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Equivalently, and slightly more formally, we just proved that the existence of ξ ∈ ''A'' such that ''f''(ξ) = ''B'' implies the following [[contradiction]]:
:<math>\begin{aligned}
\xi\
\xi \in B &\iff \xi \in f(\xi) && \text{(by assumption that }f(\xi)=B\text{)}; \\
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