Question:

Which of the following is an example of a valid First-Order Predicate Logic expression?

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A first-order formula must apply predicates to objects and quantify over object variables, not over predicate symbols.
Updated On: Jul 2, 2026
  • P → Q
  • P + Q = R
  • ∀x (Student(x) → Studies(x))
  • ∀P → Q
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The Correct Option is C

Solution and Explanation

Step 1: First-Order Predicate Logic (FOPL) uses predicates over objects, variables, and quantifiers like \( \forall \) and \( \exists \).

Step 2: Option (A), \( P \rightarrow Q \), uses only propositional symbols, so it is propositional logic, not first-order.

Step 3: Option (B), \( P + Q = R \), is an arithmetic equation, not a logic formula. Option (D) quantifies over a predicate symbol P, which is not allowed in first-order logic (that would be higher-order).

Step 4: Option (C), \( \forall x\,(\text{Student}(x) \rightarrow \text{Studies}(x)) \), quantifies over a variable x and applies predicates to it. It reads 'every student studies', a well-formed FOPL sentence.

Correct option: (C).
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