Validity and Models Quiz: Test Arguments Across Interpretations

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.

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).

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.

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 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 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.

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).

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.

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.

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.

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).

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.

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.

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.

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.

A formula is valid precisely when it is true in all models and under all interpretations. That means there is no countermodel in which the formula comes out false.

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.

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.

By definition, an unsatisfiable formula has no model in which it is true, so it is false in every interpretation. Equivalently, there is no interpretation that makes it true, which is why choices A and B are both correct and the others are not.

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.