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I have error : Undefined control sequence for this code, however I am not sure which part is wrong.

\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}

However, when I recompile the overleaf with compiler pdfLaTeX, the result comes well as follows.

result

This is the snippet of the error message.

error

So, is there anyone could suggest what is wrong with the code, please?

Thank you.

Nicholas TI
  • 331
  • Can reproduce on TeXlive 23 and Overleaf pdflatex 2022. Must have been updated - what is your local TeX version? What error do you get, the lemma error or another? Could you edit your error message into the question please – JamesT Apr 10 '23 at 12:37
  • HI @JamesT, I edited the error message into the question. Secondly, how to check the local TeX version? Right now, I write the code in www.overleaf.com. – Nicholas TI Apr 10 '23 at 12:43
  • proof should not be used inside lemma, but after it. – egreg Apr 10 '23 at 13:40

1 Answers1

3

First of all, the code you pasted works as a standalone only if you remove the lemma and proof environments.

Second, your error is the use of $\equal$ I'm guessing instead of $\equiv$. That is where the "undefined control sequence" is.

\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}

If you, instead, wanted an equal sign, then a simple = should do the trick.

\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}
Korne
  • 130