I'm trying to negate that: $\exists B \in \mathcal{B}$ such that $x,y \in B$ and $x \in C(B)$.
Looks that the negation is equivalent to: $\forall B \in \mathcal{B}(x,y \in B \implies x \not \in C(B))$. Is it right?
2
-
$\begingroup$ This is more appropriate for math.se no? Whilst it fits into the context of economics I don't see how economics would help you prove this... $\endgroup$– BrennanJan 19, 2021 at 22:21
-
1$\begingroup$ Once the appropriate meaning is attached to these letters it becomes an economics question. MWG is one of THE standard textbooks. $\endgroup$– BayesianJan 20, 2021 at 10:24
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Yes, it is. Step by step:
$\lnot \exists B \in A (x,y \in A \land x \in C(B))$ is equivalent to
$\forall B \in A \lnot(x,y \in B \land x \in C(B))$ is equivalent to
$\forall B \in A (x,y \notin B \vee x \notin C(B))$ is equivalent to
$\forall B \in A (x,y \in B \implies x \notin C(B))$.
Reference: How to prove it, by Velleman