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In the literature, claims or results are sometimes given as ‘If there exists some $x$ such that $P(x)$ is true, then (for the same $x,)\, Q(x)$ is also true.’

I am quite confused by such statements, particularly because the term 'exists' appears in the statement. Since these are propositions, interpreting them as $\exists x\,(P(x)\Longrightarrow Q(x))$ seems not to make sense.

Should they be interpreted as $\forall x\,(P(x)\Longrightarrow Q(x)),$ and the 'exists' in the statement is used to ignore the vacuous truth case?

ryang
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lychtalent
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  • Yes, the vacuously true case can be let aside, and the sentence is indeed equivalent to $\forall x,(P(x)\implies Q(x))$. Note that it can's start with $\exists x$, because it would be an affirmative statement (such an $x$ does indeed exist), while your original sentence does not state that this $x$ exists, but that if it exists, something follows. – Jean-Claude Arbaut Mar 19 '22 at 09:41
  • I doubt the sentence is a logical sentence in a strict sense because $x$ in $Q(x)$ is unbound. – Hermis14 Mar 19 '22 at 10:02

2 Answers2

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"If there exists some $x$ such that $P(x)$ is true, then for the same $\boldsymbol {x,}Q(x)$ is also true."

  1. Let's first get out of the way that the above sentence means neither $∃x\,\Big(P(x)→Q(x)\Big)\tag✘$ nor $\Big(∃x\,P(x)\Big)→Q(x),\tag✘$ which can be rewritten as $\Big(∃\color{red}z\,P(\color{red}z)\Big)→Q(x),\tag*{}$ which is logically equivalent to $∀\color{red}z\,\Big(P(\color{red}z)→ Q(x)\Big),\tag*{}$ which is an open formula with $x$ as a free variable.

  2. A coherent way to unpack the given sentence is as $∃x\,P(x) → ∀x\,\Big(P(x) → Q(x)\Big),\tag*{}$ which can be rewritten as $∃\color{red}z\,P(\color{red}z) → ∀\color{green}y\,\Big(P(\color{green}y) → Q(\color{green}y)\Big),\tag*{}$ which is logically equivalent to $∀\,x\,\Big(P(x) → Q(x)\Big).\tag✔$

  3. Alternatively: $∀z\,\bigg(∃x\,\Big(P(x) \land z=x\Big)→ Q(z)\bigg),\tag*{}$ which can be rewritten as $∀x\,\bigg(∃z\,\Big(P(z) \land x=z\Big)→ Q(x)\bigg),\tag*{}$ and is logically equivalent to $∀\,x\,\Big(P(x) → Q(x)\Big).\tag✔$

  4. Or, we can simply clarify the given sentence as

    • If P(x) is true of an arbitrary x, then Q(x) is true of (that) x,

      i.e., If P(x) is true of an x, then Q(x) is true of (that) x,

    then apply universal generalisation: $∀\,x\,\Big(P(x) → Q(x)\Big).\tag✔$

It goes without saying that the given sentence is most naturally and clearly rewritten as $$\text{Every $x$ that satisfies $P(x)$ also satisfies $Q(x)$},\tag✔$$ with its quantification explicit.

ryang
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Such statements typically have the form: if a smallest, largest, or otherwise unique object which satisfies the property $P$ exists, then this object has the additional property $Q$. For example, one could say about a monoid: if a multiplicative inverse of $x$ exists, then it satisfies a property $Q$.

This is, in theory, equivalent to: for each $y$, if $y$ is the multiplicative inverse of $x$, then $Q(y)$. In practice, this would be a slightly strange way to think about such a statement.

The statement really says: $x^{-1}$ has the property $Q$ – with the proviso that we exclude cases where the term $x^{-1}$ does not refer to anything.

Of course, terms with no denotation are not allowed in standard first-order logic, so indeed if you wanted to formalize this in first-order logic, you would have to invoke some work-around, such paraphrasing the theorem using universal quantification along the lines you suggested: for each $y$, if $x \cdot y = 1$ and $y \cdot x$, then $Q(y)$.

Pilcrow
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