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, 19:45, 5 June 2008→Current Version @ PlanetMath : TeX Format: updare
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This gives $\operatorname{E}A^\circ$ the type $(\mathbb{B}^n \times \mathbb{D}^n\ +\!\to \mathbb{B}) = (\mathbb{B}^n \times \mathbb{D}^n, \mathbb{B}^n \times \mathbb{D}^n \to \mathbb{B}).$
\begin{quote}
\begin{quote}
$\operatorname{E}A^\circ = [ \operatorname{E}\mathcal{A} ] = [ \mathcal{A}\ \cup\ \operatorname{d}\mathcal{A} ] = [ a_1, \ldots, a_n, \operatorname{d}a_1, \ldots, \operatorname{d}a_n ].$
$\operatorname{E}A^\circ = [ \operatorname{E}\mathcal{A} ] = [ \mathcal{A}\ \cup\ \operatorname{d}\mathcal{A} ] = [ a_1, \ldots, a_n, \operatorname{d}a_1, \ldots, \operatorname{d}a_n ].$
+\end{quote}
+
+This gives $\operatorname{E}A^\circ$ the type:
+
+\begin{quote}
+$[ \mathbb{B}^n \times \mathbb{D}^n ] = (\mathbb{B}^n \times \mathbb{D}^n\ +\!\to \mathbb{B}) = (\mathbb{B}^n \times \mathbb{D}^n, \mathbb{B}^n \times \mathbb{D}^n \to \mathbb{B}).$
\end{quote}
\end{quote}
\end{itemize}
\end{itemize}
−$\dots$
$\dots$
