Question:
Let \(P(x)\) be a predicate.
Which of the following statements is/are NOT valid in first-order logic?
Let \(P(x)\) be a predicate.
Which of the following statements is/are NOT valid in first-order logic?
Which of the following statements is/are NOT valid in first-order logic?
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Try a small two-element domain with mixed truth values for P; a formula is valid only if it stays true under every such assignment.
Updated On: Jul 22, 2026
- \(\forall x\, P(x) \Rightarrow \exists x\, P(x)\)
- \(\exists x\, P(x) \Rightarrow \forall x\, P(x)\)
- \(\exists x\, P(x) \Leftrightarrow \forall x\, P(x)\)
- \(\forall x\, P(x) \Rightarrow \exists x\, \neg P(x)\)
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The Correct Option is B, C, D
Solution and Explanation
Step 1: Recall what "valid" means in first-order logic.
A formula is valid if it comes out true under every possible interpretation, that is, for every non-empty domain and every way of assigning meaning to \(P\). To show a formula is NOT valid, it is enough to find just one interpretation that makes it false.
Step 2: Check statement (A), \(\forall x\, P(x) \Rightarrow \exists x\, P(x)\).
Assume the domain is non-empty, as standard in first-order logic. If \(P(x)\) is true for every element of the domain, then in particular it is true for at least one element, so \(\exists x\, P(x)\) must also be true. The implication can never be made false, so this statement is valid, and is not one of the NOT-valid answers.
Step 3: Check statement (B), \(\exists x\, P(x) \Rightarrow \forall x\, P(x)\).
Take a domain \(\{a, b\}\) with \(P(a)\) true and \(P(b)\) false. Then \(\exists x\, P(x)\) is true (because of \(a\)), but \(\forall x\, P(x)\) is false (because of \(b\)). A true antecedent with a false consequent makes the whole implication false, so this interpretation breaks the statement. It is NOT valid.
Step 4: Check statement (C), \(\exists x\, P(x) \Leftrightarrow \forall x\, P(x)\).
Using the same interpretation as Step 3, \(\exists x\, P(x)\) is true while \(\forall x\, P(x)\) is false, so the two sides disagree and the biconditional is false. This statement is NOT valid either.
Step 5: Check statement (D), \(\forall x\, P(x) \Rightarrow \exists x\, \neg P(x)\).
Take a domain where \(P(x)\) is true for every element, for example a single-element domain \(\{a\}\) with \(P(a)\) true. Then \(\forall x\, P(x)\) is true, but \(\exists x\, \neg P(x)\) is false since there is no element where \(P\) fails. A true antecedent and false consequent again make the implication false, so this statement is NOT valid. In fact, this implication is false in every interpretation where \(\forall x\, P(x)\) holds, so it can never be valid.
Final Answer:
Statements (B), (C) and (D) are NOT valid, while (A) is valid.\[ \boxed{\text{(B), (C), (D)}} \]
A formula is valid if it comes out true under every possible interpretation, that is, for every non-empty domain and every way of assigning meaning to \(P\). To show a formula is NOT valid, it is enough to find just one interpretation that makes it false.
Step 2: Check statement (A), \(\forall x\, P(x) \Rightarrow \exists x\, P(x)\).
Assume the domain is non-empty, as standard in first-order logic. If \(P(x)\) is true for every element of the domain, then in particular it is true for at least one element, so \(\exists x\, P(x)\) must also be true. The implication can never be made false, so this statement is valid, and is not one of the NOT-valid answers.
Step 3: Check statement (B), \(\exists x\, P(x) \Rightarrow \forall x\, P(x)\).
Take a domain \(\{a, b\}\) with \(P(a)\) true and \(P(b)\) false. Then \(\exists x\, P(x)\) is true (because of \(a\)), but \(\forall x\, P(x)\) is false (because of \(b\)). A true antecedent with a false consequent makes the whole implication false, so this interpretation breaks the statement. It is NOT valid.
Step 4: Check statement (C), \(\exists x\, P(x) \Leftrightarrow \forall x\, P(x)\).
Using the same interpretation as Step 3, \(\exists x\, P(x)\) is true while \(\forall x\, P(x)\) is false, so the two sides disagree and the biconditional is false. This statement is NOT valid either.
Step 5: Check statement (D), \(\forall x\, P(x) \Rightarrow \exists x\, \neg P(x)\).
Take a domain where \(P(x)\) is true for every element, for example a single-element domain \(\{a\}\) with \(P(a)\) true. Then \(\forall x\, P(x)\) is true, but \(\exists x\, \neg P(x)\) is false since there is no element where \(P\) fails. A true antecedent and false consequent again make the implication false, so this statement is NOT valid. In fact, this implication is false in every interpretation where \(\forall x\, P(x)\) holds, so it can never be valid.
Final Answer:
Statements (B), (C) and (D) are NOT valid, while (A) is valid.\[ \boxed{\text{(B), (C), (D)}} \]
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