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Given the statement : "Men who are intelligent have knowledge."

What is the correct representation of the statement in first-order predicate calculus?

- $\forall x$ man $(x) \wedge$ intelligent $(x) \rightarrow \exists y$ knowledge $(y) \wedge$ have knowledge $(x, y)$.
- $\forall x \exists y$ man $(x) \wedge$ intelligent $(x) \rightarrow$ knowledge $(y) \wedge$ have knowledge $(x, y)$
- $\forall x, y$ man $(x) \wedge$ intelligent $(x) \wedge$ knowledge $(y) \rightarrow$ have knowledge $(x, y)$
- None of the above.

8 votes

Best answer

a is correct.

b is wrong due to implication having lower precedence than quantifiers. It will be evaluated as $\left(\forall x \exists y \text{man} (x) \wedge \text{intelligent}(x)\right) \implies \left(\text{knowledge} (y) \wedge \text{have knowledge} (x,y) \right)$.

Here, for the RHS, there is no quantification for $x$ or $y$, meaning it must be true for every $x$ and $y$.

We can make (b) correct by writing it as follows:

$\forall x \exists y \left( \text{man} (x) \wedge \text{intelligent}(x) \implies \text{knowledge} (y) \wedge \text{have knowledge} (x,y) \right)$

Ref: https://en.wikipedia.org/wiki/First-order_logic#Notational_conventions