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  • 3 years ago

Why \[\forall x \in \emptyset: P(x)\] is equivalent to \[x\in \emptyset \Rightarrow P(x)\]?

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  1. anonymous
    • 3 years ago
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    it's just how mathematicians/logicians write things down. it's a matter of notation, and nothing more. they both state the same thing in a different way. they both claim that for any x in the empty set, P(x) is true another trivial example, what if we said: E = the set of all even numbers P(x) = "x + 2 is in E" and we made this claim: For all even numbers e in E, e + 2 is in E. \[\forall e \in E, P(e)\] see how this is an equivalent statement to: "if e is in E, then e + 2 is in E" \[(e \in E) \rightarrow P(e)\]

  2. UnkleRhaukus
    • 3 years ago
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    \[\begin{array}{|c|c|c|}\hline\phi&\psi&\phi\Rightarrow\psi\\\hline T&T&T\\T&F&F\\F&T&T\\F&F&T\\\hline\end{array}\]\[\begin{array}{|c|c|c|c|}\hline \forall x \in \emptyset:P(x)&x\in\emptyset &P(x)& x\in\emptyset \Rightarrow P(x)\\\hline T&T&T&T\\\hline\end{array}\]

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