anonymous
  • anonymous
Why \[\forall x \in \emptyset: P(x)\] is equivalent to \[x\in \emptyset \Rightarrow P(x)\]?
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • 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" \[(e \in E) \rightarrow P(e)\]
UnkleRhaukus
  • UnkleRhaukus
\[\begin{array}{|c|c|c|}\hline\phi&\psi&\phi\Rightarrow\psi\\\hline T&T&T\\T&F&F\\F&T&T\\F&F&T\\\hline\end{array}\]\[\begin{array}{|c|c|c|c|}\hline \forall x \in \emptyset:P(x)&x\in\emptyset &P(x)& x\in\emptyset \Rightarrow P(x)\\\hline T&T&T&T\\\hline\end{array}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.