What do ∃x (P(x) → Q) and ∃x (P(x) → Q(x)) mean?

Question

The existentially-quantified conjunction $$∃x\;(P(x) \land Q(x))$$ means that there exists at least one $x$ such that $x$ is both $P$ and $Q.$ That is, some $P$ is a $Q.$

However, what do the existentially-quantified conditionals $$∃x\:(P(x) → Q) \tag1$$ and $$∃x\:\big(P(x) → Q(x)\big)\tag2$$ mean?

Background & Motivation:

• No one: ~∃x
• Someone: ∃x
• Everyone: ∀x
• Not everyone: ~∀x

∀xA(x): For every x, x is an A: = All x’s are A’s (/ or every x is an A)

∃xA(x): There exists (at least) an x such that it is an A:
= Some x is an A (/ or some x’s are A’s)

• Domain: D: {d1, d2, …, dk}: di: objects of the domain,
• Names: N: {c1, c2, …, ck}, ci: constants/names

Quantifying using the existential and universal quantifiers:

- None (n = 0):  ~∃x
- Some (n ≥1): ∃x
- Every (n = k): ∀x
- Not every (n < k): ~∀x

where: n = # of items to be symbolized, where k: = # of items in the domain.

These quantifier terms can be used to symbolize the following:

• No A is a B: ~∃x(A(x) ^ B(x))
• Some A’s are B’s: ∃x(A(x) ^ B(x))
• All A’s are B’s: ∀x(A(x) --> B(x))
• Not all A’s are B’s: ~∀x(A(x) --> B(x))

and the following negations below:

- Some A’s are not B’s: ∃x(A(x) ^ ~B(x))
- All’s A’s are not B’s: ∀x(A(x) --> B(x))

Answer 1

1. These sentences are logically equivalent to one another:

   • $$∃x\:\Big(P(x) → Q\Big) \tag1$$
   • For some object $x,$ if $P(x)$ is true, then $Q$ is true.
   • For some object $c,$ if $c$ satisfies $P(x),$ then $Q$ is true.
   • $$∀xP(x) → Q\tag{1e}$$
   • $Q$ is true if $P(x)$ is true of every object.

2. On the other hand, $$∃x\:\Big(P(x) → Q(x)\Big)\tag2$$ means

   • For some object $x,$ if $P(x)$ is true, then $Q(x)$ is true.

   Note that

   1. if a particular object fails to satisfies $P(x),$ then sentence $(2)$ is immediately true;

   2. if some object satisfies $P(x)$ but not $Q(x),$ then sentence $(2)$ can still be true.

      If this feels unintuitive, it's because you are mixing up sentence $(2)$ with $$\forall x\:\Big(P(x) → Q(x)\Big).$$

I think sentences $(1)$ and $(2)$ are typically not useful forms, so are seldom encountered. In particular—even if $P(x)$ and $Q(x)$ stand for $\text“x$ is [property/thing P/Q]”—observe that neither sentence $(1)$ nor $(2)$ can be framed as categorical propositions (i.e.,
  Every A is B;
  Some A is B;
  No A is B;
  Some A is not B).

To be clear: an existentially-quantified conditional → is unusual, and feels unintuitive because we naturally tend to want to read its $∃$ as $∀$ instead. The → is not the problem, because rewriting the above sentences as $$∃x\:\Big(\lnot P(x) \lor Q\Big)\\∃x\:\Big(\lnot P(x) \lor Q(x)\Big)$$ does not actually make them less tricky to correctly process.

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Appendix

Proof that $(1)$ is logically equivalent to $(1\text e):$

Consider the formula \begin{gather}∃x\:\Big(Px\to Q\Big) \quad↔\quad \Big(∀x\:Px\Big)\to Q.\tag{*}\end{gather}

• If $(*)$'s LHS is false, then $∀x\:\Big(Px\land \lnot Q\Big),$ so $Px$ is universally true and $\lnot Q$ true, so $Px$ is universally true while $Q$ false, so $(*)$'s RHS is false; by contrapositive and since we have been abstractly inferring, $(*)$'s RHS logically implies $(*)$'s LHS.
• On the other hand, if $(*)$'s RHS is false, then $Px$ is universally true and $Q$ false, so $(Px\land \lnot Q)$ is universally true, so $(Px\to Q)$ is universally false, so $(*)$'s LHS is false; by contrapositive and since we have been abstractly inferring, $(*)$'s LHS logically implies $(*)$'s RHS.

Hence, \begin{gather}∃x\:\Big(Px\to Q\Big) \quad\equiv\quad \Big(∀x\:Px\Big)\to Q.\end{gather}

A verification.

Answer 2

The difference lies in the variable $x$ where $Q(x)$ is some statement involving $x$ while $Q$ is not necessarily involving $x$. For instance; in the universe of real numbers $\mathbb R$, where the element is $x$ is arbitrary

• "there exists a real number $x$ s.t. if $x$ is multiple of $6$ then $x$ is multiple of $3$" can be written as $\exists x(P(x)\implies Q(x))$ where $P(x):x$ is multiple of $6$ and $Q(x):x$ is multiple of $3$.
• "there exists a real number $x$ s.t. if $x$ is multiple of $6$ then $2$ is even" can be written as $\exists x(P(x)\implies Q)$ where $P(x):x$ is multiple of $6$ and $Q:2$ is even.

While in the first example, the truth value ($T/F$) is dependent on the complete conditional statement but in second example, the truth value of the conditional is $T$ only as long as $Q$ is $T$. If $Q$ a tautology then so is the overall conditional statement.

Answer 3

The simplest "translation" of the sentence $\exists x(P(x)\implies Q(x))$ in English, as others have already mentioned, would be "There exists some x, such that whenever $P(x)$ is true, $Q(x)$ is also true. However, you would be right in saying that the above statement does not make too much sense intuitively speaking, so I propose that we rewrite the predicate in another way:

$\exists x(P(x)\implies Q(x))$

is the same as $\exists x(\neg P(x)\lor Q(x))$

which is the same as $\exists x:\neg P(x)\lor \exists x:Q(x)$

The above predicates would be translated as "there exists some $x$, such that $\neg P(x)$ or $Q(x)$ (or both) is true".