Q.Using properties of sets, prove that for all sets $A$ and $B$, $A - (A - B) = A \cap B$.

The key idea for this proof is to express set difference in terms of intersection and complement. This allows us to leverage standard set algebra properties, such as De Morgan's laws and distributive laws, to simplify the expression.

Essential Reasoning:

1. Begin by applying the property $X - Y = X \cap Y^c$ to the outer set difference in $A - (A - B)$: $A - (A - B) = A \cap (A - B)^c$
2. Next, apply the property $X - Y = X \cap Y^c$ to the inner term $(A - B)$, then use De Morgan's Law $(X \cap Y)^c = X^c \cup Y^c$, and the double complement law $(Y^c)^c = Y$: $A \cap (A \cap B^c)^c = A \cap (A^c \cup (B^c)^c) = A \cap (A^c \cup B)$ …

To prove $A - (A - B) = A \cap B$, we need to understand what set difference means. The expression $X - Y$ represents all elements that are in set $X$ but not in set $Y$. It's like taking set $X$ and "scooping out" any elements that also happen to be in $Y$.

Let's break down the expression $A - (A - B)$:

First, consider $A - B$. This set contains all elements that are in $A$ but not in $B$.

Now, we are looking at $A - (A - B)$. This means we are taking set $A$ and removing all elements that are in the set $(A - B)$.

Think about it intuitively:

• $A$ contains elements that are unique to $A$, and elements common to $A$ and $B$.
• $A - B$ contains only the elements unique to $A$.
• When we remove $A - B$ from $A$, we are removing those elements unique to $A$.
• What's left in $A$? Only the elements that were common to $A$ and $B$. This is exactly what $A \cap B$ represents.

Let's formalize this intuition using the definition of set difference and logical equivalences.

We will start with an arbitrary element $x$ belonging to the left-hand side, $A - (A - B)$, and show that it must also belong to the right-hand side, $A \cap B$, and vice-versa. This establishes their equality.

1. Start with an element $x$ in $A - (A - B)$.

   By the definition of set difference, if $x \in A - (A - B)$, it means:

   $x \in A \text{ and } x \notin (A - B)$

2. Expand the term $x \notin (A - B)$.

   The statement $x \notin (A - B)$ means that it is not true that $x$ is in $A$ and $x$ is not in $B$.

   So, $x \notin (A - B) \iff \neg (x \in A \text{ and } x \notin B)$.

3. Apply De Morgan's Law to the negation.

   De Morgan's Law states that $\neg (P \land Q) \equiv \neg P \lor \neg Q$.

   Applying this, $\neg (x \in A \text{ and } x \notin B)$ becomes:

   $x \notin A \text{ or } x \in B$ (since $\neg (x \notin B)$ is equivalent to $x \in B$).

4. Substitute this back into the original expression.

   Now we have:

   $x \in A \text{ and } (x \notin A \text{ or } x \in B)$

5. Apply the distributive law of logic.

   The distributive law states that $P \land (Q \lor R) \equiv (P \land Q) \lor (P \land R)$.

   Applying this to our expression: …