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We are simply asked if this statement is true or false. From what I was able to gather this statement is true because an empty set has no values to make a counter argument.

Yet I do not fully understand the logic behind this specific statement so I'd be really glad if someone could explain it in more detail.

Thomas Andrews
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Wannabree
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    $\forall x\in{} (P(x))$ is true for any statement $P(x).$ Basically, because if $x\in{},$ we have a contradiction. – Thomas Andrews Apr 09 '24 at 21:08
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    There's not much more detail than what you said. If it's not true that for all $x$ in the empty set, $P(x)$, then there must be some $x$ in the empty set for which $\neg P(x)$. But even without the "$\neg P(x)$", we know that's not possible. – user469053 Apr 09 '24 at 21:10
  • By the same argument, I suppose you can conclude that this statement is false because an empty set has no values to make it true. In fact, I believe that the question is unanswerable quite frankly because the statement "$x<y$" presupposes that $x$ exists, which it does not if $x\in{}$. –  Apr 09 '24 at 21:10
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    @TobySaunders-A'Court Why do there need to be values to make a statement true? – user469053 Apr 09 '24 at 21:11
  • Why do there need to be values to prove something false? The only way to proceed is to take as an axiom that things are either true or false until the otherwise is shown. –  Apr 09 '24 at 21:12
  • Because the negation of "for all" is "there exists," which can't hold with no values. – user469053 Apr 09 '24 at 21:13
  • Your interpretation is false. $\forall x\in A: P(x)$ is actually not a valid first order logic statement, but rather a shorthand for $\forall x:(x\in A\implies P(x)).$ – Thomas Andrews Apr 09 '24 at 21:13
  • I think the question's wrong. – Alexandra Apr 09 '24 at 21:15
  • But also, $\forall x\in A: P(x)$ and $\forall x\in B: Q(x)$ lets us deduce $\forall x\in A\cap B: P(x)\land Q(x),$ while your interpretation would require an extra condition, $A\cap B\neq \emptyset.$ – Thomas Andrews Apr 09 '24 at 21:17
  • See also Vacuous Truth. https://en.wikipedia.org/wiki/Vacuous_truth – Thomas Andrews Apr 09 '24 at 21:24
  • Indeed, and it is also possible for a statement to be vacuously false. Like, "All of the people who live on Mars are Spanish". – K.defaoite Apr 09 '24 at 21:29
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    @K.defaoite: That statement is also vacuously true. It's not false (vacuously or otherwise). – mjqxxxx Apr 09 '24 at 21:53
  • If Martians are nonexistent, then "*every* Martian is Spanish" is true. Simiarly, if $x$ isn't a member of $S,$ then *every* $x$ in $S$ does satisfy $P(x),$ that is, ∀x∈S P(x) is true. – ryang Apr 10 '24 at 04:24
  • See Range of quantification as well as Guarded quantifier: the abbreviated syntax $\forall x \in A(Px)$ is more formally: $\forall x (x \in A \to Px)$ and still betetr avoiding setmembership relation and using instead a predicate: $\forall x (Ax \to Px)$. – Mauro ALLEGRANZA Apr 10 '24 at 10:12

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