Consider the sentence [math: (∃x(P(x)∧Q(x)))↔((∃xP(x))∧(∃xQ(x)))]. Which of the following is true?(If (D) is true, you must answer (D) and not (B) or (C).)

Question

BlackTom AI · Accepted Answer

Existential quantifiers do not in general distribute over conjunction: ∃x(P(x) ∧ Q(x)) implies (∃x P(x)) ∧ (∃x Q(x)), but the converse can fail because the witnesses for P and Q might be different.

1. Relevant knowledge points and concepts
- The formula is (∃x(P(x) ∧ Q(x))) ↔ ((∃x P(x)) ∧ (∃x Q(x))).  
- For a biconditional to be true in an interpretation, both directions must hold.  
- ∃x(P ∧ Q) asserts a single element satisfies both P and Q; (∃x P) ∧ (∃x Q) only asserts existence of possibly different elements for P and for Q.

2. Complete reasoning step by step
- Show the left-to-right direction: If some a satisfies P(a) ∧ Q(a), then ∃x P(x) and ∃x Q(x) are both true (use same a). So L→R is valid in every interpretation.
- Show the right-to-left direction can fail: It may be that some b satisfies P(b) and some c (different) satisfies Q(c), but no single element satisfies both P and Q. Then R is true while L is false, making the biconditional false.

3. Calculations / intermediate steps for the given options
- Option (B) interpretation: domain = integers; P(x): x ≤ 0; Q(x): x ≥ 6.
  - ∃x P(x) is true (e.g. x = 0 satisfies x ≤ 0).
  - ∃x Q(x) is true (e.g. x = 6 satisfies x ≥ 6).
  - ∃x (P(x) ∧ Q(x)) requires an integer ≤ 0 and ≥ 6 simultaneously — no such integer exists, so this is false.
  - Therefore RHS is true, LHS false, biconditional false. So the sentence is false under (B).
- Option (C) interpretation: domain = integers; P(x): x ≤ 6; Q(x): x ≥ 0.
  - ∃x P(x) is true (e.g. x = 6).
  - ∃x Q(x) is true (e.g. x = 0).
  - ∃x (P ∧ Q) asks for an integer with 0 ≤ x ≤ 6; such integers exist (e.g. x = 0), so LHS true.
  - Both sides true, biconditional true. So the sentence is NOT false under (C).

4. Why the correct answer is correct
- (B) provides a standard counterexample showing the RHS can be true while the LHS is false because the witnesses for P and Q are different and their properties are incompatible. That makes the biconditional false in that interpretation.

5. Why other options are incorrect
- (A) is wrong: the sentence is not valid (not a tautology) because a counterexample like (B) exists.  
- (C) is wrong because under the given P and Q the intersection is nonempty, so the biconditional is true, not false.  
- (D) is wrong because (C) is not a counterexample, so both (B) and (C) are not true.

6. Conclusion
The correct choice is (B). This matches the provided answer: (B) The sentence is false under the interpretation with integers, P(x): x ≤ 0 and Q(x): x ≥ 6.

Consider the sentence [math: (∃x(P(x)∧Q(x)))↔((∃xP(x))∧(∃xQ(x)))]. Which of the following is true?(If (D) is true, you must answer (D) and not (B) or (C).)Single choice

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