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When using existential quantifiers, is there a difference between using a conjunction and implication? For example, for this question:

There is an agent who sells policies only to people who are not insured:

$$ ∃x Agent(x) ∧ ∀y,z Policy(y)∧Sells(x, y, z)⇒(Person(z)∧ ¬Insured(z)) $$

(Answer taken from a textbook, not sure if it is correct though.)

If you used $\wedge$ instead of $\Rightarrow$, would there be a difference? Since it is only the existential unifier, so is it okay if there are cases where it doesn't work?

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    $\begingroup$ The answer could use more parentheses. ... $(\mathit{Policy}(y) \land \mathit{Sells}(x,y,z)) \Rightarrow (\mathit{Person}(z) \land \lnot \mathit{Insured}(z))$ makes sense, but ... $\mathit{Policy}(y) \land (\mathit{Sells}(x,y,z) \Rightarrow (\mathit{Person}(z) \land \lnot \mathit{Insured}(z)))$ is less sensible, since it claims everything is a policy. $\endgroup$
    – aschepler
    May 2, 2022 at 21:00

2 Answers 2

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Hint: changing an implication to a conjunction will change the meaning of a formula except in trivial cases: $A \land B$ is only equivalent to $A \Rightarrow B$ if $A$ and $B$ are both true (as you can check using truth tables). E.g., if $E(x)$ means $x \in \Bbb{N}$ is even and $O(x)$ means $x$ is odd, then $\exists x(E(x) \land O(x))$ is clearly false while $\exists x(E(x) \Rightarrow O(x))$ is true (any odd number provides a witness). Likewise in your example, changing the implication to a conjunction makes the statement much stronger.

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Suppose, $$∃x Agent(x) ∧ ∀y,z Policy(y)∧Sells(x, y, z)\land(Person(z)∧ ¬Insured(z)).$$ That must be an interesting agent, so let's give them a name. Let's assume the agent is named Jane, so Jane has the interesting property $$ Agent(\text{Jane}) ∧ ∀y,z Policy(y)∧Sells(\text{Jane}, y, z)\land(Person(z)∧ ¬Insured(z))$$ The second part has an allquantor, hence remains true if we replace $y$ with the Eiffel Tower.

$$ Agent(\text{Jane}) ∧ ∀z Policy(\text{Eiffel Tower})∧Sells(\text{Jane}, \text{Eiffel Tower}, z)\land(Person(z)∧ ¬Insured(z))$$

In the remaining allquantor, we are allowed to specialize to, say, the North Pole: $$ Agent(\text{Jane}) ∧ ∀z Policy(\text{Eiffel Tower})∧Sells(\text{Jane}, \text{Eiffel Tower}, \text{North Pole})\land(Person(\text{North Pole})∧ ¬Insured(\text{North Pole}))$$

Jane is an agent, who sells the Eiffel Tower (a policy) to the North Pole, which by the way is an uninsured person.

Nope.

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  • $\begingroup$ Thank you for this! But why does it still work if -> is used? $\endgroup$
    – Ashley
    May 2, 2022 at 20:19

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