Q.Match the following sets for all sets $A$, $B$ and $C$:
Column I:

The key idea for solving these problems involves applying fundamental Set Operations (union, intersection, complement, difference) and properties of the Cartesian Product, along with De Morgan's Laws and absorption laws.

Here is the step-by-step reasoning for each expression:

(i) $((A' \cup B') - A)'$

1. Apply De Morgan's Law: $(A' \cup B') = (A \cap B)'$.
2. Use the definition of set difference $X - Y = X \cap Y'$: $((A \cap B)' \cap A')'$.
3. Apply De Morgan's Law again: $(A \cap B) \cup A$.
4. Since $A \cap B$ is a subset of $A$, their union is $A$. This matches (b) $A$.

(ii) $[B' \cup (B' - A)]'$

1. Use the definition of set difference $X - Y = X \cap Y'$: $[B' \cup (B' \cap A')]'$.
2. Apply the absorption law $X \cup (X \cap Y) = X$: $[B']'$.
3. Apply the double complement law $(X')' = X$: $B$. This matches (c) $B$.

(iii) $(A - B) - (B - C)$

1. Use the definition of set difference $X - Y = X \cap Y'$: $(A \cap B') - (B \cap C')$.
2. Apply the definition again: $(A \cap B') \cap (B \cap C')'$.
3. Apply De Morgan's Law: $(A \cap B') \cap (B' \cup C)$.
4. Distribute and simplify: $(A \cap B' \cap B') \cup (A \cap B' \cap C) = (A \cap B') \cup (A \cap B' \cap C) = A \cap B'$.
5. Convert back to set difference: $A - B$. This matches (a) $A - B$.

(iv) $(A - B) \cap (C - B)$

1. Use the definition of set difference $X - Y = X \cap Y'$: $(A \cap B') \cap (C \cap B')$.
2. Rearrange and simplify: $A \cap C \cap B'$.
3. Convert back to set difference: $(A \cap C) - B$. This matches (f) $(A \cap C) - B$.

(v) $A \times (B \cap C)$

1. An element $(x, y)$ is in $A \times (B \cap C)$ if $x \in A$ and $y \in (B \cap C)$.
2. This means $x \in A$ and ($y \in B$ and $y \in C$). …

The core idea behind solving this problem is to systematically simplify each expression in Column I using the fundamental laws of set algebra and the definitions of set operations. For expressions involving set difference and complement, converting them to expressions involving only union and intersection often makes simplification easier. For Cartesian products, we rely on their definition in terms of ordered pairs and how they interact with union and intersection.

Here are the key definitions and laws we will use:

• Set Difference: $X - Y = X \cap Y'$
• De Morgan's Laws:
  • $(X \cup Y)' = X' \cap Y'$
  • $(X \cap Y)' = X' \cup Y'$
• Double Complement: $(X')' = X$
• Absorption Laws:
  • $X \cup (X \cap Y) = X$
  • $X \cap (X \cup Y) = X$
• Distributive Laws:
  • $X \cup (Y \cap Z) = (X \cup Y) \cap (X \cup Z)$
  • $X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z)$
• Cartesian Product: $X \times Y = \{ (x, y) \mid x \in X \text{ and } y \in Y \}$

Let's simplify each expression from Column I:

1. Simplify $((A' \cup B') - A)'$ We start by simplifying the expression inside the outermost complement.
   • First, apply the definition of set difference $X - Y = X \cap Y'$ to $(A' \cup B') - A$:

*   Now, we have $(A' \cup B') \cap A'$. Notice that $A'$ is common to both parts of the intersection. This is a direct application of the absorption law $X \cap (X \cup Y) = X$, where $X = A'$ and $Y = B'$.

*   Finally, we apply the outermost complement:

This matches **(b) $A$** from Column II.

2. Simplify $[B' \cup (B' - A)]'$

Again, we simplify the expression inside the outermost complement first.

* Apply the definition of set difference $X - Y = X \cap Y'$ to $(B' - A)$:

*   Substitute this back into the expression:

*   This is a direct application of the absorption law $X \cup (X \cap Y) = X$, where $X = B'$ and $Y = A'$.

*   Finally, apply the outermost complement:

This matches **(c) $B$** from Column II.

3. Simplify $(A - B) - (B - C)$

We will convert all set differences to intersections with complements.

* Convert $(A - B)$:

*   Convert $(B - C)$:

*   Substitute these back into the original expression:

*   Now, apply the definition of set difference $X - Y = X \cap Y'$ to this entire expression, where $X = (A \cap B')$ and $Y = (B \cap C')$:

*   Apply De Morgan's Law to $(B \cap C')'$:

*   Substitute this back:

*   Now, distribute $(A \cap B')$ over the union $(B' \cup C)$:

*   Since $B' \cap B' = B'$, the first term simplifies:

*   This is another application of the absorption law $X \cup (X \cap Y) = X$, where $X = (A \cap B')$ and $Y = C$.

*   Finally, convert $A \cap B'$ back to set difference:

This matches **(a) $A - B$** from Column II.

4. Simplify $(A - B) \cap (C - B)$

Convert the set differences to intersections with complements.

* Convert $(A - B)$:

*   Convert $(C - B)$:

*   Substitute these back into the original expression:

*   Since intersection is associative and commutative, we can rearrange and group terms:

*   Since $B' \cap B' = B'$:

*   Convert this back to set difference:

This matches **(f) $(A \cap C) - B$** from Column II.

5. Simplify $A \times (B \cap C)$

The Cartesian product distributes over intersection. We can prove this by using the definition of ordered pairs.

* An element $(x, y)$ belongs to $A \times (B \cap C)$ if and only if:

*   By the definition of intersection, $y \in (B \cap C)$ means $y \in B$ and $y \in C$. So:

*   We can rearrange the logical "and" statements: …