Validity and Models Quiz: Test Arguments Across Interpretations

In a nonempty domain, ∀x(P(x) ∨ Q(x)) guarantees that every element satisfies at least one of P or Q. Because at least one element exists, this implies that there is some element with P or some element with Q, i.e., ∃x P(x) ∨ ∃x Q(x).

A valid formula is true in every possible model under every interpretation — there is no countermodel in which it comes out false. Option A describes satisfiability not validity. Option B describes unsatisfiability. Option D describes a contingent formula.

If φ is valid, it is true in every model. Its negation ¬φ would then be false in every model, which means ¬φ has no model and is unsatisfiable.

An unsatisfiable formula has no model in which it is true, meaning it is false in every interpretation confirming A, and there is no interpretation that makes it true confirming B. Option C describes a contingent formula. Option D is the opposite of unsatisfiability — satisfiable formulas have at least one interpretation making them true.

For any element x in any model, the disjunction P(x) ∨ ¬P(x) is always true by the law of excluded middle. Since this holds for every x in every model, the universally quantified sentence is valid.

To show an argument is invalid, we need a specific model in which all the premises are true but the conclusion is false. Such a countermodel demonstrates that the argument is not truth-preserving in all cases.

When empty domains are allowed, ∀x P(x) is vacuously true but ∃x P(x) is false, making the implication ∀x P(x) → ∃x P(x) false in the empty model. The other formulas remain true in all models, so this is the one that is not valid.

The formula ∃x ∀y (y = x) says there is an element x such that every y in the domain equals x. This is true exactly when the domain has a single element (a singleton) and false in any domain with zero or more than one element.

If there exists an element that satisfies both P and Q, then that same element witnesses ∃x P(x) and also witnesses ∃x Q(x). Hence we can conclude ∃x P(x) ∧ ∃x Q(x), but no universal claim follows.

For any object x in the domain, we can always choose y = x, which makes x = y true. Thus the formula ∀x ∃y (x = y) holds in every structure and is therefore valid.

For each of the 3 domain elements, the unary predicate P can be either true or false. This gives 2 × 2 × 2 = 2^3 = 8 possible interpretations, one for each subset of the domain.

To prove ∀x Q(x), we consider an arbitrary element a in the domain. From ∀x P(x) we get P(a), and from ∀x(P(x) → Q(x)) we get P(a) → Q(a). Modus ponens then gives Q(a), and since a was arbitrary, ∀x Q(x) holds.

The formula ∀x(P(x) → Q(x)) states that every P is a Q, while ∀x P(x) → ∀x Q(x) only says that if everything is P then everything is Q. These are not equivalent in general, so the answer is No.

The premise ∃x P(x) requires at least one element with property P, while the conclusion ∀x P(x) requires every element to have that property. A countermodel therefore must have some elements with P and at least one element without P.

If every element satisfies P(x) ∨ Q(x) and a particular element a does not satisfy P(a), then Q(a) must be true for the disjunction to hold. This is a standard valid argument pattern (disjunctive syllogism).

If some element satisfies P and all elements fail Q, then that same P-element automatically satisfies P(x) ∧ ¬Q(x). Thus there must exist at least one element making P true and Q false.

The first conjunct says every P is a Q, while the second conjunct says there is an element that is P and not Q. These two requirements directly contradict each other, so no model can satisfy both at once.

If P(a) and P(b) are both true, then there is at least one element (for example a) such that P holds. This justifies the existential statement ∃x P(x), but it does not justify any universal conclusions.

The statement ∃x(P(x) ∧ Q(x)) says that there is at least one element in the domain that makes both P and Q true. Other elements may or may not satisfy P or Q, but at least one must satisfy both.

Negating an existential quantifier produces a universal quantifier with a negated predicate. Thus ¬∃x P(x) is equivalent to saying that for every x, P(x) is false: ∀x ¬P(x).