Let [math: A] and [math: B] be sets. If [math: p] is the proposition "[math: x∈A]" and [math: q] is the proposition "[math: x∈B]", which of the following is logically equivalent to "[math: x∈A∪B]"?

Question

BlackTom AI · Accepted Answer

Union of sets corresponds to logical disjunction: x ∈ A ∪ B exactly when x ∈ A or x ∈ B. With p = "x ∈ A" and q = "x ∈ B", the proposition for membership in the union is p ∨ q.

1. Relevant knowledge points and concepts
- Set union: A ∪ B = {x | x ∈ A or x ∈ B}.
- Logical connectives: ∨ is OR (true when at least one operand is true), ∧ is AND (true only when both true), ↔ is biconditional (true when both have same truth value), ¬ is negation.
- Truth-table method: compare truth values of p ∨ q to the condition "x is in A ∪ B".

2. Step-by-step reasoning (truth table)
List all possibilities for p and q and whether x ∈ A ∪ B should hold:
- p = T, q = T: x ∈ A and x ∈ B ⇒ x ∈ A ∪ B should be True.
- p = T, q = F: x ∈ A only ⇒ x ∈ A ∪ B should be True.
- p = F, q = T: x ∈ B only ⇒ x ∈ A ∪ B should be True.
- p = F, q = F: x ∉ A and x ∉ B ⇒ x ∉ A ∪ B should be False.

Now evaluate p ∨ q for the same rows:
- T ∨ T = T
- T ∨ F = T
- F ∨ T = T
- F ∨ F = F
These match exactly the union membership pattern, so p ∨ q is equivalent to "x ∈ A ∪ B".

3. Intermediate steps / mapping to set operations
- Intersection corresponds to p ∧ q (only T when both are true).
- Set difference A \ B corresponds to p ∧ ¬q (true when in A and not in B).

4. Why the correct answer is correct
Because the definition of union uses "or", the logical disjunction p ∨ q captures exactly the membership condition for A ∪ B.

5. Why other options are incorrect
- p ∧ q: True only when x is in both A and B (this describes A ∩ B, not A ∪ B).
- p ↔ q: True when p and q have the same truth value; it is true when both are false, but for union both-false should be false, so ↔ does not match union.
- p ∧ ¬q: True only when x ∈ A but x ∉ B (describes A \ B), not the union.
- The duplicate form p ∨ q (if another symbol variant appears) is equivalent to the same OR; the intended correct choice is the disjunction.

Conclusion: The logically equivalent proposition to "x ∈ A ∪ B" is p ∨ q, matching the provided answer [math: p∨_⁡q].

Let [math: A] and [math: B] be sets. If [math: p] is the proposition "[math: x∈A]" and [math: q] is the proposition "[math: x∈B]", which of the following is logically equivalent to "[math: x∈A∪B]"?单项选择题

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