☰

证明中未定义的控制顺序

tag-icon • tag-icon undefined
证明中未定义的控制顺序

我有错误：此代码未定义控制序列，但我不确定哪个部分是错误的。

\documentclass[journal]{IEEEtran}
\begin{document}

\begin{lemma}\label{lemma:R5_R6}
\begin{proof}

Besides place and transition which relate to the robot movement, $P^R_k$ and $T_k$, circuit $R_5(A^{\prime}_X)$ and $R_6(A^{\prime\prime}_X)$ are formed by $A^{\prime}_X$ and $A^{\prime\prime}_X$, respectively, for $X \in S$. Since $y= n$ where $n$ is an odd number, then $X \equiv \{x_1, x_2,\dots, x_n\}$ means the circuit connects all activities places for $i=1,\dots,n$. By following the Definition \ref{def:A_prime}, circuit $R_5(A^{\prime}_X)$ where $A^{\prime}_X \equal \{P_{x_1}^{\prime},  P_{x_2}^{\prime\prime},P_{x_3}^{\prime},\dots,P_{x_n}^{\prime}\}$ connects all cleaning place $c_i$ for all $i=1,\dots,n$ together with the respective $P^R_k$ and $T_k$. Similarly, according to the Definition\ref{def:A_doubleprime}, circuit $R_6(A^{\prime\prime}_X)$ where $A^{\prime\prime}_{X} \equiv \{P_{x_1}^{\prime\prime}, P_{x_2}^{\prime},P_{x_3}^{\prime\prime},\dots,P_{x_n}^{\prime\prime}\}$ connects all processing place $p_i$ for all $i=1,\dots,n$ together with the respective $P^R_k$ and $T_k$. Therefore, the Lemma \ref{lemma:R5_R6} holds. 
 
\end{proof}
\end{lemma}


\end{document}

然而，当我用编译器 pdfLaTeX 重新编译 overleaf 时，结果如下。

结果

这是错误消息的片段。

错误

那么，有人可以指出代码中存在什么问题吗？

谢谢。

答案1

首先，只有删除引理和证明环境，您粘贴的代码才能作为独立代码工作。

其次，您的错误在于使用了$\equal$我猜​​ 而不是$\equiv$。这就是“未定义的控制序列”所在。

\documentclass[journal]{IEEEtran}
\begin{document}
    

            
            Besides place and transition which relate to the robot movement, $P^R_k$ and $T_k$, circuit $R_5(A^{\prime}_X)$ and $R_6(A^{\prime\prime}_X)$ are formed by $A^{\prime}_X$ and $A^{\prime\prime}_X$, respectively, for $X \in S$. Since $y= n$ where $n$ is an odd number, then $X \equiv \{x_1, x_2,\dots, x_n\}$ means the circuit connects all activities places for $i=1,\dots,n$. By following the Definition \ref{def:A_prime}, circuit $R_5(A^{\prime}_X)$ where $A^{\prime}_X \equiv \{P_{x_1}^{\prime},  P_{x_2}^{\prime\prime},P_{x_3}^{\prime},\dots,P_{x_n}^{\prime}\}$ connects all cleaning place $c_i$ for all $i=1,\dots,n$ together with the respective $P^R_k$ and $T_k$. Similarly, according to the Definition\ref{def:A_doubleprime}, circuit $R_6(A^{\prime\prime}_X)$ where $A^{\prime\prime}_{X} \equiv \{P_{x_1}^{\prime\prime}, P_{x_2}^{\prime},P_{x_3}^{\prime\prime},\dots,P_{x_n}^{\prime\prime}\}$ connects all processing place $p_i$ for all $i=1,\dots,n$ together with the respective $P^R_k$ and $T_k$. Therefore, the Lemma \ref{lemma:R5_R6} holds. 
            

    
    
\end{document}

如果您想要一个等号，那么一个简单的=方法就可以了。

\documentclass[journal]{IEEEtran}
\begin{document}
    

            
            Besides place and transition which relate to the robot movement, $P^R_k$ and $T_k$, circuit $R_5(A^{\prime}_X)$ and $R_6(A^{\prime\prime}_X)$ are formed by $A^{\prime}_X$ and $A^{\prime\prime}_X$, respectively, for $X \in S$. Since $y= n$ where $n$ is an odd number, then $X \equiv \{x_1, x_2,\dots, x_n\}$ means the circuit connects all activities places for $i=1,\dots,n$. By following the Definition \ref{def:A_prime}, circuit $R_5(A^{\prime}_X)$ where $A^{\prime}_X = \{P_{x_1}^{\prime},  P_{x_2}^{\prime\prime},P_{x_3}^{\prime},\dots,P_{x_n}^{\prime}\}$ connects all cleaning place $c_i$ for all $i=1,\dots,n$ together with the respective $P^R_k$ and $T_k$. Similarly, according to the Definition\ref{def:A_doubleprime}, circuit $R_6(A^{\prime\prime}_X)$ where $A^{\prime\prime}_{X} \equiv \{P_{x_1}^{\prime\prime}, P_{x_2}^{\prime},P_{x_3}^{\prime\prime},\dots,P_{x_n}^{\prime\prime}\}$ connects all processing place $p_i$ for all $i=1,\dots,n$ together with the respective $P^R_k$ and $T_k$. Therefore, the Lemma \ref{lemma:R5_R6} holds. 
            

    
    
\end{document}

相关内容