4

Is there a difference between $\exists x(\phi(x) \rightarrow \forall y\phi(y))$ and $\exists x \phi(x) \rightarrow \forall y\phi(y)$?

The first one is the Drinker's paradox, which is a true in an non-empty domain (see here). But isn't the second one also true in a non-empty domain? I'm suspicious because I've never seen the Drinker's paradox written the second way.

Mauro ALLEGRANZA
  • 94,169
Doubt
  • 1,719

2 Answers2

4

Let the domain be $\{1,2\}$ and let $\phi(1) = \top$ and $\phi(2) = \bot$. The first statement is true since for $x=2$, we have $\neg\phi(x)$. The second is false since $\exists x \phi(x)$ is true (since $\phi(1) = \top$), but $\forall y \phi(y)$ is false (since $\phi(y) =\bot$).

silver
  • 794
3

Take the domain to be the prime numbers and $\phi(x)$ to mean "$x$ is even". Then $\exists x \phi(x)$ is true since $2$ is even, but $\forall y \phi(y)$ is false since $3$ is not even.

The reason the drinker's paradox holds is that you can make sure a suitable $x$ exists: if everyone is drinking, take $x$ to be arbitrary; if not everyone drinks, choose $x$ to be any non-drinker. However, for the second statement you give, when not everyone drinks, taking a non-drinker makes the antecedent true, and that makes the entire sentence false.

Ittay Weiss
  • 79,840
  • 7
  • 141
  • 236
  • I don't understand your last sentence "However...". The formula $\exists x \phi(x) \to \forall y\phi(y)$ is false when someone but not everyone is drinking. Indeed, in this case, and only in this case, the is antecedent true, while the consequent is false. The sentence $\exists x (\phi(x) \to \forall y \phi(y))$, instead, holds in any case. – Taroccoesbrocco Jun 15 '21 at 11:29