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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?

  1. $\forall x$ man $(x) \wedge$ intelligent $(x) \rightarrow \exists y$ knowledge $(y) \wedge$ have knowledge $(x, y)$.
  2. $\forall x \exists y$ man $(x) \wedge$ intelligent $(x) \rightarrow$ knowledge $(y) \wedge$ have knowledge $(x, y)$
  3. $\forall x, y$ man $(x) \wedge$ intelligent $(x) \wedge$ knowledge $(y) \rightarrow$ have knowledge $(x, y)$
  4. None of the above.
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2 Answers

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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

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Man ^ Intelligent ------> knowledge ^ has
option A and B both are same .
there is no y term in if condition so we can move y to adjacent of for all x.
∀x((Man(x)∧Intelligent (x))→∃y(Knowledge(y) ∧ has(x,y)) =
∀x∃y((Man(x)∧Intelligent (x))→Knowledge (y) ∧ has(x,y))
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Means that at least one x is there in universal set. For example, if x is a man, at least one man is there.
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Arjun Sir, both A and B are correct right?
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No. B is wrong. Because $\implies$ has lower precedence than quantifiers $\forall$ and $\exists$. So, the quantification applies to only the part before $\implies$ and the $x$ and $y$ on the RHS has no quantification meaning it is for the entire set.
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