quantifiers in logically equivalent implications

Question

I'm really confused as to why: ∃x∈D (p(x)⇒E) is logically equivalent to (∀x∈D p(x))⇒E p(x) is a generic predicate based on x and E is a predicate that is unrelated to x. ∀x∈D (p(x)⇒E) is also logically equivalent to (∃x∈D p(x))⇒E Can someone please help?

Answer 1

The sentence "$\exists x \in D(P(x) \implies E)$" says "There is some object such that if that object has property $P$, then $E$ is true." Now, suppose some $x$ doesn't have $P$. Then the statement "$P(x) \implies E$" is vacuously true (recall that "if A then B" is considered true if A is false). If, on the other hand, every $x$ has $P$, then the statement "$\exists x \in D(P(x) \implies E)$ can only be true if $E$ is, because the existential quantifier has to "pick" an $x$ for which $P(x)$ is true. So in other words, the sentence is true in exactly the following two situations: either (a) some $x$ doesn't have property $P$, or (b) every $x$ has property $P$ and $E$ is true.

The sentence $(\forall x \in D)P(x) \implies E$ is true if either (a) $(\forall x \in D)P(x)$ is false, or (b) $(\forall x \in D)P(x)$ is true and $E$ is true also. (Remember what it means for $A \implies B$ to be true.) Translating this into English, $(\forall x \in D)P(x) \implies E$ is true in exactly the following two situations: (a) not every $x$ has property $P$ (so some $x$ does not have property $P$), or (b) every $x$ has property $P$ and $E$ is true. Notice that these are exactly the two situations we had for the other sentence; so the two sentences are equivalent.

Answer 2

To explain the equivalences you are interested in, let us first establish that for any $\varphi(x)$ and $\psi$, where $\psi$ does not contain any free variable $x$ (so this is just like your $p(x) and $E$) we have that:

Ia. $\exists x \: \varphi(x) \lor \psi \Leftrightarrow \exists x (\varphi(x) \lor \psi)$

To understand this, note that an existential can be seen as kind of disjunction, that is, if $a,b,c,...$ denote the objects in your domain, then you can think of an existential like this:

$\exists x \: \varphi(x) \approx \varphi(a) \lor \varphi(b) \lor \varphi(c) \lor ...$ (I use $\approx$ since this is technically not a logical equivalence, but if you really want to prove the above equivalence, you'd need to go into formal semantics, and what you'll be doing there does follow this very idea)

So:

$\exists x \: \varphi(x) \lor \psi \approx$

$(\varphi(a) \lor \varphi(b) \lor \varphi(c) ...) \lor \psi \Leftrightarrow$ (by Association, Commutation, and since $P \Leftrightarrow P \lor P$)

$\varphi(a) \lor \psi \lor \varphi(b) \lor \psi \lor \varphi(c) \lor \psi ... \approx$

$\exists x (\varphi(x) \lor \psi)$

By thinking of a universal as a kind of conjunction over all objects over the domain, i.e using

$\forall x \: \varphi(x) \approx \varphi(a) \land \varphi(b) \land \varphi(c) \land ...$

you can likewise show that:

Ib. $\forall x \: \varphi(x) \lor \psi \Leftrightarrow \forall x (\varphi(x) \lor \psi)$ (good exercise to do for yourself!)

OK, so now we can establish your equivalences using more more important equivalence:

II. $\varphi \rightarrow \psi \Leftrightarrow \neg \varphi \lor \psi$ (just verify this with a truth-table)

For example:

$\exists x (p(x) \rightarrow E) \Leftrightarrow$ (by II)

$\exists x (\neg p(x) \lor E) \Leftrightarrow$ (by Ia)

$\neg \exists x \: p(x) \lor E \Leftrightarrow$ (Quantifier Negation ... this is why the quantifier changes!!)

$\forall x \: \neg p(x) \lor E \Leftrightarrow$ (by Ib)

$\forall x \: (\neg p(x) \lor E) \Leftrightarrow$ (by II)

$\forall x \: (p(x) \rightarrow E)$

And again, I'll leave the other one to yourself.