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Some results in literature are 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 this is a proposition, interpreting it as $\exists x(P(x)\implies Q(x))$ seems not make sense. Should it be interpreted as $\forall x(P(x)\implies Q(x))$, 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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  1. ‘If there exists some $x$ such that $P(x)$ is true, then $(\text{for the same }x,)\:Q(x)$ is also true’

    If the original phrasing is actually without the parenthetical part, then it means $$∃\color{red}z\,P(\color{red}z)→ Q(x),$$ which is logically equivalent to $$∀\color{red}z\:\Big(P(\color{red}z)→ Q(x)\Big);$$ note that that this is an open formula and that variable $x$ is free.

  2. ‘If there exists some $x$ such that $P(x)$ is true, then for the same $x,\,Q(x)$ is also true’

    On the other hand, if the sentence is indeed as above҂, then its full meaning made explicit (without which the sentence is actually illogical) must be
      ‘If there exists some $\color{red}z$ such that $P(\color{red}z)$ is true, then, for every $\color{green}y$ such that $P(\color{green}y)$ is true, $Q(\color{green}y)$ is also true’, i.e., $$∃\color{red}z\,P(\color{red}z) → ∀\,\color{green}y\Big(P(\color{green}y) → Q(\color{green}y)\Big),$$ which is logically equivalent to $$∀\,x\Big(P(x) → Q(x)\Big).$$

      ҂paraphrase: ‘If $P(x)$ is true for some value of $x,$ then $Q(x)$ is true for the same value of $x.$’

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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