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Let E be an empty set . why the assertion "$(\forall x \in E) , p(x)$ " is true ??

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    $\begingroup$ It's a vacuous truth $\endgroup$
    – J. W. Tanner
    Jun 20, 2022 at 0:47
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    $\begingroup$ What would a counterexample look like? $\endgroup$
    – Steven Landsburg
    Jun 20, 2022 at 2:47
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    $\begingroup$ Because there isn't any $x \in E$ where $p(x)$ is false $\endgroup$
    – fleablood
    Jun 20, 2022 at 3:08
  • $\begingroup$ Use the principle of vacuous truth: $A\implies (\neg A \implies B)$ (a tautology). In your example, you have $A~\equiv~ x\notin E~$ and $~B~\equiv ~ p(x)$. $\endgroup$
    – Dan Christensen
    Jun 20, 2022 at 13:16

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I think this question should be posted on the Stock Exchange.

However, the empty set is the set that does not contain any element, so the part $x \in E$ is already false, therefore both $F \to F$, and $F \to T$ are True.

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    $\begingroup$ ok thanks sir . $\endgroup$
    – منوعات
    Jun 20, 2022 at 0:44