Xiao edited introduction.tex  over 8 years ago

Commit id: efb01b3ef6543511d71ebf812bd387ff2f39062e

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Here is some sample LaTeX notation. By associativity, if $\zeta$ is combinatorially closed then $\delta = \Psi$. Since $${S^{(F)}} \left( 2, \dots,-\mathbf{{i}} \right) \to \frac{-\infty^{-6}}{\overline{\alpha}},$$ $l < \cos \left( \hat{\xi} \cup P \right)$. Thus every functor is Green and hyper-unconditionally stable. Obviously, every injective homeomorphism is embedded and\section{的}\subsection{hh}\subsubsection{cc} Clifford. Because $\mathcal{{A}} > S$, $\tilde{i}$ is not dominated by $b$. Thus ${T_{t}} > | A |$.  \subsection{Subsection Heading Here}  Subsection text here. Let's show some\centerline{T some  \centerline{T  = \begin{bmatrix} $\Partial$^{2}U $\Partial$^{2}  & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 11 & -11 \\ 21 & 20 \end{bmatrix}} more LaTeX: Obviously, ${W_{\Xi}}$ is composite. Trivially, there exists an ultra-convex and arithmetic independent, multiply associative equation. So $\infty^{1} > \overline{0}$. It is easy to see that if ${v^{(W)}}$ is not isomorphic to $\mathfrak{{l}}$ then there exists a reversible and integral convex, bounded, hyper-Lobachevsky point. One can easily see that $\hat{\mathscr{{Q}}} \le 0$. Now if $\bar{\mathbf{{w}}} > h' ( \alpha )$ then ${z_{\sigma,T}} = \nu$. Clearly, if $\| Q \| \sim \emptyset$ then every dependent graph is pseudo-compactly parabolic, complex, quasi-measurable and parabolic. This completes the proof.  \subsubsection{Subsubsection Heading Here}