) $

Solution 1:

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, $x\in (A\setminus B)\cup(A \setminus C)$.

Solution 2:

You write,

Can we say that:

$x∈A$ and $x∉B$ and $x∈C$ OR
$x∈A$ and $x∉C$ and $x∈B$

In short, the answer is no, we cannot say that, because you've not told us what $x$ is. You've got to tell us what this $x$ is before you start talking about it.

To show that sets $S$ and $T$ are equal, a typical strategy is to show $S \subseteq T$ and $T\subseteq S$. In order to do this, you're going to let $x$ be some arbitrary member of $S$ and show that this implies that it is in $T$, and then let $y$ be some arbitrary member of $T$ and show that this implies that it is in $S$. Once that's done, you've won.

In your example, $A\setminus (B\cap C)$ is your $S$. So begin by explicitly writing:

"Let $x\in A\setminus (B\cap C)$."

Now what do we know about $x$? Well, if $x\in A\setminus (B\cap C)$ then $x\in A$ but $x\notin (B\cap C)$. Unpacking the second part of that last sentence, if $x\notin (B\cap C)$ then either $x\notin B$ or $x\notin C$—otherwise $x$ would be in their intersection. So this gives us two cases: either $x\notin B$ or $x\notin B$. Suppose that $x\notin B$. We know that $x\in A$ so this means that $x\in A\setminus B$, and so $x\in (A\setminus B)\cup (A\setminus C)$. In the second case, we know that $x\in A$ and $x\notin C$, which means $x\in A\setminus C$, which further implies that $x\in (A\setminus B)\cup (A\setminus C)$. Since these are the only two possible cases, we conclude that $x\in (A\setminus B)\cup (A\setminus C)$.

To go to the other direction, again, begin by explicitly writing

"Let $y\in (A\setminus B) \cup (A\setminus C)$"

And from there, determine what you can say about $y$. I'll let you finish the other direction.