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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?

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  • $\begingroup$ 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... $\endgroup$
    – Brennan
    Jan 19 '21 at 22:21
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    $\begingroup$ Once the appropriate meaning is attached to these letters it becomes an economics question. MWG is one of THE standard textbooks. $\endgroup$
    – Bayesian
    Jan 20 '21 at 10:24
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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

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