From 861ee760da5372e8f049b65bd71b3da2a56c8e97 Mon Sep 17 00:00:00 2001
From: Robbert Krebbers <mail@robbertkrebbers.nl>
Date: Sun, 10 Dec 2017 13:33:56 +0100
Subject: [PATCH] Docs: put Banach and America/Rutten in a theorem environment.
---
docs/algebra.tex | 17 +++++++++++++----
1 file changed, 13 insertions(+), 4 deletions(-)
diff --git a/docs/algebra.tex b/docs/algebra.tex
index 0a9c8687d..dc5dfad27 100644
--- a/docs/algebra.tex
+++ b/docs/algebra.tex
@@ -83,14 +83,23 @@ COFEs are \emph{complete OFEs}, which means that we can take limits of arbitrary
\end{defn}
The function space $\ofe \nfn \cofeB$ is a COFE if $\cofeB$ is a COFE (\ie the domain $\ofe$ can actually be just an OFE).
+$\SProp$ as defined above is complete, \ie it is a COFE.
Completeness is necessary to take fixed-points.
-For once, every contractive function $f : \cofe \to \cofe$ where $\cofe$ is a COFE and inhabited has a \emph{unique} fixed-point $\fix(f)$ such that $\fix(f) = f(\fix(f))$.
-This also holds if $f^k$ is contractive for an arbitrary $k$.
-Furthermore, by America and Rutten's theorem~\cite{America-Rutten:JCSS89,birkedal:metric-space}, every contractive (bi)functor from $\COFEs$ to $\COFEs$ has a unique\footnote{Uniqueness is not proven in Coq.} fixed-point.
-$\SProp$ as defined above is complete, \ie it is a COFE.
+\begin{thm}[Banach's fixed-point]
+\label{thm:banach}
+Given an inhabited COFE $\ofe$ and a contractive function $f : \ofe \to \ofe$, there exists a unique fixed-point $\fixp_T f$ such that $f(\fixp_T f) = \fixp_T f$.
+\end{thm}
+
+The above theorem also holds if $f^k$ is contractive for an arbitrary $k$.
+\begin{thm}[America and Rutten~\cite{America-Rutten:JCSS89,birkedal:metric-space}]
+\label{thm:america_rutten}
+Let $1$ be the discrete COFE on the unit type: $1 \eqdef \Delta \{ () \}$.
+Given a locally contractive bifunctor $G : \COFEs^{\textrm{op}} \times \COFEs \to \COFEs$, and provided that \(G(1, 1)\) is inhabited,
+then there exists a unique\footnote{Uniqueness is not proven in Coq.} COFE $\ofe$ such that $G(\ofe^{\textrm{op}}, \ofe) \cong \ofe$ (\ie the two are isomorphic in $\COFEs$).
+\end{thm}
\subsection{RA}
--
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