From 69696cb76d8c5874b78b74a1d32e5a0bd33d2036 Mon Sep 17 00:00:00 2001 From: Leon Vatthauer Date: Wed, 6 Mar 2024 16:29:43 +0100 Subject: [PATCH] Continue working on proofs concerning K --- thesis/main.tex | 52 ++++----------- thesis/src/04_iteration.tex | 123 ++++++++++++++++++++++++++++-------- 2 files changed, 109 insertions(+), 66 deletions(-) diff --git a/thesis/main.tex b/thesis/main.tex index 7619561..8fa05e7 100644 --- a/thesis/main.tex +++ b/thesis/main.tex @@ -99,51 +99,18 @@ \usepackage{noto-mono} -\usepackage{mathpazo} \usepackage{unicode-math} +% \setmainfont{STIX-Regular} +% \setmathfont{STIX Two Math Regular} + \usepackage{mathrsfs} \usepackage[colorinlistoftodos,prependcaption,textsize=tiny]{todonotes} \usepackage{xargs} \usepackage{xstring} -%\usepackage{fontspec} -%\setmonofont{Noto Sans Mono} -\usepackage{pict2e} - -% \newcommand{\lbparen}{% -% \mathopen{% -% \sbox0{$()$}% -% \setlength{\unitlength}{\dimexpr\ht0+\dp0}% -% \raisebox{-\dp0}{% -% \begin{picture}(.32,1) -% \linethickness{\fontdimen8\textfont3} -% \roundcap -% \put(0,0){\raisebox{\depth}{$($}} -% \polyline(0.32,0)(0.1,0)(0.1,1)(0.32,1) -% \end{picture}% -% }% -% }% -% } - -% \newcommand{\rbparen}{% -% \mathclose{% -% \sbox0{$()$}% -% \setlength{\unitlength}{\dimexpr\ht0+\dp0}% -% \raisebox{-\dp0}{% -% \begin{picture}(.32,1) -% \linethickness{\fontdimen8\textfont3} -% \roundcap -% \put(0.02,0){\raisebox{\depth}{$)$}} -% \polyline(0.1,0)(0.32,0)(0.32,1)(0.1,1) -% \end{picture}% -% }% -% }% -% } - -\newcommand{\lbparen}{〖} % https://unicodeplus.com/U+3016 - -\newcommand{\rbparen}{〗} +\newcommand*{\lbparen}{〖} +\newcommand*{\rbparen}{〗} % category C \newcommand*{\C}{\ensuremath{\mathscr{C}}} @@ -160,10 +127,15 @@ \newcommand*{\strongpreelgot}[1]{\ensuremath{\mathit{StrongPreElgot}(#1)}} \newcommand*{\setoids}{\ensuremath{\mathit{Setoids}}} % free objects -\newcommand*{\freee}[1]{\ensuremath{#1^\bigstar}} +\newcommand*{\freee}[1]{\ensuremath{#1^\star}} \newcommand*{\free}[1]{ \ensuremath{ - \IfSubStr{#1}{\circ}{{\freee{(#1)}}}{\freee{#1}} + \IfSubStr{#1}{\circ} + {{\freee{(#1)}}} + {\IfSubStr{#1}{\;} + {\freee{(#1)}} + {\freee{#1}} + } } } % right stability diff --git a/thesis/src/04_iteration.tex b/thesis/src/04_iteration.tex index 7a00d81..57fef40 100644 --- a/thesis/src/04_iteration.tex +++ b/thesis/src/04_iteration.tex @@ -305,7 +305,7 @@ A symmetrical variant of the previous definition is sometimes useful. \\=\;&\ls{f \circ swap} \circ (h^{\#_y} \times id) \circ swap \\=\;&((\ls{f \circ swap} + id) \circ dstr \circ (id \times h))^{\#_a} \circ swap \\=\;&(((\ls{f \circ swap} \circ swap) + id) \circ dstl \circ (id \times h))^{\#_a}\tag{\ref{law:uniformity}} - \\=\;&(((\rs{f} + id) \circ dstl \circ (id \times h))^{\#_a} + \\=\;&(((\rs{f} + id) \circ dstl \circ (id \times h))^{\#_a}, \end{alignat*} for any $Z \in \obj{\C}, h : Z \rightarrow KY + Z$. @@ -318,7 +318,7 @@ A symmetrical variant of the previous definition is sometimes useful. \\=\;&(\ls{f \circ swap} + id) \circ dstr \circ (h \times id) \circ swap. \end{alignat*} - The diagram commutes, since + The requisite diagram commutes, since \begin{alignat*}{1} &\rs{f} \circ (id \times \eta) \\=\;&\ls{f \circ swap} \circ swap \circ (id \times \eta) @@ -342,19 +342,20 @@ A symmetrical variant of the previous definition is sometimes useful. \\=\;&((g + id) \circ dstl \circ (id \times h))^{\#_a} \circ swap \\=\;&(((g \circ swap) + id) \circ dstr \circ (h \times id))^{\#_a},\tag{\ref{law:uniformity}} \end{alignat*} + for any $Z \in \obj{\C}, h : Z \rightarrow KY + Z$. - where the application of \ref{law:uniformity} is justified by + The application of \ref{law:uniformity} is justified by \begin{alignat*}{1} &(id + swap) \circ ((g \circ swap) + id) \circ dstr \circ (h \times id) \\=\;&(g + id) \circ (swap + swap) \circ dstr \circ (h \times id) \\=\;&(g + id) \circ dstl \circ swap \circ (h \times id) - \\=\;&(g + id) \circ dstl \circ (id \times h) \circ swap + \\=\;&(g + id) \circ dstl \circ (id \times h) \circ swap. \end{alignat*} - This concludes one direction of the proof that left stability and right stability imply each other, the other direction follows by a symmetrical proof. + This concludes one direction of the proof that left stability and right stability imply each other, the other direction follows symmetrically. \end{proof} -The following theorem illustrates that stability of $KX$ indeed simulates the behavior of $KX$ in a cartesian closed category. +In a less general setting stability indeed follows automatically. \todo[inline]{Fix proof, replace free object operator, fix references.} @@ -362,31 +363,70 @@ The following theorem illustrates that stability of $KX$ indeed simulates the be In a cartesian closed category every free Elgot algebra is stable. \end{theorem} \begin{proof} - Let $X \in \obj{\C}$ and $((KX, (-)^\#), \llbracket - \rrbracket)$ be a free Elgot algebra on $X$. + Let $\C$ be cartesian closed and let $(KX, (-)^\#, \free{(-)})$ be a free Elgot algebra on some $X \in \obj{\C}$. - We will show that $KX$ is left-stable, i.e. given $X \in \obj{\C}, A \in \vert\elgotalgs{\C}\vert$ and $f : Y \times X \rightarrow A$ we define - $\ls{f} := eval \circ (\llbracket curry\;f \rrbracket \times id)$. - Note that we are using the universal property of a free object over $\textit{Elgot}(\C)$ which when spelled out requires us to show that $A^X$ is an Elgot algebra, for that we reference lemma~\ref{lem:elgotexp}. - \ref{sharpl1} and \ref{sharpl2} then follow by properties of the exponential and of distributive categories, uniqueness is more interesting: - Let $g : KY \times X \rightarrow A$ be a morphism satisfying \ref{sharpl1} and \ref{sharpl2}, we need to show that $g = eval \circ (\llbracket curry\;f \rrbracket \times id)$. We use that fact that $curry$ is an injective mapping, i.e. it suffices to show that: - \[curry\; g = \llbracket curry\;f \rrbracket = curry (eval \circ (\llbracket curry\;f \rrbracket \times id))\] - Where the second step holds for any exponential and the first step is proven by the universal property of free objects, i.e. we need to show that $curry\;g \circ \eta = curry\;f$ which follows by \ref{sharpl1}, and we need to check that $curry\;g$ is left iteration preserving which follows from \ref{sharpl2}. + To show left stability of $KX$ we define $\ls{f} := eval \circ (\free{curry\;f} \times id)$ for any $X \in \obj{\C}, A \in \vert\elgotalgs{\C}\vert$, and $f : Y \times X \rightarrow A$. + We will now verify that this does indeed satisfy the requisite properties, i.e. + \begin{alignat*}{1} + &eval \circ (\free{curry\;f} \times id) \circ (\eta \times id) + \\=\;&eval \circ ((\free{curry\;f} \circ \eta) \times id) + \\=\;&eval \circ (curry\;f) \times id) + \\=\;&f + \end{alignat*} + and for any $Z \in \obj{\C}, h : Z \rightarrow KY + Z$: + \begin{alignat*}{1} + &eval \circ (\free{curry\;f} \times id) \circ (h^{\#_a} \times id) + \\=\;&eval \circ ((\free{curry\;f} \circ h^{\#_a}) \times id) + \\=\;&eval \circ (curry(((eval + id) \circ dstr \circ (((\free{curry\;f} + id) \circ h) \times id))^{\#_b})) \times id) + \\=\;&((eval + id) \circ dstr \circ (((\free{curry\;f} + id) \circ h) \times id))^{\#_b} + \\=\;&((eval + id) \circ dstr \circ ((\free{curry\;f} + id) \times id) \circ (h \times id))^{\#_b} + \\=\;&((eval + id) \circ ((\free{curry\;f} \times id) + id) \circ dstr \circ (h \times id))^{\#_b} + \\=\;&(((eval \circ (\free{curry\;f} \times id)) + id) \circ dstr \circ (h \times id))^{\#_b}. + \end{alignat*} + + \unsure[inline]{Maybe prove that curry is injective in preliminaries.} + + Lastly, we need to check uniqueness of $\ls{f}$. Let us consider a left iteration preserving morphism $g : KY \times X \rightarrow A$ that satisfies $g \circ (\eta \times id) = f$. Since $curry$ is an injective mapping it suffices to show that + \begin{alignat*}{1} + &curry\;\ls{f} + \\=\;&curry(eval \circ (\free{curry\;f} \times id)) + \\=\;&\free{curry\;f} + \\=\;&curry\;g. + \end{alignat*} + Where the last step is the only non-trivial one. Since $\free{curry\;f}$ is a unique iteration preserving morphism satisfying $\free{curry\;f} \circ \eta = curry\;f$, we are left to show that $g$ is also iteration preserving and satisfies the same property. + + \unsure[inline]{Maybe prove helper lemmas about exponentials in preliminaries, like subst, beta and eta.} + + Indeed, + \begin{alignat*}{1} + &curry\;g \circ h^{\#} + \\=\;&curry\;(g \circ (h^{\#} \times id)) + \\=\;&curry\;(((g + id) \circ dstr \circ (h \times id))^{\#_a}) + \\=\;&curry\;((((eval \circ (curry\; g \times id)) + id) \circ dstr \circ (h \times id))^{\#_a}) + \\=\;&curry\;(((eval + id) \circ ((curry\; g \times id) + id) \circ dstr \circ (h \times id))^{\#_a}) + \\=\;&curry\;(((eval + id) \circ dstr \circ ((curry\;g + id) \times id) \circ (h \times id))^{\#_a}) + \\=\;&curry\;(((eval + id) \circ dstr \circ (((curry\;g + id) \circ h) \times id))^{\#_a}) + \end{alignat*} + for any $Z \in \obj{\C}, h : Z \rightarrow KY + Z$, and + \begin{alignat*}{1} + &curry\;g \circ \eta + \\=\;&curry(g \circ (\eta \times id)) + \\=\;&curry\;f + \end{alignat*} + + This completes the proof. \end{proof} -For the rest of this chapter we will assume every $KX$ to exist and be stable to show that it is an equational lifting monad and in fact the initial strong pre-Elgot monad. +For the rest of this chapter we will assume every $KX$ to exist and be stable. Under these assumptions we show that $\mathbf{K}$ is an equational lifting monad and in fact the initial strong pre-Elgot monad. -Before proving strength, we will introduce a proof principle similar to remark~\ref{rem:coinduction}. - -\begin{remark}[Proof by stability]~\label{rem:proofbystability} - Given two morphisms $g, h : X \times KY \rightarrow A$ where $X, Y \in \obj{\C}, A \in \obj{\elgotalgs{\C}}$ to show that $g = h$ it suffices to find a morphism $f : X \times Y \rightarrow A$ such that $g$ and $h$ satisfy \ref{sharpr1} and \ref{sharpr2}. +Let us first introduce a proof principle similar to \autoref{rem:coinduction}. +\begin{remark}[Proof by right-stability]~\label{rem:proofbystability} + Given two morphisms $g, h : X \times KY \rightarrow A$ where $X, Y \in \obj{\C}, A \in \obj{\elgotalgs{\C}}$ to show that $g = h$ it suffices to show that $g$ and $h$ are right iteration preserving and there exists a morphism $f : X \times Y \rightarrow A$ such that $g \circ (id \times \eta) = f$ and $h \circ (id \times \eta) = f$. \end{remark} - -Of course there is also a symmetric version of this: - +Of course there is also a symmetric version of this. \begin{remark}[Proof by left-stability]~\label{rem:proofbyleftstability} - Given two morphisms $g, h : KY \times X \rightarrow A$ where $X, Y \in \obj{\C}, A \in \obj{\elgotalgs{\C}}$ to show that $g = h$ it suffices to find a morphism $f : Y \times X \rightarrow A$ such that $g$ and $h$ satisfy \ref{sharpl1} and \ref{sharpl2}. + Given two morphisms $g, h : X \times KY \rightarrow A$ where $X, Y \in \obj{\C}, A \in \obj{\elgotalgs{\C}}$ to show that $g = h$ it suffices to show that $g$ and $h$ are left iteration preserving and there exists a morphism $f : Y \times X \rightarrow A$ such that $g \circ (\eta \times id) = f$ and $h \circ (\eta \times id) = f$. \end{remark} - \todo[inline]{Maybe do a helper proposition first that characterizes tau with the three laws in the file} \begin{theorem} @@ -397,7 +437,7 @@ Of course there is also a symmetric version of this: \change[inline]{Maybe put complete proof} - To check naturality and the strength laws we will use remark~\ref{rem:proofbystability} and for brevity only state the needed unifying morphism by pasting \ref{sharpl1} into the required diagram. The proofs of \ref{sharpr1} and \ref{sharpr2} can then be looked up in the formalization. + To check naturality and the strength laws we will use \autoref{rem:proofbystability} and for brevity only state the needed unifying morphism by pasting \ref{sharpl1} into the required diagram. The proofs of \ref{sharpr1} and \ref{sharpr2} can then be looked up in the formalization. For naturality of $\tau$ we use: % https://q.uiver.app/#q=WzAsNSxbMSwwLCJBIFxcdGltZXMgS0IiXSxbMCwzLCJBIFxcdGltZXMgQiJdLFsxLDIsIksoQSBcXHRpbWVzIEIpIl0sWzMsMiwiSyhYIFxcdGltZXMgWSkiXSxbMywwLCJYIFxcdGltZXMgS1kiXSxbMSwwLCJpZCBcXHRpbWVzIFxcZXRhIiwwLHsiY3VydmUiOi0yfV0sWzAsMiwiXFx0YXUiLDJdLFsyLDMsIksoZlxcdGltZXMgZykiLDJdLFswLDQsImYgXFx0aW1lcyBLZyJdLFs0LDMsIlxcdGF1Il0sWzEsMywiXFxldGEgXFxjaXJjIChmIFxcdGltZXMgZykiLDIseyJjdXJ2ZSI6Mn1dXQ== @@ -507,7 +547,38 @@ Of course there is also a symmetric version of this: \end{theorem} \begin{proof} - \todo[inline]{Proof that K is equational lifting} + We need to show that $\tau \circ \Delta = K \langle \eta , id \rangle$. Note that $K \langle \eta , id \rangle = \free{\eta \circ \langle \eta , id \rangle}$, which is a unique Elgot algebra morphism satisfying $K \langle \eta , id \rangle \circ \eta = \eta \circ \langle \eta , id \rangle$. It thus suffices to show that $\tau \circ \Delta$ satisfies the same identity and is iteration preserving. + + The identity follows easily + + \todo[inline]{Reference $\tau-\eta$ in last step:} + + \begin{alignat*}{1} + &\tau \circ \Delta \circ \eta + \\=\;&\tau \circ \langle \eta , \eta \rangle + \\=\;&\tau \circ (id \times \eta) \circ \langle \eta , id \rangle + \\=\;&\eta \circ \langle \eta , id \rangle. + \end{alignat*} + + For iteration preservation of $\tau \circ \Delta$ consider $Z \in \obj{\C}$ and $h : Z \rightarrow KX + Z$, then + \begin{alignat*}{1} + &\tau \circ \Delta \circ h^{\#} + \\=\;&\tau \circ \langle h^{\#} , h^{\#} \rangle + \\=\;&\tau \circ (id \times h^{\#}) \circ \langle h^{\#} , id \rangle + \\=\;&((\tau + id) \circ dstl \circ (id \times f))^\# \circ \langle h^{\#} , id \rangle + \\=\;&(((\tau \circ \Delta) + id) \circ f)^\#.\tag{\ref{law:uniformity}} + \end{alignat*} + + Note that by monicity of $dstl^{-1}$ and by \ref{law:fixpoint} + \[(\Delta + \langle f^\# , id \rangle) \circ f = dstl \circ \langle f^\# , f \rangle. \] + The application of \ref{law:uniformity} is then justified by + \begin{alignat*}{1} + &(id + \langle f^\# , id \rangle) \circ ((\tau \circ \Delta) + id) \circ f + \\=\;&((\tau \circ \Delta) + \langle f^\# , id \rangle) \circ f + \\=\;&(\tau + id) \circ (\Delta + \langle f^\# , id \rangle) \circ f + \\=\;&(\tau + id) \circ dstl \circ \langle f^\# , f \rangle + \\=\;&(\tau + id) \circ dstl \circ (id \times f) \circ \langle f^\# , id \rangle. + \end{alignat*} \end{proof} \begin{theorem}