![tikz海报:\,\多图错误](https://linux22.com/image/381455/tikz%E6%B5%B7%E6%8A%A5%EF%BC%9A%5C%2C%5C%E5%A4%9A%E5%9B%BE%E9%94%99%E8%AF%AF.png)
我在 Share Latex 上收到以下错误消息
<argument> \,\multimap
\,
l.241 \end{exe}}
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
当我使用以下代码创建文件时:
\documentclass[25pt,a1paper]{tikzposter}
\usepackage{graphicx}
\graphicspath{{/home/Darrin/Downloads}}
\usetheme{Rays}
\usepackage{lipsum}
\usepackage{multicol}
\usepackage{subfig}
\usetheme{Madrid}
\usepackage{comment}
\newenvironment{variableblock}[3]{%
\setbeamercolor{block body}{#2}
\setbeamercolor{block title}{#3}
\begin{block}{#1}}{\end{block}}
\usepackage{tikz-qtree}
\usepackage{booktabs}
\usepackage{multicol}
\usepackage{amsfonts,amsmath,braket}
\newcommand{\interp}[2][]{
\(
\left\llbracket\,\text{#2}\,\right\rrbracket^{#1}
\)
}
\newcommand{\den}[1]{
\(
\left[\,\text{#1}\,\right]
\)
}
\newcommand{\argum}[1]{
\(
\left(\,\text{#1}\,\right)
\)
}
\newcommand{\ceil}[2][]{
\(
\left\lceil\,\text{#2}\,\right\rceil^{#1}
\)
}
\newcommand{\wraptext}[2][1in]{
\begin{varwidth}{#1}{\RaggedRight#2}\end{varwidth}
}
%\usepackage{prooftree} % proofs
\usepackage{tikz-qtree-compat}
\newcommand{\formula}[1]{\ensuremath{\mathit{#1}}}
\newcommand{\vmformula}[3][]{%
\begin{array}[b]{@{}l}
\mbox{\textbf{#1}}\\
\formula{#2:}\\%[-0.5ex]
\formula{#3}
\end{array}
}
\newcommand{\mformula}[2]{%
\begin{array}[b]{@{}l}
\formula{#1:}\\%[-0.5ex]
\formula{#2}
\end{array}
}
\newcommand{\sig}{\mbox{$_\sigma\,$}} % sigma, with space after
\newcommand{\sigb}{\mbox{$_\sigma$}} % sigma, no space after
\newcommand{\upsig}. {\mbox{\ensuremath{\uparrow\hspace{-0.35em}_\sigma\,}}}
\newcommand{\Upsig}. {\mbox{\ensuremath{\uparrow\hspace{-0.35em}_\sigma}}}
\newcommand{\upsigb}{\mbox{\ensuremath{\uparrow\hspace{-0.35em}_\sigma}}}
\newcommand{\linimp}{\mbox{\ensuremath{\,\multimap\,}}}
\newcommand{\linimpE}{\mbox{\small\ensuremath{\multimap_{\mathcal{E}}}}}
\newcommand{\linimpIi}[1]{\mbox{\small\ensuremath{\multimap_{{\mathcal{I}},#1}}}}
\newcommand{\tensor}{\mbox{\ensuremath{\,\otimes\,}}}
\newcommand{\letE}{\mbox{\small\ensuremath{\mathit{\beta\mbox{-}reduction}}}}
\newcommand{\tensorEij}[2]{\mbox{\small$\otimes_{{\mathcal{E}},#1,#2}$}}
\newcommand{\llet}[3]{\ensuremath{\mathsf{let~}{#1}\mathsf{~be~}{#2}\mathsf{~in~}{#3}}}
\newcommand{\betared}{\ensuremath{\Rightarrow_\beta}}
\newcommand{\type}[1]{\mbox{\ensuremath{\mathit{#1}}\/}} % type name
\newcommand{\forallE}{\mbox{\small$\forall_{{\mathcal{E}}}$}}
\usetikzlibrary{decorations.pathreplacing,shapes.misc}
\tikzset{
show control points/.style={
decoration={show path construction, curveto code={
\draw [blue, dashed]
(\tikzinputsegmentfirst) -- (\tikzinputsegmentsupporta)
node [at end, cross out, draw, solid, red, inner sep=2pt]{};
\draw [blue, dashed]
(\tikzinputsegmentsupportb) -- (\tikzinputsegmentlast)
node [at start, cross out, draw, solid, red, inner sep=2pt]{};
}
},
postaction=decorate
},
}
\usepackage[normalem]{ulem}
\usepackage{stmaryrd}
\usepackage{xcolor,mdframed}
%\setbeamertemplate{theorems}[numbered]
\newtheorem{axiom}{Axiom}
\usepackage{gb4e}
\begin{document}
\title{HMMMMMMMM}
\author{hmmmm}
\maketitle
\begin{columns}
%COLUMN1
\column{.65}
%block.a
\block{Oh my }{
Oh my
\bigskip \\
\coloredbox{\begin{itemize}
\item Oh my
\item Oh my
\end{itemize}}
\lipsum[2]
\bigskip
\innerblock{oh my }
{\begin{center}
$1 +1 = 2$
\end{center}
}
}
%block.b
\block{Oh my } {\lipsum[1]
}
%COLUMN2
\column{.35}
\block{Oh my }
{%\includegraphics[width=\linewidth]{chairs}
}
\end{columns}
\block{Oh}{\begin{exe}
\ex
\begin{tikzpicture}[baseline=(current bounding box.center), every tree node/.style={align=center,anchor=north}, level distance = 18ex, scale= .5]
\Tree [.{$Prove (h, \mathbf{\neg Inductive_{h} (the.primes_{h})} : t $} [.{$\sigma: i$} ] [.{$\eta \,(Prove (h, \mathbf{\neg Inductive_{h} (the.primes_{h}))}: \Diamond t $} \edge node[auto=left] {$\eta$}; [. {$Prove (h, \neg \lambda j. \left\{
\begin{array}{ll}
\mathbf{Inductive_{\mathtt{h}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right.h\Bigg)}& \text{if }j =\mathtt{h} \\
\mathbf{Inductive_{\mathtt{\sigma}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right.\sigma \Bigg)}& \text{if } j=\sigma
\end{array}
\right. h) : t$} [.{$h: e$} ] [.{$\lambda x. Prove(x, \neg \lambda j. \left\{
\begin{array}{ll}
\mathbf{Inductive_{\mathtt{h}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right.h \Bigg)}& \text{if }j =\mathtt{h} \\
\mathbf{Inductive_{\mathtt{\sigma}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right.\sigma \Bigg)}& \text{if } j=\sigma
\end{array}
\right. \kappa(x)): e \linimp t$} [.{$\lambda p. \lambda x. Prove(x, p \hspace{0.2cm}\kappa(x))\,$ \\ $\Diamond t \linimp e \linimp t$} ] [.{$\neg \lambda j \left\{
\begin{array}{ll}
\mathbf{Inductive_{\mathtt{h}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right.h \Bigg)}& \text{if }j =\mathtt{h} \\
\mathbf{Inductive_{\mathtt{\sigma}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right. \sigma \Bigg)}& \text{if } j=\sigma
\end{array}
\right.: \Diamond t$} [.{$\lambda p. (\neg p)$ \\ $\Diamond t \linimp \Diamond t$} ] [.{$\lambda j. \left\{
\begin{array}{ll}
\mathbf{Inductive_{\mathtt{h}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right. h \Bigg)}& \text{if }j =\mathtt{h} \\
\mathbf{Inductive_{\mathtt{\sigma}}\Bigg(\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right. \sigma \Bigg)}& \text{if } j=\sigma
\end{array}
\right.: \Diamond t$} [.{$\lambda x, \lambda j. \left\{
\begin{array}{ll}
\mathbf{Inductive_{\mathtt{h}}(x_h)}& \text{if }j =\mathtt{h} \\
\mathbf{Inductive_{\mathtt{\sigma}}(x_{\sigma})}& \text{if } j=\sigma
\end{array}
\right.:$ \\ $\Diamond e \linimp \Diamond t$} ] [.{$\lambda i. \left\{
\begin{array}{ll}
\mathbf{the.primes}_{\mathtt{h}}& \text{if }i =\mathtt{h} \\
\mathbf{the.primes}_{\sigma}& \text{if } i=\sigma
\end{array}
\right. : \Diamond e$} ] ] ] ] ] ] ] ]
\end{tikzpicture}
\end{exe}}
\end{document}
有谁知道我该如何解决这个问题?