Advanced Existential Reasoning and Quantifier Interactions Quiz

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| Questions: 15 | Updated: Dec 1, 2025
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1) ∃ x (P(x) ∧ Q(x)) implies (∃ x P(x)) ∧ (∃ x Q(x)).

Explanation

If there exists an x such that both P(x) and Q(x) are true, then certainly there exists an x such that P(x) is true, and there exists an x such that Q(x) is true. Therefore, the implication holds, and the conjunction is true. 

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About This Quiz
Advanced Existential Reasoning And Quantifier Interactions Quiz - Quiz

In this quiz, you’ll dive deeper into what it really means to say “there exists” in different domains. You’ll test statements like “There exists a largest integer,” see why some quantifier claims are always true (tautologies), and analyze how things change in special cases like a domain with only one... see moreelement. Along the way, you’ll compare stronger and weaker forms such as ∃x(P(x) ∧ Q(x)) versus ∃x P(x) ∧ ∃x Q(x), and use De Morgan’s laws to negate existential and universal statements. You’ll also focus on proof techniques for existence: giving explicit witnesses, and using proof by contradiction to show that ¬∃x P(x) can’t hold. By the end, you’ll be more fluent in reading, negating, and proving existential statements, and better able to reason about how quantifiers depend on the underlying domain. see less

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2) The statement "There exists a largest integer" is:

Explanation

In the domain of integers, for any integer n, there is always a larger integer n+1. Therefore, there is no largest integer, so the statement is false. This holds regardless of the specific infinite nature of integers.

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3) Which values are correct witnesses for ∃x (x² = 9) in integers?

Explanation

A witness is any element that makes the formula true. Since both 3² = 9 and (-3)² = 9, both are witnesses. The existential claim only requires at least one, but multiple witnesses don't invalidate it. When asked "which is a witness," any correct value is acceptable, but the best answer acknowledges all possibilities.
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4) In a domain with one element {a}, which is equivalent to ∃x P(x)?

Explanation

With only one element a in the domain, saying "there exists an x such that P(x)" is equivalent to saying "P(a) is true." This is also equivalent to "for all x, P(x)" since there's only one element to consider. In singleton domains, existential and universal quantifiers collapse to the same meaning.
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5) In any nonempty domain, ∃ x (x = x) is always true.

Explanation

The statement ∃ x (x = x) means that there exists an x that is equal to itself. Since every object is equal to itself, and in first-order logic, domains are typically non-empty, there is always at least one x, so this statement is true in any non-empty domain. If domains are allowed to be empty, it would be false, but standard logic assumes non-empty domains.
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6) ∃ x (P(x) ∧ Q(x)) is stronger than ∃ x P(x) ∧ ∃ x Q(x).

Explanation

∃ x (P(x) ∧ Q(x)) requires that there is a single x that satisfies both P and Q, while ∃ x P(x) ∧ ∃ x Q(x) only requires that there exists some x satisfying P and some x satisfying Q, which could be different x's. Therefore, ∃ x (P(x) ∧ Q(x)) implies ∃ x P(x) ∧ ∃ x Q(x), but not vice versa, making it stronger.
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7) What is the logical negation of ∃x P(x)?

Explanation

The negation of "there exists an x where P holds" is "for all x, P does not hold." This is quantifier duality: ¬∃x φ ≡ ∀x ¬φ. Option D is just the negation symbol moved, not a simplified form. Option B incorrectly keeps the existential quantifier. 
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8) Which is equivalent to ¬(∀ x P(x) ∧ ∀ x Q(x))?

Explanation

By De Morgan's law, ¬(∀ x P(x) ∧ ∀ x Q(x)) is equivalent to ¬∀ x P(x) ∨ ¬∀ x Q(x). Then, by quantifier duality, ¬∀ x P(x) is equivalent to ∃ x ¬P(x), and similarly for Q. Therefore, it is equivalent to ∃ x ¬P(x) ∨ ∃ x ¬Q(x). 

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9) True or False: ∃ x ∀ y P(x,y) → ∀ y ∃ x P(x,y) is always true.

Explanation

If there exists an x such that for all y, P(x,y) is true, then for each y, there exists some x (namely, the same x) such that P(x,y) is true. Therefore, the implication ∃ x ∀ y P(x,y) → ∀ y ∃ x P(x,y) is always true. The converse is not always true.
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10) The minimal domain size to make ∃ x ∃ y (x ≠ y) true is:

Explanation

The statement ∃ x ∃ y (x ≠ y) requires that there exist two distinct elements x and y. Therefore, the domain must have at least two elements. With one element, x and y would be the same, so x ≠ y is false. With zero elements, there are no x or y, so the statement is false.
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11) True or False: ∃ x P(x) ∧ ∃ x ¬P(x) can be true in same model.

Explanation

This conjunction means that there exists some x with P(x) true and there exists some x with P(x) false. This can be true if the domain has at least two elements, one satisfying P and one not satisfying P. Therefore, it is true in models with multiple elements.
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12) Which is equivalent to ∃ x P(x) ∨ ∀ x ¬P(x)?

Explanation

This disjunction is always true because either there exists some x with P(x) true, or for all x, P(x) is false (which is equivalent to ¬∃ x P(x)). Thus, it covers all possible cases and is a tautology.
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13) Which is a VALID way to prove ∃x P(x)?

Explanation

All three are valid proof strategies. Option (A) is a constructive proof since it directly exhibits a witness (c). Option (B) gives a proof by contradiction by showing that the negation (∀x ¬P(x)) is impossible. Option (C) is equivalent to Option (B) because ¬∃x P(x) is the same as ∀x ¬P(x), so contradicting it proves ∃x P(x).
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14) Which condition makes ∃x P(x) true?

Explanation

The existential quantifier ∃ means "there exists at least one element in the domain." The formula ∃x P(x) is satisfied when you can find any single element a in your domain where P(a) is true. It does not require P to be true for all elements (that's ∀x P(x)), nor does it require multiple witnesses. One is enough.
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15) “There exists at least one x where P(x) and Q(x)” translates to:

Explanation

The phrase requires finding an x that satisfies both P and Q simultaneously, which is correctly captured by the existential quantifier with conjunction. Option C (∃x(P(x) → Q(x)))) would be true if there exists any x where either P(x) is false or Q(x) is true, which is much weaker than requiring an x that satisfies both properties. This is why conjunction (not implication) is needed when translating 'and' in existential statements.
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∃ x (P(x) ∧ Q(x)) implies (∃ x P(x)) ∧ (∃ x Q(x)).
The statement "There exists a largest integer" is:
Which values are correct witnesses for ∃x (x² = 9) in integers?
In a domain with one element {a}, which is equivalent to ∃x P(x)?
In any nonempty domain, ∃ x (x = x) is always true.
∃ x (P(x) ∧ Q(x)) is stronger than ∃ x P(x) ∧ ∃ x Q(x).
What is the logical negation of ∃x P(x)?
Which is equivalent to ¬(∀ x P(x) ∧ ∀ x Q(x))?
True or False: ∃ x ∀ y P(x,y) → ∀ y ∃ x P(x,y) is always...
The minimal domain size to make ∃ x ∃ y (x ≠ y) true is:
True or False: ∃ x P(x) ∧ ∃ x ¬P(x) can be true in same model.
Which is equivalent to ∃ x P(x) ∨ ∀ x ¬P(x)?
Which is a VALID way to prove ∃x P(x)?
Which condition makes ∃x P(x) true?
“There exists at least one x where P(x) and Q(x)” translates to:
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