Question

Show \(\not \models \exists x \varphi \wedge \exists x \psi \rightarrow \exists x(\varphi \wedge \psi)\).

Short answer

The implication is not universally valid; a counterexample exists.

Step-by-step solution

Understand the Problem

We need to show that the logical statement \( \exists x \varphi \wedge \exists x \psi \rightarrow \exists x(\varphi \wedge \psi) \) is not universally valid. This means that it is possible to find a structure in which the premise is true but the conclusion is false.

Analyze the Premise

The premise of the implication is \( \exists x \varphi \wedge \exists x \psi \), meaning there is some element \( a \) in the domain such that \( \varphi(a) \) is true, and there is potentially another element \( b \) such that \( \psi(b) \) is true. These elements \( a \) and \( b \) do not have to be the same.

Analyze the Conclusion

The conclusion of the implication is \( \exists x(\varphi \wedge \psi) \), which means that there should exist at least one element \( c \) in the domain such that both \( \varphi(c) \) and \( \psi(c) \) are true. This is a stronger requirement than the premise.

Find a Counterexample

Consider a structure where the domain consists of two elements \( \{a, b\} \). Let \( \varphi(x) \) be true for \( a \) and \( \psi(x) \) be true for \( b \). Here, \( \exists x \varphi \) and \( \exists x \psi \) are both true since \( \varphi(a) \) and \( \psi(b) \) are true. However, there is no single element where both \( \varphi \) and \( \psi \) are true, hence \( \exists x(\varphi \wedge \psi) \) is false.

Conclude the Argument

Since we found a structure where the premise is true but the conclusion is false, the implication \( \exists x \varphi \wedge \exists x \psi \rightarrow \exists x(\varphi \wedge \psi) \) is not universally valid. Therefore, it is demonstrated that \( ot \models \exists x \varphi \wedge \exists x \psi \rightarrow \exists x(\varphi \wedge \psi) \).

Key concepts

Universal Validity

Universal validity in predicate logic is a concept that implies a statement holds true in every possible case and every possible interpretation. It means that no matter how you interpret the components of the formula, the result will always be true. This is a very strong condition because it demands that the formula be true in every conceivable scenario.

For example, if a statement is universally valid, it would apply to every possible domain and for every possible assignment of values to variables. However, if you can find even one case where the statement does not hold true, it is not universally valid. In the context of the original exercise, determining whether the statement is universally valid involves finding out if there exists any scenario where the premise holds, but the conclusion does not. If such a scenario exists, the statement is not universally valid, as we saw in the solved exercise where a counterexample proved that the statement was not true in all cases.

Existential Quantifier

The existential quantifier, denoted by the symbol \( \exists \), is used in predicate logic to indicate that there is at least one element for which a certain property or statement holds true. It essentially says, "there exists" some element in your set or domain that satisfies a given condition.

In terms of logical formulas, the existential quantifier can be used to express statements like "There exists an \( x \) such that \( \varphi(x) \) is true." This means there is some specific element in the domain that makes \( \varphi \) true.

Within the original task, the existential quantifier is used to search for elements where certain conditions are met. For instance, \( \exists x \varphi \) asserts that there is an element \( a \) with property \( \varphi(a) \) true, and \( \exists x \psi \) asserts that there is possibly another element \( b \) such that \( \psi(b) \) is true. Importantly, existential quantifiers don't require the same element to satisfy different conditions, as seen in the distinction between elements \( a \) and \( b \).

Logical Implication

Logical implication is a fundamental concept whereby one statement leads to another. In logic, the statement \( A \rightarrow B \) implies that if \( A \) is true, then \( B \) must also be true. If the premise \( A \) holds, the conclusion \( B \) should follow logically; however, it does not state anything about what happens when \( A \) is false.

In the exercise, \( \exists x \varphi \wedge \exists x \psi \rightarrow \exists x(\varphi \wedge \psi) \) was given as the implication. Here, the premise was the existence of some elements \( x \) that satisfy either \( \varphi \) or \( \psi \), while the conclusion demands that there is one \( x \) for which both are true simultaneously. The key lesson from exploring logical implications is understanding that they can frequently appear valid unless carefully analyzed, and this apparent validity might be challenged or disproven using a counterexample.

Counterexample in Logic

A counterexample in logic is a sample that disproves a statement or proposition. Finding a counterexample is a powerful method because it involves demonstrating that a statement isn't universally valid or true.

In logical terms, providing one counterexample is sufficient to show that the universal claim fails because even the failure in one case means it's not universally true.

In the provided exercise, a counterexample was used to disprove the implication \( \exists x \varphi \wedge \exists x \psi \rightarrow \exists x(\varphi \wedge \psi) \). By considering a domain of elements \( \{a, b\} \), with \( \varphi(a) \) being true for \( a \) and \( \psi(b) \) being true for \( b \), it was shown that while the premise was true, the conclusion failed. Since these conditions held true, finding just this single scenario underscored the statement’s lack of universal validity.