Add lecture

week2/20-01-2022/20-01-2022.tex

@@ -28,6 +28,46 @@
 \maketitle
 \pagenumbering{arabic}
 
+Multiple mixed quantifiers are different.
 
+\begin{equation}
+    \forall x\forall y P(x,y)=\forall y\forall x P(x,y)
+\end{equation}
+
+However\dots
+\begin{equation}
+    \forall x\exists y P(x,y) \neq \exists y\forall x P(x,y)
+\end{equation}
+The first says, \say{For all x, there is some y such that $x+y=0$}.
+The second says, \say{For some y, all x fit the statement $x+y=0$}.
+
+How do you negate multiple quantified statements? You follow DeMorgans, repeatedly.
+
+\section{Inferencing}%
+\label{sec:Inferencing}
+
+\subsection{Arguments}%
+\label{sub:Arguments}
+
+All statements in an argument are premises (except the last, which is the conclusion).
+The argument is valid if the conclusion is true when all premises are true.
+\emph{Modus Ponens}
+\begin{equation}
+    \begin{split}
+        q\to p\\
+        q\\
+        \therefore p
+    \end{split}
+\end{equation}
+
+\emph{Modus Tollens}
+\begin{equation}
+    \begin{split}
+        p\to q\\
+        \lnot q\\
+        \therefore \lnot p
+    \end{split}
+\end{equation}
+Contrapositive of modus ponens.
 
 \end{document}