# What is the difference between using a conjunction and implication with existential quantifiers?

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?

• 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. May 2, 2022 at 21:00

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.

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.

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