Question

Consider the predicate calculus expression, \(\forall x\, P(x) \rightarrow \exists x\, P(x)\). Which of the following English sentences with respect to a club called Sunshine is/are equivalent to this predicate calculus expression?

• A. \textbf{If somebody in Sunshine plays cricket, then everybody in Sunshine plays cricket.}
• B. \textbf{Even though nobody plays billiards in Sunshine, somebody in Sunshine sometimes plays billiards.}
• C. \textbf{If everybody plays hockey in Sunshine, then somebody plays hockey in Sunshine.}
• D. \textbf{Nobody plays football in Sunshine, while all play cricket in Sunshine.}

Hint:

For statements with quantifiers, watch for \textbf{converse} traps: \(\forall x\,P(x) \rightarrow \exists x\,P(x)\) is true, but its converse \(\exists x\,P(x) \rightarrow \forall x\,P(x)\) is generally false.

Answer 1

The Correct Option is C

Step 1: Parse the logic form
The schema \(\forall x\, P(x) \rightarrow \exists x\, P(x)\) reads: If everyone has property \(P\), then someone has property \(P\). This is a valid logical implication because universal truth entails existential truth. Step 2: Match each statement to a logical form
\(\bullet\) (A) says \(\exists x\, P(x) \rightarrow \forall x\, P(x)\) (if \emph{somebody} plays, then \emph{everybody} plays). This is the converse of the target; not equivalent.
\(\bullet\) (B) asserts \(\neg \exists x\, P(x)\) and \(\exists x\, P(x)\) simultaneously (a contradiction), not an implication of the target form.
\(\bullet\) (C) says \(\forall x\, P(x) \rightarrow \exists x\, P(x)\) (if \emph{everybody} plays, then \emph{somebody} plays) — exactly matches the target schema.
\(\bullet\) (D) combines \(\neg \exists x\, P_f(x)\) with \(\forall x\, Q(x)\) — unrelated to the form \(\forall \rightarrow \exists\). Step 3: Conclusion
Only option (C) is equivalent to \(\forall x\, P(x) \rightarrow \exists x\, P(x)\). \[ \boxed{Correct Answer: (C)} \]