Question

The negation of a statement '$x \in A \cap B \rightarrow (x \in A \text{ and } x \in B)$' is

• A. $x \in A \cap B \rightarrow (x \in A \text{ or } x \in B)$
• B. $x \in A \cap B \text{ and } (x \notin A \text{ or } x \notin B)$
• C. $x \in A \cap B \text{ or } (x \in A \text{ or } x \in B)$
• D. $x \notin A \cap B \text{ and } (x \in A \text{ and } x \in B)$

Hint:

Remember that the negation of an "If P, then Q" statement is always "P happened, and NOT Q occurred". This pattern instantly tells you that the first part must remain unchanged, which easily leaves option (B) as the correct choice!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the formal negation of a conditional mathematical logic statement containing set theory components.

Step 2: Key Formula or Approach:
Let's represent the individual statements with logical variables:

• Let $p$ be the proposition: $x \in A \cap B$

• Let $q$ be the proposition: $(x \in A \text{ and } x \in B)$
The given statement is a conditional implication: $p \rightarrow q$. The standard rule for negating a conditional statement is: $$ \sim(p \rightarrow q) \equiv p \wedge \sim q $$

Step 3: Detailed Explanation:
We need to find $\sim q$, which requires negating a conjunction statement using De Morgan's laws: $$ \sim (x \in A \text{ and } x \in B) \equiv (x \notin A \text{ or } x \notin B) $$ Now, combining it with our conditional negation rule $p \wedge \sim q$: $$ x \in A \cap B \text{ and } (x \notin A \text{ or } x \notin B) $$

Step 4: Final Answer:
The exact logical negation corresponds to option (B).