跳转至

集合论和其他符号⚓︎

约 127 个字 1 行代码 预计阅读时间不到 1 分钟

\[\def\arraystretch{2}
   \begin{array}{c|c|c|c|c|c|c|c}
   \hline
   \hline
   \text{原型} & 符号 &  \text{原型} & 符号 & \text{原型} & 符号 & \text{原型} & 符号  \cr \hline
   \text{forall} & \forall   &   \text{exists}  &       \exists   &  \text{nexists} & \nexists  & \text{in} & \in  \cr \hline    
   \text{empty} & \empty   &   \text{varnothing}  &       \varnothing   &  \text{notin} & \notin  & \text{subset} & \subset  \cr \hline    
   \text{land} & \land   &   \text{lor}  &       \lor   &  \text{because} & \because  & \text{therefore} & \therefore  \cr \hline    
   \text{implies} & \implies   &   \text{impliedby}  &       \impliedby   &  \text{neg} & \neg  & \text{leftrightarrow} & \leftrightarrow  \cr \hline    
   \end{array}
\]
  • 上下对齐的公式:
\[f(n)=\begin{dcases}  x_{11} &+ & x_{12} &+ & x_{13} & = &1, & i = 1,2...m \cr & &x_{12} & & &  =& 2 ,& j = 1,2,...n \end{dcases}\]
$$f(n)=\begin{dcases}  x_{11} &+ & x_{12} &+ & x_{13} & = &1, & i = 1,2...m \cr & &x_{12} & & &  =& 2 ,&  \end{dcases}$$