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\noindent From the text, pg. 29: \\
\textbf{Rule C.P.:} \textit{If we can derive $S$ from $R$ and a set of premises, then we may derive $R \rightarrow S$ from the premises alone.} \\
\textbf{Question from 14 February class:} when Rule C.P. is invoked and when $R$ is a premise, do we list the line corresponding to $R$ when invoking rule C.P.? \\
\textbf{Answer:} The answer is no, see examples of this on line (8) of the deduction on pg. 29 and line (9) in the example on pg. 30. The key fact about \textbf{Rule C.P.} is that it is the only rule we have at this point that \textit{allows you to remove premises from the left column}. Practically, this means, in some sense, that we may introduce a new premise, use it to ``do something", and then make it ``go away". \\
\hrule \vspace{10pt}
\noindent\textbf{A ``practical" use of Rule C.P.:} can you deduce $P \rightarrow R$ from the following premises?:\\
\begin{tabular}{lll}
$\{1\}$ & (1) $P \rightarrow Q$ & Premise \\
$\{2\}$ & (2) $\neg (\neg Q) \rightarrow R$ & Premise
\end{tabular} \\
It seems ``obvious" that we should be able to conclude $P \rightarrow R$ from this, but it also seems difficult to deduce formally because the sentence $Q \rightarrow S$ does \textit{not} follow from the Law of Double Negation (it doesn't match the ``correct form"). This is an instance where exploiting \textbf{Rule P} along with \textbf{Rule C.P.} is the easiest way to proceed. Recall what \textbf{Rule P} says (pg. 28): \\
\textbf{Rule P:} \textit{We may introduce a premise at any point of the derivation.} \\
\textbf{The strategy:} We will introduce a new premise $Q$ and then use it to conclude $\neg \neg Q$, from which we may deduce $R$. This shows that ``\textit{we can derive $R$ from $Q$ and a set of premises}", and so \textbf{Rule C.P.} tells us that ``\textit{we may derive $Q \rightarrow R$ from the premises alone}" \\
\begin{tabular}{lll}
$\{1\}$ & (1) $P \rightarrow Q$ & Premise \\
$\{2\}$ & (2) $\neg (\neg Q) \rightarrow R$ & Premise \\
$\{3\}$ & (3) $Q$ & Premise \\
$\{3\}$ & (4) $\neg \neg Q$ & 3 T (Law of Double Negation) \\
$\{2,3\}$ & (5) $R$ & 2,4 T (Law of Detachment) \\
$\{2\}$ & (6) $Q \rightarrow R$ & 3, 5 C.P. \\
$\{1,2\}$ & (7) $P \rightarrow R$ & 1,6 T (Law of Hypothetical Syllogism)
\end{tabular} \\
\hrule\vspace{10pt}
\noindent\textbf{Challenge:} Can you deduce $P \rightarrow R$ from premises (1) and (2) without using \textbf{Rule C.P.}? \\
\hrule\vspace{10pt}
\noindent\textit{Additional comment}: I just want to point out again that there are logics out there which do not accept the ``Law of Double Negation" -- {\color{blue}\href{https://en.wikipedia.org/wiki/Intuitionistic_logic}{intuitionistic logic}} is probably the most common (read the ``syntax" section on its wikipedia page!)
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