Negating and Translating Universal Statements Quiz

The original statement “All chickens can fly” asserts that every single chicken possesses the property of being able to fly. To negate a universal statement, logic requires asserting that there exists at least one counterexample in which the property fails. The statement “Some chickens cannot fly” says exactly this: it claims that at least one chicken does not fly. Because the logical negation of “all P are Q” always has the form “there exists a P that is not Q,” choice a expresses the correct logical negation.

The phrase “For all x” explicitly signals universal quantification. The property being asserted of every x is simply “x is greater than zero.” Therefore the symbolic representation must combine the universal quantifier ∀x with the specific inequality x > 0. Only choice d expresses precisely that every object in the domain satisfies the condition of being greater than zero. The other choices either remove the quantifier, replace the predicate with an unspecified P(x), or change the quantifier type altogether.

While both ¬∀x P(x) and ∃x ¬P(x) are logically equivalent representations of 'Not all numbers are prime,' the direct translation of the English negation produces ¬∀x P(x). In formal practice, we would typically transform this to ∃x ¬P(x) using quantifier negation rules, as this form explicitly identifies the existence of a counterexample.

The sentence can be represented as either ¬∃x(P(x) ∧ ¬H(x)) or the logically equivalent ∀x(P(x) → H(x)). While both are formally correct, ¬∃x(P(x) ∧ ¬H(x)) more directly mirrors the structure of the English statement 'There is no student who is not happy' by explicitly denying the existence of counterexamples.

The statement “No dogs bark” means that for every object in the domain, if it is a dog then it does not bark. This corresponds to a universally quantified implication: ∀x (P(x) → ¬Q(x)). While ¬∃x (P(x) ∧ Q(x)) is also logically equivalent, the form using an implication more directly captures the intended meaning that being a dog guarantees not barking. The other options misrepresent the condition or weaken it.

To deny a universal claim, you must assert the existence of a counterexample. The universal claim being negated is “All birds can fly,” which would be formalized as ∀x (P(x) → Q(x)). Its negation requires identifying at least one bird that cannot fly, represented by ∃x (P(x) ∧ ¬Q(x)). This states that there exists an object such that it is a bird and it does not have the ability to fly. This is the exact logical contradiction of the original universal.

The statement “Every student passed” asserts that if any individual is a student, then this individual must have passed. This matches the structure of a universally quantified implication from Student(x) to Passed(x). The existential forms are too weak, and the conjunction under a universal incorrectly states that everything in the domain is both a student and passed. The correct form is ∀x (Student(x) → Passed(x)).

In a finite domain, a universal quantifier expands into a conjunction over each element in the domain. If the domain contains only a and b, then ∀xP(x) means exactly that P(a) is true and P(b) is true. This is captured only by P(a) ∧ P(b). A disjunction is not strong enough, and the conditional and biconditional do not ensure that both P(a) and P(b) hold.

The expression x = x reflects the reflexive property of equality: every object is identical to itself. In first-order logic with identity, reflexivity is an axiom, so x = x holds in every structure and every domain. The universally quantified statement ∀x(x = x) is therefore true in all interpretations.

The first statement guarantees that whenever P(x) holds, Q(x) follows. The second guarantees that whenever Q(x) holds, R(x) follows. By combining these two implications, we conclude that whenever P(x) is true, R(x) must also be true. Because the original statements apply to all x, the resulting implication must also apply universally. So we conclude ∀x(P(x) → R(x)).

This equivalence shows the effect of moving a negation across a quantifier. Negating a universal statement creates an existential statement in which the predicate is negated, following a quantifier-level version of De Morgan’s Laws. The transformation from ∀xP(x) to ¬∃x¬P(x) expresses that the universal claim is true exactly when there does not exist a counterexample. This structural realignment exemplifies De Morgan’s Law applied to quantifiers.

A universal statement is false precisely when at least one counterexample exists. The statement “Every student passed” requires every student to have passed. It is negated by showing that at least one student did not pass. The option “Some student failed” states exactly that. Other options do not correctly capture the minimal necessary condition for falsifying the universal claim.

Universal statements require verifying the claim for every element in the specified domain. The statement asserts that all natural numbers are greater than or equal to zero. Checking only some natural numbers provides no guarantee that the condition holds universally. Negative numbers are not natural numbers and therefore irrelevant to the implication. Correct verification requires showing that it holds for all natural numbers.

Universal statements are disproven by supplying a single counterexample. The claim “All swans are white” requires that every swan is white. Once a single swan was discovered that was not white, the universal statement was definitively falsified. Observing many white swans does not prove the universal, because universal claims demand absolute consistency, and one counterexample suffices to break them.

Applying the quantifier-negation rule ¬∀x φ(x) ≡ ∃x ¬φ(x) to φ(x) ≡ (x² ≥ 0) yields ∃x ¬(x² ≥ 0), which simplifies to ∃x(x²