What do $\exists x (P(x) \rightarrow Q)$ and $\exists x (P(x) → Q(x))$ 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 two sentences are logically equivalent:

   • There is something such that if it satisfies $P(x),$ then it satisfies $Q(x)$

     $∃x\,(Px → Qx)\tag1$

   • If everything satisfies $\boldsymbol{P(x)},$ then something satisfies $\boldsymbol{Q(x)}$

     $(∀x\,Px) \to ∃x\,Qx.\tag{1e}$

2. And these two sentences are logically equivalent as long as the discourse domain is nonempty:

   • There is something such that if it satisfies $P(x),$ then $Q$ is true

     $∃x\,(Px → Q) \tag2$

   • If everything satisfies $\boldsymbol{P(x)},$ then $\boldsymbol{Q}$ is true

     $(∀x\,Px) → Q.\tag{2e}$

The existentially-quantified conditional is a weak and rarely useful assertion: it is true whenever the universe contains any object that satisfies $Q(x)$ or $Q$ and vacuously true whenever it contains any object that fails to satisfy $P(x),$ and it doesn't say much about the relationship between its antecedent and consequent or about the discourse domain.

This is in contrast to the categorical propositions \begin{array}{l l l} \textsf{Some P is Q} && \exists x\,(Px\land Qx) \\ \textsf{Some P is not Q} && \exists x\,(Px\land \lnot Qx) \\ \textsf{Every P is Q} &(\textsf{No P is not Q}) & \forall x\,(Px\to Qx) \\ \textsf{Every P is not Q} &(\textsf{No P is Q}) & \forall x\,(Px\to\lnot Qx). \end{array}

It is easy to mistake sentence $(1\Big/2)$ for the lookalikes $\exists x\,(Px \color\red\land \;Qx\Big/Q)\tag I$ $\text{Something satisfies both $P(x)$ and $Q(x)\Big/Q$}\tag*{}$ or $\color\red{\forall} x\,(Px →\; Qx\Big/Q)\tag A$ $\text{Everything that satisfies $P(x)$ also satisfies $Q(x)\Big/Q$},\tag*{}$ which parse more naturally and intuitively.

However, the categorical propositions $(I)$ and $(A)$ are each logically stronger assertions than sentence $(1\Big/2):$ the former logically entails the latter but not vice versa. For example, the universe $\text{\{quotable non-politician, non-quotable politician\}}\tag*{}$ is correctly described by $\exists x\,\big(\text{Politician}(x) \to \text{Quotable}(x)\big)\tag1$ but neither $\exists x\,\big(\text{Politician}(x) \land \text{Quotable}(x)\big)\tag I$ nor $\forall x\,\big(\text{Politician}(x) \to \text{Quotable}(x)\big).\tag A$

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Appendix: Proof that $(2)$ is equivalent to $(2\text e)$

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

• If $(*)$'s LHS is false, then $∀x\,(Px\land \lnot Q),$ 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\,(Px\to Q) \quad\equiv\quad (∀x\,Px)\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".