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Hello all im fairly new to LaTeX so any help would be appreciated. I have my code below and for some reason, I cannot get it to run due to many errors (mostly due to error files and undefined control sequences). I have attached the error codes to this post as well.enter image description here If you could run my code and see what is wrong that would be greatly appreciated!!!

\documentclass[10pt]{article}
\usepackage[usenames]{color} %used for font color
\usepackage{amssymb} %maths
\usepackage{amsmath} %maths
\usepackage{graphicx}
\usepackage{booktabs}
\graphicspath{/} 

\section{1.6}
  \paragraph{6.} \mbox{} \\ \\ \\ \\ \\ \\

\section{1.7}
  \paragraph{8} Yes. $ \\ \\ \null \quad \quad \overline{A \cup B} $ \mbox{}: \\ \\ \null \quad \includegraphics{vennd} \\ \\ \null \quad \quad $\overline{A} \cap \overline{B} $ : \\ \null \quad \includegraphics{vennd}

\section{1.8}
  \paragraph{6a} $ [0,2] \cup [0,3] \cup [0,4] \cup \dots = [0,\infty) $
  \paragraph{6b} $ [0,2] \cap [0,3] \cap [0,4] \cap \dots = [0,2] $

\section{2.1}
  \paragraph{6} Statement is true.
  \paragraph{14} Not a Statement

\section{2.2}
  \paragraph{8.} $ P = (x = 0) \\ \null \quad \quad Q = (y = y) \\ \null \quad \quad P \lor Q $

\section{2.3}
  \paragraph{2} If a function is continuous, then it is differentiable.

\section{2.4}
  \paragraph{4} $ a \in \mathbb{Q} \iff 5a \in \mathbb{Q} $

\section{2.5}
  \paragraph{4}
    \begin{center}
      \begin{tabular}{|c|c|c|c|c|c|}
        $P$ & $Q$ & $P \or Q$ & $\lnot{(P \lor Q)}$ & $\lnot{P}$ & $\not{(P \lor \lnot{P}$ \\ \midrule 
T & T & T & F & F & F \\
T & F & T & F & F & F \\
F & T & T & F & T & T \\
F & F & F & T & T & T \\
\end{tabular}
\end{center}

\paragraph{8}
  \begin{center}
    \begin{tabular}{cccccc}
      $P$ & $Q$ & $R$ & $\lnot{R}$ & $Q \land \lnot{R}$ & $P \lor )Q \land \lnot{R})$ \\midrule
T & T & T & F & F & T
T & T & F & T & T & T
T & F & T & F & F & F
T & F & F & T & F & T
F & T & T & F & F & F
F & T & F & T & T & T
F & F & T & F & F & F
F & F & F & T & F & F
\end{tabular}
\end{center}
\paragraph{10} Suppose  $((P \land Q) \lor R) \imlies (R \lor S)$ is false. \\ Then, $ R = true must be false, therefore $$ R = false, S = false $. \\ Also, $ ((P \land Q) \lor R)$ must be t $ R = false $ then $ (P \land Q) $ must be true. Therefore $ P = true, Q = true $

\section{2.6}
  \paragraph{2}
    \begin{center}
      \begin{tabular}{cccccc}
        $P$ & $Q$ & $R$ $Q\land R$ & $P \lor (Q \land R)$ & $P \lor Q$ & $(P \lor R) $ \\ \midrule
      \end{tabular}
      \end{center}
  \paragraph{10}
      Yes. \\
      \begin{equation}
        \begin{split}
         (P \implies Q) \lor R & \stackrel{?}{=} \lnot((O \land \lnot{Q}) \land \lnot{R}) \stackrel{?}{=} (\lnot (P \land \lnot{Q}) \lor R) \\ & \stackrel{?}{=} ((\lnot{P} \lor R) \\ (\lnot P \lorQ) \lor R & = ((\lnot{P} \lor Q) \lor R)
         \end{split}
      \end{equation}

      Since $ P\implies Q $ is logically equivalent to $ \lnot P \lor Q $ :
      \begin{center}
        \begin{tabular}{|c|c|c|c|c|}
          $P$ & $Q$ & $P \implies Q$ & $\lnot P$ & $ \lnot P \lor Q $ \\ \midrule
          T & T & T & F & T \\
          T & F & F & F & F \\
          F & T & T & T & T \\
          F & F & T & T & T \\

        \end{tabular}
       \end{center}
\end{document}  
  • 1
    welcome to tex.sx. in the second tabular you don't have any \\ to end the rows. that's what the complaint is about. you don't have "extra alignment tab"s except in a technical sense -- the absence of the end-of-row marker means that parsing continues and another & is what is found next. – barbara beeton Oct 13 '17 at 16:31
  • 2
    You've got several errors. I suggest to tackle them one at a time, in the right order: You are missing \begin{document}; the filed vennd apparently does not exist; the extra \or is the consequence of a misspelled \lor; in the second tabular you forgot to end the lines with \\; the undefined control sequences are also simply misspelled (\imlies instead of \implies, \lorQ isntead of \lor Q, ...); and so on... – campa Oct 13 '17 at 16:31
  • 1
    apart from the error messages the markup is very strange, there should never be \\ like \\ \\ \\ \\ \\ \\ that presumably makes warnings about underfull hboxes, and \section is a numbered section so you will get things like 2 1.8 for the second section which is numbered 2 but has title 1.8 which is very confusing. paragraph is a level 4 section heading so only intended to be used after \subsection and \subsubsection, don't use \\ before a displayed equation such as Yes. \\ – David Carlisle Oct 13 '17 at 16:35
  • Do not use LaTeXit for this. It is intended only for single equations and the like. Use for example TeXshop, or TeXworks, which is actually intended for complete documents. – Torbjørn T. Oct 13 '17 at 18:51
1

In some cases it wasn't clear what the intended output was but this produces the output that I think you intended, with no errors or warnings. I put comments inline in the code where I made changes.

\documentclass[10pt]{article}
\usepackage[usenames]{color} %used for font color
\usepackage{amssymb} %maths
\usepackage{amsmath} %maths
\usepackage{graphicx}
\usepackage{booktabs}

% only if your images are at the root of the filesystem
% \graphicspath{/} 



% \begin{document} was missing
\begin{document}

%  \\ \\ \\ \\ \\ \\
% make badness 10000 warnings (which is the maximum badness)

% \section{1.6} makes 1 1.6 which looks weird
\setcounter{section}{1}
\setcounter{subsection}{6}
\subsection{}
\subsubsection*{6.}

\subsection{}
  \subsubsection*{8} Yes. 

$\overline{A \cup B} $\\
 \includegraphics[height=1em]{example-image} % missing file {vennd}


$\overline{A} \cap \overline{B} $\\
\includegraphics[height=1em]{example-image} % missing file {vennd}

\stepcounter{section}
\subsection{}
  \subsubsection*{6a} $ [0,2] \cup [0,3] \cup [0,4] \cup \dots = [0,\infty) $
  \subsubsection*{6b} $ [0,2] \cap [0,3] \cap [0,4] \cap \dots = [0,2] $

\subsection{}
  \subsubsection*{6} Statement is true.
  \subsubsection*{14} Not a Statement

\subsection{}
  \subsubsection*{8.} $ P = (x = 0) \\ \null \quad \quad Q = (y = y) \\ \null \quad \quad P \lor Q $

\subsection{}
  \subsubsection*{2} If a function is continuous, then it is differentiable.

\subsection{}
  \subsubsection*{4} $ a \in \mathbb{Q} \iff 5a \in \mathbb{Q} $

\subsection{}
  \subsubsection*{4}
    \begin{center}
      \begin{tabular}{|c|c|c|c|c|c|}
        $P$ & $Q$ & $P 
% \lor not \or
\lor
% spurious {
Q$ & $\lnot{(P \lor Q)}$ & $\lnot{P}$ & $\not(P \lor \lnot{P}$ \\ \midrule 
T & T & T & F & F & F \\
T & F & T & F & F & F \\
F & T & T & F & T & T \\
F & F & F & T & T & T \\
\end{tabular}
\end{center}

\subsubsection*{8}
  \begin{center}
    \begin{tabular}{cccccc}
      $P$ & $Q$ & $R$ & $\lnot{R}$ & $Q \land \lnot{R}$ & $P \lor )Q \land \lnot{R})$ \\midrule
T & T & T & F & F & T \\
T & T & F & T & T & T \\
T & F & T & F & F & F \\
T & F & F & T & F & T \\
F & T & T & F & F & F \\
F & T & F & T & T & T \\
F & F & T & F & F & F \\
F & F & F & T & F & F 
\end{tabular}
\end{center}
\subsubsection*{10} Suppose  $((P \land Q) \lor R) 
% \implies not \imlies 
\implies (R \lor S)$ is false. \\ Then, $ R = true must be false, therefore $$ R = false, S = false $. \\ Also, $ ((P \land Q) \lor R)$ must be t $ R = false $ then $ (P \land Q) $ must be true. Therefore $ P = true, Q = true $

\subsection{}
  \subsubsection*{2}
    \begin{center}
      \begin{tabular}{cccccc}
        $P$ & $Q$ & $R$ $Q\land R$ & $P \lor (Q \land R)$ & $P \lor Q$ & $(P \lor R) $ \\ \midrule
      \end{tabular}
      \end{center}
  \subsubsection*{10}
      Yes.% never use \\ before a display math or at the end of a paragraph
      \begin{equation}
        \begin{split}
         (P \implies Q) \lor R & \stackrel{?}{=} \lnot((O \land \lnot{Q})
\land \lnot{R}) \stackrel{?}{=} (\lnot (P \land \lnot{Q}) \lor R) \\
& \stackrel{?}{=} ((\lnot{P} \lor R) \\ (\lnot P 
% \\lor Q not \lorQ
\lor Q) \lor R & = ((\lnot{P} \lor Q) \lor R)
         \end{split}
      \end{equation}

      Since $ P\implies Q $ is logically equivalent to $ \lnot P \lor Q $ :
      \begin{center}
        \begin{tabular}{|c|c|c|c|c|}
          $P$ & $Q$ & $P \implies Q$ & $\lnot P$ & $ \lnot P \lor Q $ \\ \midrule
          T & T & T & F & T \\
          T & F & F & F & F \\
          F & T & T & T & T \\
          F & F & T & T & T \\

        \end{tabular}
       \end{center}
\end{document} 

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