\documentclass[12pt]{article}
\usepackage[paper=letterpaper, left=1in, right=1in, top=1in, bottom=1in]{geometry}

\usepackage[parfill]{parskip}
\usepackage{amssymb}
\usepackage{amsmath}
\pagestyle{empty}

\begin{document}

\begin{center}
{\large CS 291}\\
Homework 1
\end{center}

\begin{flushright}
Milo Minderbinder\\
mmm123@truman.edu
\end{flushright}

\textbf{Section 6.2, problem 7b.}  Use Quine's method to show that the
wff $(A \rightarrow B) \land (B \rightarrow \neg A) \rightarrow A$ is
a contingency.

\emph{Answer:} First we compute the wff $W(A/\text{true})$.

\begin{center}
\begin{eqnarray*}
W(A/\text{true}) & \equiv & (\text{true} \rightarrow B) \land (B
\rightarrow \text{false}) \rightarrow \text{true} \\ 
& \equiv & B \land \neg B \rightarrow \text{true} \\ 
& \equiv & \text{true}
\end{eqnarray*}
\end{center}

Therefore $W(A/\text{true})$ is a tautology.  Next we check $W(A/\text{false})$.

\begin{align*}
W(A/\text{false}) & \equiv (\text{false} \rightarrow B) \land (B 
\rightarrow \text{true}) \rightarrow \text{false} \\
& \equiv \text{true} \land \text{true} \rightarrow \text{false} \\
& \equiv \text{false}
\end{align*}

This produces a contradiction. Thus the wff is a contingency.

\textbf{Section 6.2, problem 9a.}  Prove the equivalence $(A
\rightarrow B) \land (A \lor B) \equiv B$.

\emph{Answer:} We start with the left hand side and derive the right,
using the equivalences given in Table 6.1 on page 420.

\begin{align*}
(A \rightarrow B) \land (A \lor B) 
  & \equiv (\neg A \lor B) \land (A \lor B)  && \text{1st conversion} \\
  & \equiv (\neg A \land A) \lor B && \text{distributive property}\\
  & \equiv \text{false} \lor B && \text{conjunction of opposites} \\
  & \equiv B && \text{disjunctive identitiy}
\end{align*}

\end{document}
