Work on proofs

thesis/.vscode/ltex.hiddenFalsePositives.en-US.txt

@@ -7,3 +7,12 @@
 {"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QWe call a morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q guarded if there exists an \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q such that the following diagram commutes: &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
 {"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QIt suffices to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, because then follows: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We prove this by coinduction using: [inline]Change name of morphism &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
 {"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Q[inline]Make this more explicit The first step in both equations can be proven by monicity of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and then using \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and the dual diagram for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q the last step holds generally for any \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QFirst we need to show naturality of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, i.e. we need to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q The needed coalgebra is shown in this diagram: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Next we check the strength laws: [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] To show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q we do coinduction using the following coalgebra: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] We don't need coinduction to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, but we need a small helper lemma: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QFirst we need to show naturality of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, i.e. we need to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q The coalgebra for coinduction is: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We write \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q to abbreviate the rather long coalgebra, i.e. in this diagram \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Q&&&&&&& &&&&&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q then follows from this diagram.\\E$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QSince \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q is a final coalgebra we get the following proof principle: Given two morphisms \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q it suffices to show that there exists a coalgebra structure \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q such that the following diagrams commute: &&&&&& &&&&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Uniqueness of the coalgebra morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q then gives us that indeed \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QFirst we need to show naturality of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, i.e. we need to show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q The coalgebra for coinduction is: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We write \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q to abbreviate the used coalgebra, i.e. in this diagram \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QNext we check the strength laws: [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] To show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q we do coinduction using the following coalgebra: &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q [\\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q] We don't need coinduction to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, but we need a small helper lemma: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QIt suffices to show \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q, because then follows: \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q We prove this by coinduction using: &&& &&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\Q[inline]Move this remark to the beginning of proof The first step in both equations can be proven by monicity of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and then using \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q and the dual diagram for \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q which is a direct consequence of \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q: &&&& &&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+$"}
+{"rule":"ENGLISH_WORD_REPEAT_RULE","sentence":"^\\QTo show that \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q it suffices to show that there exists a coalgebra structure \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q such that the following diagrams commute: &&&&&& &&&&&& \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q Uniqueness of the coalgebra morphism \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q then results in \\E(?:Dummy|Ina|Jimmy-)[0-9]+\\Q.\\E$"}

thesis/main.pyg

@@ -0,0 +1,7 @@
+mutual
+data Delay (A : Set) : Set where
+  now   : A        → Delay A
+  later : Delay′ A → Delay A
+record Delay′ (A : Set) : Set where
+  coinductive
+  field force : Delay A

thesis/src/01_preliminaries.tex

@@ -381,7 +381,7 @@ For programmers a second equivalent definition is more useful:
   A Kleisli triple on a category $\C$ is a triple $(F, \eta, (-)^*)$, where $F : \obj{C} \rightarrow \obj{C}$ is a mapping on objects, $(\eta_X : X \rightarrow FX)_{X\in\obj{C}}$ is a family of morphisms and for every morphism $f : X \rightarrow FY$ there exists a morphism $f^* : FX \rightarrow FY$ called the Kleisli lifting, where the following laws hold:
   \begin{alignat*}{2}
     &\eta_X^* &&= id_{FX} \tag*{(K1)}\label{K1}\\
-    &\eta_X \circ f^* &&= f \tag*{(K2)}\label{K2}\\
+    &f^* \circ \eta_X &&= f \tag*{(K2)}\label{K2}\\
     &f^* \circ g* &&= (f \circ g^*)^* \tag*{(K3)}\label{K3}
   \end{alignat*}
 \end{definition}

thesis/src/03_partiality-monads.tex

@@ -180,25 +180,26 @@ Since $DX$ is defined as a final coalgebra, we can define morphisms via corecurs
   We will make use of the fact that every $DX$ is a final coalgebra:
   \begin{itemize}
     \item[\ref{D1}] This follows by definition of $now$.
-    \item[\ref{D2}] We define $f^* = \;! \circ i_1$, where $!$ is the unique coalgebra morphism in this diagram:
+    \item[\ref{D2}] We define $f^* = \;\coalg{\alpha} \circ i_1$, where $\coalg{\alpha}$ is the unique coalgebra morphism in this diagram:
-    \todo[inline]{Use other name than '!' for unique morphism => semantic brackets!}
 
-      % https://q.uiver.app/#q=WzAsNSxbMCwxLCJEWCArIERZIl0sWzcsMSwiWSArIChEWCArIERZKSJdLFswLDAsIkRYIl0sWzAsMiwiRFkiXSxbNywyLCJZICsgRFkiXSxbMCwxLCJbIFsgWyBpXzEgLCBpXzIgXFxjaXJjIGlfMiBdIFxcY2lyYyAob3V0IFxcY2lyYyBmKSAsIGlfMiBcXGNpcmMgaV8xIF0gXFxjaXJjIG91dCAsIChpZCArIGlfMikgXFxjaXJjIG91dCBdIl0sWzIsMCwiaV8xIl0sWzAsMywiISIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDQsIm91dCJdLFsxLDQsImlkICsgISIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==
+          % https://q.uiver.app/#q=WzAsNSxbMCwxLCJEWCArIERZIl0sWzcsMSwiWSArIChEWCArIERZKSJdLFswLDAsIkRYIl0sWzAsMiwiRFkiXSxbNywyLCJZICsgRFkiXSxbMCwxLCJcXGFscGhhIDo9IFsgWyBbIGlfMSAsIGlfMiBcXGNpcmMgaV8yIF0gXFxjaXJjIChvdXQgXFxjaXJjIGYpICwgaV8yIFxcY2lyYyBpXzEgXSBcXGNpcmMgb3V0ICwgKGlkICsgaV8yKSBcXGNpcmMgb3V0IF0iXSxbMiwwLCJpXzEiXSxbMCwzLCJcXGNvYWxne1xcYWxwaGF9IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzMsNCwib3V0Il0sWzEsNCwiaWQgKyBcXGNvYWxne1xcYWxwaGF9IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d
-      \[\begin{tikzcd}
+          \[\begin{tikzcd}[ampersand replacement=\&]
               DX \\
-        {DX + DY} &&&&&&& {Y + (DX + DY)} \\
+              {DX + DY} \&\&\&\&\&\&\& {Y + (DX + DY)} \\
-        DY &&&&&&& {Y + DY}
+              DY \&\&\&\&\&\&\& {Y + DY}
-        \arrow["{[ [ [ i_1 , i_2 \circ i_2 ] \circ (out \circ f) , i_2 \circ i_1 ] \circ out , (id + i_2) \circ out ]}", from=2-1, to=2-8]
+              \arrow["{\alpha := [ [ [ i_1 , i_2 \circ i_2 ] \circ (out \circ f) , i_2 \circ i_1 ] \circ out , (id + i_2) \circ out ]}", from=2-1, to=2-8]
               \arrow["{i_1}", from=1-1, to=2-1]
-        \arrow["{!}", dashed, from=2-1, to=3-1]
+              \arrow["{\coalg{\alpha}}", dashed, from=2-1, to=3-1]
               \arrow["out", from=3-1, to=3-8]
-        \arrow["{id + !}", dashed, from=2-8, to=3-8]
+              \arrow["{id + \coalg{\alpha}}", dashed, from=2-8, to=3-8]
             \end{tikzcd}\]
           $out \circ f^* = [ out \circ f , i_2 \circ f^* ] \circ out$ then follows from this diagram.
 
+          \improvement{Details of why it is coalgebra morphism}
+
           We are left to check uniqueness. Let $g : DX \rightarrow DY$ and $out \circ g = [ out \circ f , i_2 \circ g ] \circ out$.
-      Then $[ g , id ] : DX + DY \rightarrow DY$\improvement{details of why it is coalgebra morphism} is a coalgebra morphism and thus $[ g , id ] =\;!$ by uniqueness of $!$. 
+          Then $[ g , id ] : DX + DY \rightarrow DY$ is a coalgebra morphism and thus $[ g , id ] =\;\coalg{\alpha}$ by uniqueness of $\coalg{\alpha}$.
-      It follows: \[g = [ g , id ] \circ i_1 =\;! \circ i_1 = f^*\]
+          It follows: \[g = [ g , id ] \circ i_1 =\;\coalg{\alpha} \circ i_1 = f^*\]
     \item[\ref{D3}] This follows immediately since $\tau$ is the unique coalgebra morphism from $(X \times DY, dstl \circ (id \times out))$ into the terminal coalgebra $(D(X \times Y) , out)$.
   \end{itemize}
 \end{proof}
@@ -211,23 +212,32 @@ Since $DX$ is defined as a final coalgebra, we can define morphisms via corecurs
   \begin{itemize}
     \item[\ref{K1}]
           By uniqueness of $now^*$ it suffices to show that $out \circ id = [ out \circ now , i_2 \circ id ] \circ out$ which instantly follows by \ref{D1}.
-    \item[\ref{K2}] Let $f : X \rightarrow DY$. We use the fact that $out$ is monic and the following equation:
+    \item[\ref{K2}] Let $f : X \rightarrow DY$. We proceed by monicity of $out$:
-    
+    \begin{alignat*}{1}
-    \[out \circ f^* \circ now \overset{\ref{D2}}{=} [ out \circ f , i_2 \circ f^* ] \circ out \circ now \overset{\ref{D1}}{=} out \circ f \]
+      &out \circ f^* \circ now
-    
+      \\=\;&[ out \circ f , i_2 \circ f^* ] \circ out \circ now\tag*{\ref{D2}}
-    \item[\ref{K3}] Using uniqueness of $(h^* \circ g)^*$ we need to show $out \circ h^* \circ g^* = [ out \circ h^* \circ g , i_2 \circ h^* \circ g^* ] \circ out$ which follows by multiple uses of \ref{D2}.
+      \\=\;&[ out \circ f , i_2 \circ f^* ] \circ i_1\tag*{\ref{D1}}
-    \todo[inline]{More details probably for both K2 and K3}
+      \\=\;&out \circ f
+    \end{alignat*}
+    \item[\ref{K3}] Let $f : Y \rightarrow Z, g : X \rightarrow Z$ to show $f^* \circ g^* = (f^* \circ g)^*$ by uniqueness of $(f^* \circ g)^*$ it suffices to show:
+    \begin{alignat*}{1}
+      &out \circ f^* \circ g^*
+      \\=\;&[ out \circ f , i_2 \circ f^* ] \circ out \circ g^*\tag*{\ref{D2}} 
+      \\=\;&[ out \circ f , i_2 \circ f^* ] \circ [ out \circ g , i_2 \circ g^* ] \circ out\tag*{\ref{D2}} 
+      \\=\;&[ [ out \circ f , i_2 \circ f^* ] \circ out \circ g , i_2 \circ f^* \circ g^* ] \circ out
+      \\=\;&[ out \circ f^* \circ g , i_2 \circ f^* \circ g^* ] \circ out\tag*{\ref{D2}}
+    \end{alignat*}
   \end{itemize}
 \end{proof}
 
-Since $(DX, out)$ is a final coalgebra we get the following proof principle:
+Finality of the coalgebras $(DX, out : DX \rightarrow X + DX)_{X \in \obj{\C}}$ yields the following proof principle:
 \begin{remark}[Proof by coinduction]
   \label{rem:coinduction}
-  Given two morphisms $f, g : X \rightarrow DY$, to show that $f = g$ it suffices to show that there exists a coalgebra structure $\alpha : X \rightarrow Y + X$ such that the following diagrams commute:
+  Given two morphisms $f, g : X \rightarrow DY$. To show that $f = g$ it suffices to show that there exists a coalgebra structure $\alpha : X \rightarrow Y + X$ such that the following diagrams commute:
-  % https://q.uiver.app/#q=WzAsOCxbMCwwLCJYIl0sWzAsMSwiRFkiXSxbMiwxLCJZICsgRFkiXSxbMiwwLCJZICsgWCJdLFs0LDAsIlgiXSxbNCwxLCJEWSJdLFs2LDAsIlkgKyBYIl0sWzYsMSwiWSArIERZIl0sWzEsMiwib3V0Il0sWzAsMywiXFxhbHBoYSJdLFswLDEsImYiXSxbMywyLCJpZCArIGYiXSxbNCw2LCJcXGFscGhhIl0sWzQsNSwiZyJdLFs2LDcsImlkICsgZyJdLFs1LDcsIm91dCIsMl1d
+  % https://q.uiver.app/#q=WzAsOCxbMCwwLCJYIl0sWzAsMSwiRFkiXSxbMiwxLCJZICsgRFkiXSxbMiwwLCJZICsgWCJdLFs0LDAsIlgiXSxbNCwxLCJEWSJdLFs2LDAsIlkgKyBYIl0sWzYsMSwiWSArIERZIl0sWzEsMiwib3V0Il0sWzAsMywiXFxhbHBoYSJdLFswLDEsImYiXSxbMywyLCJpZCArIGYiXSxbNCw2LCJcXGFscGhhIl0sWzQsNSwiZyJdLFs2LDcsImlkICsgZyJdLFs1LDcsIm91dCJdXQ==
-  \[\begin{tikzcd}
+  \[\begin{tikzcd}[ampersand replacement=\&]
-    X && {Y + X} && X && {Y + X} \\
+      X \&\& {Y + X} \&\& X \&\& {Y + X} \\
-    DY && {Y + DY} && DY && {Y + DY}
+      DY \&\& {Y + DY} \&\& DY \&\& {Y + DY}
       \arrow["out", from=2-1, to=2-3]
       \arrow["\alpha", from=1-1, to=1-3]
       \arrow["f", from=1-1, to=2-1]
@@ -235,9 +245,9 @@ Since $(DX, out)$ is a final coalgebra we get the following proof principle:
       \arrow["\alpha", from=1-5, to=1-7]
       \arrow["g", from=1-5, to=2-5]
       \arrow["{id + g}", from=1-7, to=2-7]
-    \arrow["out"', from=2-5, to=2-7]
+      \arrow["out", from=2-5, to=2-7]
     \end{tikzcd}\]
-  Uniqueness of the coalgebra morphism $! : (X, \alpha) \rightarrow (DY, out)$ then gives us that indeed $f = g$.
+  Uniqueness of the coalgebra morphism $\coalg{\alpha} : (X, \alpha) \rightarrow (DY, out)$ then results in $f = g$.
 \end{remark}
 
 \begin{theorem}
@@ -248,29 +258,31 @@ Since $(DX, out)$ is a final coalgebra we get the following proof principle:
 
   First we need to show naturality of $\tau$, i.e. we need to show that
   \[\tau \circ (f \times (now \circ g)^*) = (now \circ (f \times g))^* \circ \tau\]
-  The needed coalgebra is shown in this diagram:
+  The coalgebra for coinduction is:
-  % https://q.uiver.app/#q=WzAsNixbMCwwLCJYIFxcdGltZXMgRFkiXSxbMCwyLCJEKEFcXHRpbWVzIEIpIl0sWzQsMiwiQVxcdGltZXMgQiArIEQoQSBcXHRpbWVzIEIpIl0sWzEsMCwiWCBcXHRpbWVzIChZICsgRFkpIl0sWzIsMCwiWCBcXHRpbWVzIFkgKyBYIFxcdGltZXMgRFkiXSxbNCwwLCJBIFxcdGltZXMgQiArIFggXFx0aW1lcyBEWSJdLFsxLDIsIm91dCJdLFswLDMsImlkIFxcdGltZXMgb3V0Il0sWzMsNCwiZHN0bCJdLFswLDEsIiEiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNCw1LCJmIFxcdGltZXMgZyArIGlkIl0sWzUsMiwiaWQgKyAhIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d
+  % https://q.uiver.app/#q=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
-  \[\begin{tikzcd}
+  \[\begin{tikzcd}[ampersand replacement=\&]
-    {X \times DY} & {X \times (Y + DY)} & {X \times Y + X \times DY} && {A \times B + X \times DY} \\
+      {X \times DY} \& {X \times (Y + DY)} \& {X \times Y + X \times DY} \&\& {A \times B + X \times DY} \\
       \\
-    {D(A\times B)} &&&& {A\times B + D(A \times B)}
+      {D(A\times B)} \&\&\&\& {A\times B + D(A \times B)}
       \arrow["out", from=3-1, to=3-5]
       \arrow["{id \times out}", from=1-1, to=1-2]
       \arrow["dstl", from=1-2, to=1-3]
-    \arrow["{!}", dashed, from=1-1, to=3-1]
+      \arrow["{\coalg{-}}", dashed, from=1-1, to=3-1]
       \arrow["{f \times g + id}", from=1-3, to=1-5]
-    \arrow["{id + !}", dashed, from=1-5, to=3-5]
+      \arrow["{id + \coalg{-}}", dashed, from=1-5, to=3-5]
     \end{tikzcd}\]
+  We write $\coalg{-}$ to abbreviate the used coalgebra, i.e. in this diagram $\coalg{-} = \coalg{(f\times g + id) \circ dstl \circ (id \times out)}$
+
   Next we check the strength laws:
   \begin{itemize}
     \item[\ref{S1}] To show that $(now \circ \pi_2)^* \circ \tau = \pi_2$ we do coinduction using the following coalgebra:
-      % https://q.uiver.app/#q=WzAsNixbMCwwLCIxIFxcdGltZXMgRFgiXSxbMCwxLCJEWCJdLFszLDAsIlggKyAxIFxcdGltZXMgRFgiXSxbMywxLCJYICsgRFgiXSxbMSwwLCIxIFxcdGltZXMgWCArIERYIl0sWzIsMCwiMSBcXHRpbWVzIFggKyAxIFxcdGltZXMgRFgiXSxbMSwzLCJvdXQiXSxbMCwxLCIhIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzIsMywiaWQgKyAhIl0sWzAsNCwiaWQgXFx0aW1lcyBvdXQiXSxbNCw1LCJkc3RsIl0sWzUsMiwiXFxwaV8yICsgaWQiXV0=
+          % https://q.uiver.app/#q=WzAsNixbMCwwLCIxIFxcdGltZXMgRFgiXSxbMCwxLCJEWCJdLFszLDAsIlggKyAxIFxcdGltZXMgRFgiXSxbMywxLCJYICsgRFgiXSxbMSwwLCIxIFxcdGltZXMgWCArIERYIl0sWzIsMCwiMSBcXHRpbWVzIFggKyAxIFxcdGltZXMgRFgiXSxbMSwzLCJvdXQiXSxbMCwxLCJcXGNvYWxney19IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzIsMywiaWQgKyBcXGNvYWxney19Il0sWzAsNCwiaWQgXFx0aW1lcyBvdXQiXSxbNCw1LCJkc3RsIl0sWzUsMiwiXFxwaV8yICsgaWQiXV0=
-      \[\begin{tikzcd}
+          \[\begin{tikzcd}[ampersand replacement=\&]
-        {1 \times DX} & {1 \times X + DX} & {1 \times X + 1 \times DX} & {X + 1 \times DX} \\
+              {1 \times DX} \& {1 \times X + DX} \& {1 \times X + 1 \times DX} \& {X + 1 \times DX} \\
-        DX &&& {X + DX}
+              DX \&\&\& {X + DX}
               \arrow["out", from=2-1, to=2-4]
-        \arrow["{!}", dashed, from=1-1, to=2-1]
+              \arrow["{\coalg{-}}", dashed, from=1-1, to=2-1]
-        \arrow["{id + !}", from=1-4, to=2-4]
+              \arrow["{id + \coalg{-}}", from=1-4, to=2-4]
               \arrow["{id \times out}", from=1-1, to=1-2]
               \arrow["dstl", from=1-2, to=1-3]
               \arrow["{\pi_2 + id}", from=1-3, to=1-4]
@@ -287,29 +299,29 @@ Since $(DX, out)$ is a final coalgebra we get the following proof principle:
             =\; & out^{-1} \circ (id + \tau) \circ dstl \circ (id \times i_1)\tag*{\ref*{helper}} \\
             =\; & now
           \end{alignat*}
-      where the last step follows since $dstl \circ (id \times i_1) = i_1$.
+          where the last step follows by $dstl \circ (id \times i_1) = i_1$.
     \item[\ref{S3}] We need to check $\tau^* \circ \tau = \tau \circ (id \times id^*)$, the coalgebra for coinduction is:
-      % https://q.uiver.app/#q=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
+          % https://q.uiver.app/#q=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
-      \[\begin{tikzcd}
+          \[\begin{tikzcd}[ampersand replacement=\&]
-        {X \times DDY} & {X \times (DY + DDY)} & {X \times DY + X \times DDY} \\
+              {X \times DDY} \& {X \times (DY + DDY)} \& {X \times DY + X \times DDY} \\
-        && {X \times Y + X \times DDY} \\
+              \&\& {X \times Y + X \times DDY} \\
-        {D(X\times Y)} && {X \times Y + D(X \times Y)}
+              {D(X\times Y)} \&\& {X \times Y + D(X \times Y)}
-        \arrow["{!}", dashed, from=1-1, to=3-1]
+              \arrow["{\coalg{-}}", dashed, from=1-1, to=3-1]
               \arrow["{id \times out}", from=1-1, to=1-2]
               \arrow["dstl", from=1-2, to=1-3]
               \arrow["out", from=3-1, to=3-3]
-        \arrow["{id + \;!}"', from=2-3, to=3-3]
+              \arrow["{id + \coalg{-}}"', from=2-3, to=3-3]
               \arrow["{[ (id + (id \times now)) \circ dstl \circ (id \times out) , i_2 ]}"', from=1-3, to=2-3]
             \end{tikzcd}\]
     \item[\ref{S4}] To show $D\alpha \circ \tau = \tau \circ (id \times \tau) \circ \alpha$ by coinduction we take the coalgebra:
-    % https://q.uiver.app/#q=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
+          % https://q.uiver.app/#q=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
-    \[\begin{tikzcd}
+          \[\begin{tikzcd}[ampersand replacement=\&]
-      {(X \times Y) \times DZ} & {(X \times Y) \times (Z+ DZ)} & {(X\times Y) \times Z + (X \times Y) \times DZ} \\
+              {(X \times Y) \times DZ} \& {(X \times Y) \times (Z+ DZ)} \& {(X\times Y) \times Z + (X \times Y) \times DZ} \\
-      && {X \times Y  \times Z + (X \times Y) \times DZ} \\
+              \&\& {X \times Y  \times Z + (X \times Y) \times DZ} \\
-      {D(X \times Y \times Z)} && {X \times Y \times Z + D(X \times Y \times Z)}
+              {D(X \times Y \times Z)} \&\& {X \times Y \times Z + D(X \times Y \times Z)}
-      \arrow["{!}", dashed, from=1-1, to=3-1]
+              \arrow["{\coalg{-}}", dashed, from=1-1, to=3-1]
               \arrow["out", from=3-1, to=3-3]
-      \arrow["{id +\;!}"', from=2-3, to=3-3]
+              \arrow["{id +\coalg{-}}"', from=2-3, to=3-3]
               \arrow["{id \times out}", from=1-1, to=1-2]
               \arrow["dstl", from=1-2, to=1-3]
               \arrow["{\langle \pi_1 \circ \pi_1 , \langle \pi_2 \circ \pi_1 , \pi_2 \rangle \rangle + id}"', from=1-3, to=2-3]
@@ -317,7 +329,7 @@ Since $(DX, out)$ is a final coalgebra we get the following proof principle:
   \end{itemize}
 \end{proof}
 
-To show that $\mathbf{D}$ is commutative we will use another proof principle previously called the \textit{Solution Theorem}~\cite{sol-thm} or \textit{Parametric Corecursion}~\cite{param-corec}. In our setting this takes the following form:
+To prove that $\mathbf{D}$ is commutative we will use another proof principle previously called the \textit{Solution Theorem}~\cite{sol-thm} or \textit{Parametric Corecursion}~\cite{param-corec}. In our setting this takes the following form:
 
 \begin{definition}
   We call a morphism $g : X \rightarrow D (Y + X)$ \textit{guarded} if there exists an $h : X \rightarrow Y + D(Y+X)$ such that the following diagram commutes:
@@ -342,26 +354,78 @@ To show that $\mathbf{D}$ is commutative we will use another proof principle pre
   It suffices to show $[ now , f ]^* = [ now , i ]^*$, because then follows:
   \[f = [ now , f ]^* \circ g = [ now , i ]^* \circ g = i\]
   We prove this by coinduction using:
-  \change[inline]{Change name of morphism}
+  % https://q.uiver.app/#q=WzAsNSxbMCwxLCJEWSJdLFszLDEsIlkgKyBEWSJdLFswLDAsIkQoWSArIFgpIl0sWzEsMCwiKFkgKyBYKSArIEQoWStYKSJdLFszLDAsIlkgKyBEKFkrWCkiXSxbMCwxLCJvdXQiXSxbMiwzLCJvdXQiXSxbMyw0LCJbIFsgaV8xICwgaCBdICwgaV8yIF0iXSxbNCwxLCJpZCArIFxcY29hbGd7LX0iXSxbMiwwLCJcXGNvYWxney19IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d
-  % https://q.uiver.app/#q=WzAsNSxbMCwxLCJEWSJdLFszLDEsIlkgKyBEWSJdLFswLDAsIkQoWSArIFgpIl0sWzEsMCwiKFkgKyBYKSArIEQoWStYKSJdLFszLDAsIlkgKyBEKFkrWCkiXSxbMCwxLCJvdXQiXSxbMiwzLCJvdXQiXSxbMyw0LCJbIFsgaV8xICwgaCBdICwgaV8yIF0iXSxbNCwxLCJpZCArIFxcOyEiXSxbMiwwLCIhIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d
   \[\begin{tikzcd}[ampersand replacement=\&]
       {D(Y + X)} \& {(Y + X) + D(Y+X)} \&\& {Y + D(Y+X)} \\
       DY \&\&\& {Y + DY}
       \arrow["out", from=2-1, to=2-4]
       \arrow["out", from=1-1, to=1-2]
       \arrow["{[ [ i_1 , h ] , i_2 ]}", from=1-2, to=1-4]
-    \arrow["{id + \;!}", from=1-4, to=2-4]
+      \arrow["{id + \coalg{-}}", from=1-4, to=2-4]
-    \arrow["{!}", dashed, from=1-1, to=2-1]
+      \arrow["{\coalg{-}}", dashed, from=1-1, to=2-1]
     \end{tikzcd}\]
 \end{proof}
 
+Let us record some facts that we will use to prove commutativity of $\mathbf{D}$:
+
+\begin{corollary}
+  The following two properties hold:
+  \begin{alignat*}{1}
+    &out \circ Df = (f + Df) \circ out
+    \\&later \circ f^* = (later \circ f)^* = f^* \circ later
+  \end{alignat*}
+
+  As well as these properties of $\tau$ and $\sigma$:
+  \begin{alignat*}{2}
+     & out \circ \tau &  & = (id + \tau) \circ dstl \circ (id \times out)\tag*{($\tau_1$)}\label{tau1}
+    \\&out \circ \sigma &&= (id + \sigma) \circ dstr \circ (out \times id)\tag*{($\sigma_1$)}\label{sigma1}
+    \\&\tau \circ (id \times out^{-1}) &&= out^{-1} \circ (id + \tau) \circ dstl\tag*{($\tau_2$)}\label{tau2}
+    \\& \sigma \circ (out^{-1} \times id) &&= out^{-1} \circ (id + \sigma) \circ dstr\tag*{($\sigma_2$)}\label{sigma2}
+  \end{alignat*}
+\end{corollary}
+
+\begin{proof}
+  Let us start with the first two properties:
+
+  \begin{alignat*}{1}
+    &out \circ Df
+    \\=\;&out \circ (now \circ f)^*
+    \\=\;&[ out \circ now \circ f , i_2 \circ (now \circ f)^* ] \circ out\tag*{\ref{D2}}
+    \\=\;&(f + Df) \circ out\tag*{\ref{D1}}
+  \end{alignat*}
+
+  \begin{alignat*}{1}
+    &(later \circ f)^*
+    % ... TODO
+    \\=\;&later \circ f^*
+    % ... TODO
+    \\=\;&f^* \circ later
+  \end{alignat*}
+
+  \begin{itemize}
+    \item[\ref{tau1}] This is just \ref{D3} restated.
+    \item[\ref{sigma1}]
+          \todo[inline]{What to do about out-law?}
+          \begin{alignat*}{1}
+             & out \circ Dswap \circ \tau \circ swap
+            \\=\;&(swap + Dswap) \circ out \circ \tau \circ swap
+            \\=\;&(swap + Dswap) \circ (id + \tau) \circ dstl \circ (id \times out) \circ swap \tag*{\ref{tau1}}
+            \\=\;&(swap + Dswap) \circ (id + \tau) \circ dstl \circ swap \circ (out \times id)
+            \\=\;&(swap + Dswap) \circ (id + \tau) \circ (swap + swap) \circ dstr \circ (out \times id)
+            \\=\;&(id + \sigma) \circ dstr \circ (out \times id)
+          \end{alignat*}
+    \item[\ref{tau2}]
+    \item[\ref{sigma2}]
+  \end{itemize}
+\end{proof}
+
 \begin{theorem}
   $\mathbf{D}$ is commutative.
 \end{theorem}
 \begin{proof}
   Using corollary~\ref{cor:solution} it suffices to show that both $\tau^* \circ \sigma$ and $\sigma^* \circ \tau$ are solutions of some guarded morphism $g$.
 
-  Let $w = (dstr + dstr) \circ dstl \circ (out \times out)$ and 
+  Let $w = (dstr + dstr) \circ dstl \circ (out \times out)$ and take
   \[g = out^{-1} \circ [ i_1 + D i_1 \circ \sigma , i_2 \circ [ D i_1 \circ \tau , later \circ now \circ i_2 ] ] \circ w\]
   $g$ is trivially guarded by $[ id + D i_1 \circ \sigma , i_2 \circ [ D i_1 \circ \tau , later \circ now \circ i_2 ] ] \circ w$.
 
@@ -370,19 +434,65 @@ To show that $\mathbf{D}$ is commutative we will use another proof principle pre
   \[\tau^* \circ \sigma = out^{-1} \circ [ id + \sigma , i_2 \circ [ \tau , later \circ \tau^* \circ \sigma ] ] \circ w = [ now , \tau^* \circ \sigma]^* \circ g \]
   \[\sigma^* \circ \tau = out^{-1} \circ [ id + \sigma , i_2 \circ [ \tau , later \circ \sigma^* \circ \tau ] ] \circ w = [ now , \sigma^* \circ \tau]^* \circ g \]
 
-  \change[inline]{Make this more explicit}
+  The last step in both equations holds generally for any $f : DX \times DY \rightarrow D(X \times Y)$ by monicity of $out$:
+  \begin{alignat*}{1}
+        & out \circ [ now , f ]^* \circ out^{-1} \circ [ i_1 + D i_1 \circ \sigma , i_2 \circ [ D i_1 \circ \tau , later \circ now \circ i_2 ] ] \circ w                             \\
+    =\; & [ out \circ [ now , f ] , i_2 \circ [ now , f ]^* ] \circ [ i_1 + D i_1 \circ \sigma , i_2 \circ [ D i_1 \circ \tau , later \circ now \circ i_2 ] ] \circ w\tag*{\ref{D2}} \\
+    =\; & [ id + \sigma , i_2 \circ [ now , f]^* \circ [ D i_1 \circ \tau , later \circ now \circ i_2 ] ] \circ w\tag*{\ref{D1}}                                                     \\
+    =\; & [ id + \sigma , i_2 \circ [ \tau , [ now , f]^* \circ later \circ now \circ i_2 ] ] \circ w                                                                                \\
+    =\; & [ id + \sigma , i_2 \circ [ \tau , [ later \circ now , later \circ f]^* \circ now \circ i_2 ] ] \circ w                                                                    \\
+    =\; & [ id + \sigma , i_2 \circ [ \tau , later \circ f ] ] \circ w
+  \end{alignat*}
+
+  \todo[inline]{The second to last step uses that $f^* \circ later = (later \circ f)^*$ this should be proven somewhere and referenced here.}
+  \todo[inline]{The fact that $f^* \circ Dg = (f \circ g)^*$ should be mentioned somewhere (in preliminaries!)}
+
+  \change[inline]{Move this remark to the beginning of proof}
   The first step in both equations can be proven by monicity of $out$ and then using \ref{D3} and the dual diagram for $\sigma$ which is a direct consequence of \ref{D3}:
-  % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIFxcdGltZXMgRFkiXSxbMCwxLCJEKFggXFx0aW1lcyBZKSJdLFs0LDAsIlggXFx0aW1lcyBZICsgWCBcXHRpbWVzIERZIl0sWzQsMSwiWCBcXHRpbWVzIFkgK0QoWCBcXHRpbWVzIFkpIl0sWzIsMCwiWCBcXHRpbWVzIChZICsgRFkpIl0sWzAsMSwiXFxoYXR7XFx0YXV9IiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzEsMywib3V0Il0sWzIsMywiaWQgKyBcXGhhdHtcXHRhdX0iLDJdLFswLDQsImlkIFxcdGltZXMgb3V0Il0sWzQsMiwiZHN0bCJdXQ==
+  % https://q.uiver.app/#q=WzAsNSxbMCwwLCJYIFxcdGltZXMgRFkiXSxbMCwxLCJEKFggXFx0aW1lcyBZKSJdLFs0LDAsIlggXFx0aW1lcyBZICsgWCBcXHRpbWVzIERZIl0sWzQsMSwiWCBcXHRpbWVzIFkgK0QoWCBcXHRpbWVzIFkpIl0sWzIsMCwiWCBcXHRpbWVzIChZICsgRFkpIl0sWzAsMSwiXFxzaWdtYSJdLFsxLDMsIm91dCJdLFsyLDMsImlkICsgXFxoYXR7XFx0YXV9IiwyXSxbMCw0LCJpZCBcXHRpbWVzIG91dCJdLFs0LDIsImRzdGwiXV0=
   \[\begin{tikzcd}[ampersand replacement=\&]
       {X \times DY} \&\& {X \times (Y + DY)} \&\& {X \times Y + X \times DY} \\
       {D(X \times Y)} \&\&\&\& {X \times Y +D(X \times Y)}
-    \arrow["{\sigma}", dashed, from=1-1, to=2-1]
+      \arrow["\sigma", from=1-1, to=2-1]
       \arrow["out", from=2-1, to=2-5]
-    \arrow["{id + \sigma}"', from=1-5, to=2-5]
+      \arrow["{id + \hat{\tau}}"', from=1-5, to=2-5]
       \arrow["{id \times out}", from=1-1, to=1-3]
       \arrow["dstl", from=1-3, to=1-5]
     \end{tikzcd}\]
-  the last step holds generally for any $f : DX \times DY \rightarrow D(X \times Y)$.
+
+  Let us now check the first step in the equation for $\sigma^* \circ \tau$, the same step for $\tau^* \circ \sigma$ is then symmetric.
+
+  Again we proceed by monicity of $out$:
+  \begin{alignat*}{1}
+     & out \circ \sigma^* \circ \tau
+    \\=\;&[ out \circ \sigma , i_2 \circ \sigma^* ] \circ out \circ \tau\tag*{\ref{D2}}
+    \\=\;&[ out \circ \sigma , i_2 \circ \sigma^* ] \circ (id + \tau) \circ dstl \circ (id \times out)\tag*{\ref{D3}}
+    \\=\;&[ (id + \sigma) \circ dstr \circ (out \times id) , i_2 \circ \sigma^* \circ \tau ] \circ dstl \circ (id \times out)\tag*{(outsigma)}
+    \\=\;&[ (id + \sigma) \circ dstr \circ (out \times id) , i_2 \circ \sigma^* \circ \tau ] \circ ((out^{-1} \times id) + (out^{-1} \times id)) \circ dstl \circ (out \times out)
+    \\=\;&[ (id + \sigma) \circ dstr, i_2 \circ \sigma^* \circ \tau \circ (out^{-1} \times id)] \circ dstl \circ (out \times out)
+    \\=\;&[ (id + \sigma) \circ dstr, i_2 \circ \sigma^* \circ D(out^{-1} \times id) \circ \tau] \circ dstl \circ (out \times out)
+    \\=\;&[ (id + \sigma) \circ dstr, i_2 \circ (\sigma \times (out^{-1} \times id))^* \circ \tau] \circ dstl \circ (out \times out)
+    \\=\;&[ (id + \sigma) \circ dstr, i_2 \circ (out^{-1} \circ (id + \sigma) \circ dstr)^* \circ \tau] \circ dstl \circ (out \times out)\tag*{($\sigma$)}
+    \\=\;&[ (id + \sigma) \circ dstr, i_2 \circ (out^{-1} \circ (id + \sigma))^* \circ Ddstr \circ \tau] \circ dstl \circ (out \times out)
+    \\=\;&[ (id + \sigma) \circ dstr, i_2 \circ (out^{-1} \circ (id + \sigma))^* \circ [D i_1 \circ \tau , D i_2 \circ \tau] \circ dstr] \circ dstl \circ (out \times out)\tag*{\ref{Dcomm-helper}}
+    \\=\;&[ (id + \sigma), i_2 \circ (out^{-1} \circ (id + \sigma))^* \circ [D i_1 \circ \tau , D i_2 \circ \tau]] \circ (dstr + dstr) \circ dstl \circ (out \times out)
+    \\=\;&[ (id + \sigma), i_2 \circ [(out^{-1} \circ i_1)^* \circ \tau , (out^{-1} \circ i_2 \circ \sigma)^* \circ \tau]] \circ w
+    \\=\;&[ (id + \sigma), i_2 \circ [ \tau , (later \circ \sigma)^* \circ \tau]] \circ w \tag*{\ref{K1}}
+    \\=\;&[ (id + \sigma), i_2 \circ [ \tau , later \circ \sigma^* \circ \tau]] \circ w \tag*{(later)}
+  \end{alignat*}
+
+  where
+  \[Ddstr \circ \tau = [ Di_1 \circ \tau , Di_2 \circ \tau ] \circ dstr \tag*{(*)}\label{Dcomm-helper}\]
+  follows by the fact that $dstr^{-1}$ is epic:
+
+  \begin{alignat*}{1}
+     & Ddstr \circ \tau \circ dstr^{-1}
+    \\=\;&[ Ddstr \circ \tau \circ (i_1 \times id) , Ddstr \circ \tau \circ (i_2 \times id) ]
+    \\=\;&[ Ddstr \circ D(i_1 \times id) \circ \tau , Ddstr \circ D(i_2 \times id) \circ \tau ]
+    \\=\;&[ Di_1 \circ \tau , Di_2 \circ \tau ]
+  \end{alignat*}
+  \todo[inline]{Add $dstl-i_1$ etc. to preliminaries as corollary!}
+
 \end{proof}
 
 \improvement[inline]{Elaborate more?}

thesis/src/04_iteration.tex

@@ -111,6 +111,7 @@ In this section we will study the monad that arises from existence of all free E
   If every $X \in \obj{\C}$ extends to a free Elgot algebra $KX$ we get a monad $\mathbf{K}$.
 \end{lemma}
 \begin{proof}
+  \todo[inline]{Maybe this can even be proven without adjunction, show that free objects yield a Kleisli triple!}
   \todo[inline]{Proof in preliminaries}
   Existence of all free objects in $\elgotalgs{\C}$ yields an adjunction between $\C$ and $\elgotalgs{\C}$ that in turn gives us a monad on $\C$.
 \end{proof}