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\section*{Computability and Logic Homework 9}

\noindent
\textbf{Due: } Thursday, November 17, 2005 

\paragraph{1.} Recall that a formula $A$ is {\bf\em valid} if $\models
A$; we say $A$ is {\bf\em satisfiable} if there exists some truth
assignment $\eta$ such that $\mathcal{T}_\eta(A) = \TT$.  For each of
the following, prove or give a counterexample as appropriate:
\begin{itemize}
\item[a.] If $A \supset B$ is valid and $A$ is valid, then $B$ is
  valid.
\item[b.] If $A \supset B$ is satisfiable and $A$ is satisfiable, then
  $B$ is satisfiable.
\item[c.] If $A \vee B$ is satisfiable, then $A$ is satisfiable or $B$
  is satisfiable.
\item[d.] If $A \wedge B$ is valid, then $A$ is valid and $B$ is valid.
\end{itemize}

\paragraph{2.} Give derivations for each of the following judgments.
You may present your derivations in either list form or tree form as
described in the logic handout.
\begin{itemize}
\item[a.] $(A \supset B),(B \supset C) \vdash A \supset C$
\item[b.] $(A \vee B), (\neg B \vee C) \vdash A \vee C$
\item[c.] $\vdash (A \vee (A \wedge B)) \supset (A \wedge (A \vee B))$
\end{itemize}

\paragraph{3. Negation}  Give deductions for each of the following.
(Or, prove that deductions exist using deductions you've shown to
exist earlier together with structural properties and/or Cut and/or
inference rules.)
\begin{itemize}
\item[a.] $A \vdash \neg \neg A$
\item[b.] $\neg \neg \neg A \vdash \neg A$ ({\em without} using the
  $\mathit{RAA}$ rule.)
\item[c.] $\neg \neg \neg A \vdash \neg A$ ({\em using} the
  $\mathit{RAA}$ rule to produce a shorter deduction than in part
  (b).)
\item[d.] $\vdash \neg \neg (A \vee \neg A)$
\end{itemize}

\paragraph{4.} Consider this alternative formulation of the $\neg I$
rule:
\[
\infer[(\neg I')]{
  \Gamma \vdash \neg A
}{
  \Gamma,A \vdash \neg A
}
\]
In proving the following, do not appeal to the completeness theorem.
Prove that the requested derivations exist by showing how to construct
them directly.
\begin{itemize}
\item[a.] Show that for any $\Gamma$ and $C$, if there is a derivation
  of $\Gamma \vdash C$ that uses this new rule, there is a derivation
  of the same judgment that uses the original rule instead.
\item[b.] Show that for any $\Gamma$ and $C$, if there is a derivation
  of $\Gamma \vdash C$ that uses the original $\neg I$ rule, there is
  a derivation of the same judgment that uses the $\neg I'$ rule
  instead.
\end{itemize}

\paragraph{5. De Morgan's Laws}
Give derivations:
\begin{itemize}
\item[a.] $A \vee B \vdash \neg (\neg A \wedge \neg B)$
\item[b.] $\neg (\neg A \vee \neg B) \vdash A \wedge B$
\item[c.] $\neg (A \wedge \neg B) \vdash A \supset B$
\end{itemize}

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