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

The correct options that are NOT valid in first-order logic are:

1. $\forall xP\left(x\right)\Rightarrow \exists x\neg P\left(x\right)$

2. $\exists xP\left(x\right)\Rightarrow \forall xP\left(x\right)$

3. $\exists xP\left(x\right)\iff \forall xP\left(x\right)$

Understanding Validity in First-Order Logic
In first-order logic, a statement is considered valid if it evaluates to true under every possible interpretation over any non-empty domain of discourse. If we can find at least one interpretation (a counterexample) where the statement evaluates to false, then the statement is not valid (invalid).

Let us analyze each of the given statements step-by-step:

1. Analyzing the Statement: $\forall xP\left(x\right)\Rightarrow \exists x\neg P\left(x\right)$
This statement claims that if the predicate $P\left(x\right)$ is true for all elements in the domain, then there must exist some element for which $P\left(x\right)$ is false.
Let us construct a counterexample:
- Let the domain be $D=\left\{1\right\}$.
- Let $P\left(1\right)$ be true.
Under this interpretation, the antecedent $\forall xP\left(x\right)$ is true because $P\left(x\right)$ holds for every element in the domain. However, the consequent $\exists x\neg P\left(x\right)$ is false because there is no element in the domain for which $P\left(x\right)$ is false. The implication becomes:
$\text{True}\Rightarrow \text{False}$
which evaluates to false. Thus, this statement is NOT valid.

2. Analyzing the Statement: $\exists xP\left(x\right)\Rightarrow \forall xP\left(x\right)$
This statement claims that if there exists at least one element for which $P\left(x\right)$ is true, then $P\left(x\right)$ must be true for all elements in the domain.
Let us construct a counterexample:
- Let the domain be $D=\{1,2\}$.
- Let $P\left(1\right)$ be true and $P\left(2\right)$ be false.
Under this interpretation, the antecedent $\exists xP\left(x\right)$ is true (since $P\left(1\right)$ is true). However, the consequent $\forall xP\left(x\right)$ is false (since $P\left(2\right)$ is false). The implication becomes:
$\text{True}\Rightarrow \text{False}$
which evaluates to false. Thus, this statement is NOT valid.

3. Analyzing the Statement: $\exists xP\left(x\right)\iff \forall xP\left(x\right)$
An equivalence $A\iff B$ is valid only if $A\Rightarrow B$ and $B\Rightarrow A$ are both valid. Since we have already shown that the direction $\exists xP\left(x\right)\Rightarrow \forall xP\left(x\right)$ is invalid, the equivalence is also NOT valid.

4. Note on the Valid Option: $\forall xP\left(x\right)\Rightarrow \exists xP\left(x\right)$
Under standard first-order logic, the domain of discourse is assumed to be non-empty. If a property $P\left(x\right)$ holds for all elements in a non-empty domain, it must hold for at least one element. Therefore, this implication is always true, making it a valid statement.