Question

Let P(x) be a predicate. Which of the following is NOT valid in first-order logic?

• A. $\forall x P(x) \Rightarrow \exists x \neg P(x)$
• B. $\forall x P(x) \Rightarrow \exists x P(x)$
• C. $\exists x P(x) \Rightarrow \forall x P(x)$
• D. $\exists x P(x) \Leftrightarrow \forall x P(x)$

Hint:

To test the validity of a logical statement, try to construct a simple counterexample. A small domain with one or two elements and a simple predicate is often sufficient to show a statement is not valid. If you can't find a counterexample, it's likely valid.

Answer 1

The Correct Option is A, C, D

Step 1: Understanding the Question:
We need to determine which of the given logical statements are "not valid". A statement is valid (a tautology) if it is true for every possible interpretation (every domain of discourse and every definition of the predicate P(x)). A statement is not valid if we can find at least one counterexample where it is false.
Step 2: Detailed Explanation:
Let's analyze each statement. We'll use a simple domain, like the set of integers, and a predicate like P(x): "x is even".
- (A) $\forall x P(x) \Rightarrow \exists x \neg P(x)$: This translates to "If all x have property P, then there exists an x that does not have property P." This is self-contradictory.
- {Counterexample:} Let the domain be \{2, 4\}. Here, $\forall x P(x)$ is true ("all numbers in the domain are even"). But $\exists x \neg P(x)$ is false ("there exists a number that is not even"). Since we have a case where (True $\Rightarrow$ False), the entire implication is false. Thus, the statement is NOT valid.
- (B) $\forall x P(x) \Rightarrow \exists x P(x)$: This translates to "If all x have property P, then there exists at least one x that has property P."
- This is always true, provided the domain of discourse is non-empty (which is a standard assumption in first-order logic).
If everyone has a property, then it's certainly true that someone has that property. This statement IS valid.
- (C) $\exists x P(x) \Rightarrow \forall x P(x)$: This translates to "If there exists an x with property P, then all x must have property P."
- {Counterexample:} Let the domain be \{2, 3\}. Here, $\exists x P(x)$ is true ("there exists an even number", which is 2). But $\forall x P(x)$ is false ("all numbers are even"). Since we have a case where (True $\Rightarrow$ False), the implication is false. Thus, the statement is NOT valid.
- (D) $\exists x P(x) \Leftrightarrow \forall x P(x)$: This biconditional statement is true only if both sides have the same truth value. As shown in (C), it's possible for $\exists x P(x)$ to be true while $\forall x P(x)$ is false. In that case, the biconditional (True $\Leftrightarrow$ False) is false. Thus, the statement is NOT valid.
Step 3: Final Answer:
The statements that are not universally valid are (A), (C), and (D).