131 lines
3.6 KiB
TeX
131 lines
3.6 KiB
TeX
\section{Logical Algebra}
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\definition{}
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Odds are, you are familiar with \textit{logical symbols}. \par
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In this handout, we'll use the following:
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\begin{itemize}
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\item $\lnot$: not
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\item $\land$: and
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\item $\lor$: or
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\item $\rightarrow$: implies
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\item $()$, parenthesis.
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\end{itemize}
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The function of these is defined by \textit{truth tables}:
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\begin{center}
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\begin{tabular}{ c | c | c }
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\multicolumn{3}{ c }{and} \\
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\hline
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$A$ & $B$ & $A \land B$ \\
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\hline
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F & F & F \\
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F & T & F \\
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T & F & F \\
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T & T & T
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\end{tabular}
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\hfill
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\begin{tabular}{ c | c | c }
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\multicolumn{3}{ c }{or} \\
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\hline
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$A$ & $B$ & $A \lor B$ \\
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\hline
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F & F & F \\
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F & T & T \\
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T & F & T \\
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T & T & T
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\end{tabular}
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\hfill
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\begin{tabular}{ c | c | c }
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\multicolumn{3}{ c }{implies} \\
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\hline
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$A$ & $B$ & $A \rightarrow B$ \\
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\hline
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F & F & T \\
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F & T & T \\
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T & F & F \\
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T & T & T
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\end{tabular}
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\hfill
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\begin{tabular}{ c | c }
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\multicolumn{2}{ c }{not} \\
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\hline
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$A$ & $\lnot A$ \\
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\hline
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T & F \\
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F & T \\
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~ & ~ \\
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~ & ~ \\
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\end{tabular}
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\end{center}
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\vspace{2mm}
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$A \land B$ is only true if both $A$ and $B$ are true. $A \lor B$ is true when $A$ or $B$ (or both) are true. \par
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$\lnot A$ is the opposite of $A$, which is why it looks like a \say{negative} sign. \par
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\vspace{2mm}
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$A \rightarrow B$ is a bit harder to understand. Read aloud, this is \say{$A$ implies $B$.} \par
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The only time $\rightarrow$ is false is when $T \rightarrow F$. This may seem counterintuitive, but it makes sense. Think about it. \par
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\problem{}
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Evaluate the following.
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\begin{itemize}
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\item $(T \land F) \lor T$
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\item $(\lnot (F \lor \lnot T) ) \rightarrow T$
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\item $(F \rightarrow T) \rightarrow (\lnot F \lor \lnot T)$
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\end{itemize}
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\vfill
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\pagebreak
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\begin{instructornote}
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After the class has done a few definable set problems, you can try to provide some intuition for $\rightarrow$ with the following example.
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\vspace{2mm}
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Say we have the sentence $\forall x ~ (a \rightarrow b)$. \par
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For example, take $\varphi = \forall x ~ ([x \geq 0] \rightarrow [\exists y ~ y^2 = x])$. \par
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$\varphi$ holds whenever any positive $x$ has a square root.
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\vspace{2mm}
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If $(\text{F} \rightarrow *)$ returned false, statements like the above would be hard to write. \par
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If $x$ is negative, $\varphi$ doesn't care whether or not it has a root. In this case, $\text{F} \rightarrow *$ must be true to avoid making whole $\forall$ false.
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\vspace{2mm}
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You can think of $[x \geq 0] \rightarrow b$ as a \say{sanity check} in a program: if $x$ isn't the kind of object we care about, return true and check the next one. If $x$ \textit{is} the kind of object we care about and $b$ is false, we have a counterexample to $[x \geq 0] \rightarrow b$, and thus $T \rightarrow F$ must be false.
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\end{instructornote}
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\problem{}
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Evaluate the following.
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\begin{itemize}
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\item $A \rightarrow T$ for any $A$
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\item $(\lnot (A \rightarrow B)) \rightarrow A$ for any $A, B$
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\item $(A \rightarrow B) \rightarrow (\lnot B \rightarrow \lnot A)$ for any $A, B$
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\end{itemize}
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\vfill
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\problem{}
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Show that $\lnot (A \rightarrow \lnot B)$ is equivalent to $A \land B$. \par
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That is, show that these give the same result for the same $A$ and $B$.
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\hint{Use a truth table}
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\vfill
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\problem{}
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Can you express $A \lor B$ using only $\lnot$, $\rightarrow$, and $()$?
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\begin{solution}
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$((\lnot A) \rightarrow B)$
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\end{solution}
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\vfill
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Note that both $\land$ and $\lor$ can be defined using the other logical symbols. \par
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The only logical symbols we \textit{need} are $\lnot$, $\rightarrow$, and $()$. \par
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We include $\land$ and $\lor$ to simplify our logical expressions.
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\pagebreak |