## anonymous 3 years ago Why $\forall x \in \emptyset: P(x)$ is equivalent to $x\in \emptyset \Rightarrow P(x)$?

1. anonymous

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

$\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}$