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From the book, pg. 40: \\
\textbf{Derived Rule for Indirect Proofs R.A.A.}: \textit{If a contradiction is derivable from a set of premises and the negation of a formula $S$, then $S$ is derivable from the set of premises alone.} \\
In other words, to use this rule to make a conclusion $C$, we $\ldots$
\begin{enumerate}
\item Introduce $\neg C$ as a premise,
\item Use that premise to derive a contradiction $P \wedge \neg P$ (for any sentence $P$ that you can find),
\item Reject $\neg C$ as valid (it led to a contradiction), and hence
\item Conclude $C$
\end{enumerate}
\hrule
\noindent From the text, pg. 42: \\
``If a declaration of war is a sound strategy, then either fifty divisions are poised at the border or twenty wings of long-range bombers are ready to strike. However, fifty divisions are not poised at the border. Therefore, if twenty wings of long-range bombers are not ready to strike, then either a declaration of war is not a sound strategy or new secret weapons are available."

Let 
\begin{flushleft}
$D=$ ``A declaration of war is a sound strategy." \\
$F=$ ``Fifty divisions are poised at the border." \\
$T=$ ``Twenty wings of long-range bombers are ready to strike." \\
$S=$ ``New secret weapons are available." 
\end{flushleft}

\hrule\vspace{10pt}

\noindent The argument has the following premises: \\
\begin{tabular}{lll}
$\{1\}$ & (1) $D \rightarrow F \vee T$ & Premise \\
$\{2\}$ & (2) $\neg F$ & Premise \\
\end{tabular} \\
and wants to conclude the sentence $\neg T \rightarrow (\neg D \vee S)$. We will introduce the negation of the conclusion and then derive a contradiction from it and premises alone. In trying to do that, I will introduce the premise $D$ on line (8) and use reductio ad absurdum to conclude $\neg D$ and use that sentence to make the final application of reductio ad absurdum in line (15). \\
\begin{tabular}{lll}
$\{1\}$ & (1) $D \rightarrow F \vee T$ & Premise \\
$\{2\}$ & (2) $\neg F$ & Premise \\
$\{3\}$ & (3) $\neg(\neg T \rightarrow \neg D \vee S)$ & Premise \\
$\{3\}$ & (4) $\neg T \wedge \neg ( \neg D \vee S)$ & 3 T Law of Negation for Implication \\
$\{3\}$ & (5) $\neg T$ & 4 T Law of Simplification \\
$\{3\}$ & (6) $\neg ( \neg D \vee S) \wedge \neg T$ & 4 T Commutative law for $\wedge$ \\
$\{3\}$ & (7) $\neg(\neg D \vee S)$ & 6 T Law of Simplification \\
$\{8\}$ & (8) $D$ & Premise \\
$\{1,8\}$ & (9) $F \vee T$ & 1, 8 T Law of Detachment \\
$\{1,2,8\}$ & (10) $T$ & 2, 9 T Modus tollendo ponens \\
$\{1,2,3,8\}$ & (11) $T \wedge \neg T$ & 5, 10 T Law of Adjunction \\
$\{1,2,3\}$ & (12) $\neg D$ & 8, 11 R.A.A. \\
$\{1,2,3\}$ & (13) $\neg D \vee S$ & 12 T Law of Addition \\
$\{1,2,3\}$ & (14) $\neg(\neg D \vee S) \wedge (\neg D \vee S)$ & 7, 13 T Law of Adjunction \\
$\{1,2\}$ & (15) $\neg T \rightarrow \neg D \vee S$ & 3, 14 T R.A.A.
\end{tabular}


\newpage
\noindent\textbf{{\texttt{\#}8(a)} from {\color{blue}\href{http://tomcuchta.com/teach/classes/2017/MATH-2200-Spring-2017-FairmontState/exams/EXAM1(solution)-MATH2200-Spring-2017-FairmontState.pdf}{Exam 1}} (via reductio ad absurdum):} We are trying to conclude $\neg P$ from premises (1) and (2). We will do so by admitting $\neg(\neg P)$ as a premise and deriving a contradiction, and then applying \textbf{R.A.A.} to conclude $\neg P$. \\
\begin{tabular}{|l|l|l|}
\hline
$\{1\}$ & (1) $P \rightarrow Q$ & Premise \\
$\{2\}$ & (2) $\neg Q$ & Premise \\
$\{3\}$ & (3) $\neg(\neg P)$ & Premise \\
$\{3\}$ & (4) $P$ & 3 T (Law of Double Negation)\\
$\{1,3\}$ & (5) $Q$ & 4 T (Law of Detachment) \\
$\{1,2,3\}$ & (6) $Q \wedge \neg Q$ & 2, 5 T (Law of Adjunction) \\
$\{1,2,\}$ & (7) $\neg P$ & 3, 6 T R.A.A. \\
\hline
\end{tabular}
\vspace{5pt} \\
\textbf{{\texttt{\#}8(b)} from {\color{blue}\href{http://tomcuchta.com/teach/classes/2017/MATH-2200-Spring-2017-FairmontState/exams/EXAM1(solution)-MATH2200-Spring-2017-FairmontState.pdf}{Exam 1}} (via reductio ad absurdum):} We are trying to prove $S$ from using premises (1) through (4). We will do so by admitting $\neg S$ as a premise and deriving a contraction, and then applying \textbf{R.A.A.} to conclude $S$. \\
\begin{tabular}{|l|l|l|}
\hline
$\{1\}$ & (1) $P\rightarrow Q$ & Premise \\
$\{2\}$ & (2) $Q \rightarrow R$ & Premise \\
$\{3\}$ & (3) $S \vee (\neg R)$ & Premise \\
$\{4\}$ & (4) $P$ & Premise \\
$\{5\}$ & (5) $\neg S$ & Premise \\
$\{3\}$ & (6) $(\neg R) \vee S$ & 3 T (Commutativity of $\vee$) \\
$\{3\}$ & (7) $R \rightarrow S$ & 6 T (Law of Equivalence of Implication and Disjunction) \\ 
$\{3\}$ & (8) $\neg S \rightarrow \neg R$ & 7 T (Law of Contraposition) \\
$\{3,5\}$ & (9) $\neg R$ & 5, 8 T (Law of Detachment) \\
$\{1\}$ & (10) $\neg Q \rightarrow \neg P$ & 1 T (Law of Contraposition) \\
$\{2\}$ & (11) $\neg R \rightarrow \neg Q$ & 2 T (Law of Contraposition) \\
$\{2,3,5\}$ & (12) $\neg Q$ & 9, 11 T (Law of Detachment) \\
$\{1,2,3,5\}$ & (13) $\neg P$ & 10, 12 T (Law of Detachment) \\
$\{1,2,3,4,5\}$ & (14) $P \wedge (\neg P)$ & 4, 13 T (Law of Adjunction) \\
$\{1,2,3,4\}$ & (15) $S$ & 5, 14 T R.A.A. \\
\hline
\end{tabular}
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