Understanding Set Theory Identities

The question asks us to find the equivalent expression for the set operation $A \cup (B \cap C)$ using standard set notation. This involves understanding how set operations like union ($\cup$) and intersection ($\cap$) interact with each other.

Analyzing the Given Expression: $A \cup (B \cap C)$

The expression $A \cup (B \cap C)$ represents the union of set A with the intersection of sets B and C. This means an element belongs to this resulting set if it is in set A OR it is in the intersection of set B and set C (which means it is in both B AND C).

The Distributive Property of Set Operations

Set operations follow certain laws, similar to arithmetic operations. One important law is the distributive property. There are two forms of the distributive property in set theory:

1. Union distributes over intersection: $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
2. Intersection distributes over union: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

The given expression, $A \cup (B \cap C)$, directly matches the left side of the first distributive property listed above.

Applying the Distributive Property

Using the distributive property of union over intersection, we can rewrite the expression $A \cup (B \cap C)$ as $(A \cup B) \cap (A \cup C)$.

Let's break down why $(A \cup B) \cap (A \cup C)$ is equivalent:

• An element is in $(A \cup B) \cap (A \cup C)$ if it is in $(A \cup B)$ AND it is in $(A \cup C)$.
• Being in $(A \cup B)$ means the element is in A OR in B.
• Being in $(A \cup C)$ means the element is in A OR in C.
• So, an element is in $(A \cup B) \cap (A \cup C)$ if (it is in A OR in B) AND (it is in A OR in C).
• This logical statement is equivalent to (the element is in A) OR (the element is in B AND in C).
• In set notation, this is equivalent to $A \cup (B \cap C)$.

This confirms the identity $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Comparing with the Options

Now, let's look at the provided options:

1. (A U B) ∩ (A ∩ C): This option has $A \cap C$ on the right side, which does not match the distributive property.
2. (A ∩  B) U  (A ∩ C): This option represents $(A \cap B) \cup (A \cap C)$, which is the distributive property of intersection over union (the second form), not the first one we need.
3. (A U B) U (A U C): This option simplifies to $(A \cup B) \cup C$, which is not equivalent to $A \cup (B \cap C)$.
4. (A U B) ∩ (A U C): This option matches $(A \cup B) \cap (A \cup C)$, which is the result of applying the distributive property to $A \cup (B \cap C)$.

Therefore, the expression $A \cup (B \cap C)$ is equivalent to $(A \cup B) \cap (A \cup C)$.

Revision Table: Key Set Identity

Additional Information: Proving Set Identities

Set identities like the distributive property can be formally proven using element-wise proofs or visually demonstrated using Venn diagrams.

Element-wise Proof Sketch:

To prove $LHS = RHS$ ($A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$):

1. Show $LHS \subseteq RHS$: Assume an element $x \in A \cup (B \cap C)$. Then $x \in A$ or $x \in (B \cap C)$.
   • Case 1: $x \in A$. If $x \in A$, then $x \in (A \cup B)$ and $x \in (A \cup C)$. Thus, $x \in (A \cup B) \cap (A \cup C)$.
   • Case 2: $x \in (B \cap C)$. If $x \in (B \cap C)$, then $x \in B$ and $x \in C$. If $x \in B$, then $x \in (A \cup B)$. If $x \in C$, then $x \in (A \cup C)$. Since $x$ is in both $(A \cup B)$ and $(A \cup C)$, $x \in (A \cup B) \cap (A \cup C)$.
   In both cases, $x \in (A \cup B) \cap (A \cup C)$. So, $A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$.
2. Show $RHS \subseteq LHS$: Assume an element $x \in (A \cup B) \cap (A \cup C)$. Then $x \in (A \cup B)$ and $x \in (A \cup C)$.
   • $x \in (A \cup B)$ means $x \in A$ or $x \in B$.
   • $x \in (A \cup C)$ means $x \in A$ or $x \in C$.
   So, $(x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \in C)$. This logical statement is equivalent to $x \in A \text{ or } (x \in B \text{ and } x \in C)$. In set notation, this means $x \in A \cup (B \cap C)$. So, $(A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$.

Since $LHS \subseteq RHS$ and $RHS \subseteq LHS$, we conclude $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

Venn Diagram Demonstration:

Drawing Venn diagrams for both $A \cup (B \cap C)$ and $(A \cup B) \cap (A \cup C)$ visually shows that they represent the same region, thereby demonstrating the identity.

Understanding these fundamental set theory identities is crucial for solving problems involving sets and logic.