Edits

Advanced/Definable Sets/parts/0 logic.tex

@@ -81,13 +81,33 @@ Evaluate the following.
 	\item $\texttt{F} \lor \texttt{T}$
 	\item $\texttt{T} \land \texttt{T}$
 	\item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
-	\item $(\texttt{T} \land \texttt{F}) \lor \texttt{T}$
-	\item $(\lnot (\texttt{F} \lor \lnot \texttt{T}) ) \rightarrow \texttt{T}$
+	\item $(\lnot (\texttt{F} \lor \lnot \texttt{T}) ) \rightarrow \lnot \texttt{T}$
 	\item $(\texttt{F} \rightarrow \texttt{T}) \rightarrow (\lnot \texttt{F} \lor \lnot \texttt{T})$
 \end{itemize}
 
+\begin{solution}
+	\texttt{F}
+	\texttt{T}
+	\texttt{T}
+	\texttt{T}
+	\texttt{F}
+	\texttt{T}
+\end{solution}
+
 \vfill
 \pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
 \begin{instructornote}
 	We can also think of $[x \geq 0] \rightarrow b$ as follows:
 	if $x$ isn't the kind of object we care about, we evaluate true and
@@ -117,8 +137,18 @@ Evaluate the following.
 	\item $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ for any $A, B$
 \end{itemize}
 
+\begin{instructornote}
+	Note that the last formula is the contrapositive of $A \rightarrow B$.
+\end{instructornote}
+
+\begin{solution}
+	All are true.
+\end{solution}
+
 \vfill
 
+% Show that A -> B ^ B -> A = T iff A = B
+
 \problem{}
 Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
 That is, show that these expressions always evaluate to the same value given