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.
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}
Best Answer
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.