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Update class_3.tex
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Fix typos.
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saona-raimundo authored Oct 24, 2023
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Expand Up @@ -116,12 +116,12 @@ \chapter{Class 3}
A few remarks are in place.
\begin{itemize}
\item
Inductively defined sets are countable and consists of finite elements.
Inductively defined sets are countable and consist of finite elements.
\item
Inductively defined sets can be written as rules $x \implies f(x)$ meaning that, if $x \in X$, then $f(x) \in X$.
\item
Inductively defined sets allow proof by induction.
Consider prove that for all $x \in X$ we have $G(x)$.
Consider proving that for all $x \in X$ we have $G(x)$.
This can be proven by showing
\begin{enumerate}
\item $G(\bottom)$
Expand All @@ -137,7 +137,7 @@ \chapter{Class 3}
}

In the definition of balanced binary sequences, we consider the complete lattice $(\Sigma^\omega, \subseteq)$ and the function on sets given by $f(X) \defas 01X \cup 10X$.
Then, balanced binary sequences corresponds to $\gfp f$.
Then, balanced binary sequences correspond to $\gfp f$.

\Definition[Interval {$[0, 1]$}]{
Define the set $S$ as the largest set $X$ such that
Expand All @@ -149,15 +149,15 @@ \chapter{Class 3}
A few remarks are in place.
\begin{itemize}
\item
Coinductively defined sets are uncountable and consists of infinite elements.
Coinductively defined sets are uncountable and consist of infinite elements.
\item
Coinductively defined sets can be written as rules $x \implied f(x)$ meaning that, for all $y \in X$, there exists $x$ such that $y = f(x)$ and $x \in X$.
\item
Coinductively defined sets allow proof by induction.
Consider prove that for all $x \in X$ we have $G(x)$.
Coinductively defined sets allow proof by coinduction.
Consider proving that for all $x$, if $G(x)$, then $x \in X$.
This can be proven by showing
\begin{enumerate}
\item For all $x$ and $i$, if $G(f_i(x))$, then $G(x)$
\item For all $x$ and $i$, if $G(f_i(x))$, then $G(x)$ \,,
\end{enumerate}
where $\{f_1, \ldots, f_n\}$ is the set of rules that define the set $X$.
\end{itemize}
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