← rota-baxter-winning-games
DiffCatRIMS__AI-generated__DiffCat_redrawn.tex
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%AI suggestion: update affiliation, venue, and date on the title slide if this source is reused for another talk.
\title{A Rota--Baxter Equation for Winning Games}
\subtitle{Games as recursive coalgebras and differential invariants}
\newcommand{\visualsubtitle}{Games as recursive \dterm{coalgebras} and \dterm{differential} invariants}
\author{Ryuya Hora}
\institute{Differentiation in category theory and program semantics}
\date{Kyoto University, April 6, 2026}
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\begin{document}
\begin{frame}[plain]
\titlepage
\vspace{0.18cm}
\begin{center}
\scriptsize
Partially based on joint work in progress with Ryo Suzuki.\\[0.1em]
Keywords: \iterm{Rota--Baxter} equation, combinatorial games, recursive \dterm{coalgebra}, \dterm{Leibniz} rule.
\end{center}
\end{frame}
\section[Front matter]{Front matter}
\begin{frame}{Table of Contents}
\small
\tableofcontents
\end{frame}
%AI suggestion: this opening slide can be shortened further if you want to get to Nim earlier.
\begin{frame}{Where I come from}
\begin{columns}[T,onlytextwidth]
\column{0.55\textwidth}
\begin{itemize}
\item I have been thinking about three languages at once:
\begin{itemize}
\item toposes and semantics,
\item impartial games and recursion,
\item \dterm{coalgebras} and their evaluation maps.
\end{itemize}
\item This talk is about a place where these languages accidentally agree.
\item The slogan is simple: recursion in games already wants both a \dterm{differential} and an \iterm{integral} story.
\end{itemize}
\column{0.45\textwidth}
\begin{figure}
\centering
\begin{tikzpicture}[x=1cm,y=1cm]
\node[wordnode,text=dcol] (topos) at (0,1.8) {toposes};
\node[wordnode] (games) at (-1.8,0) {games};
\node[wordnode,text=dcol] (coal) at (1.8,0) {\dterm{coalgebras}};
\draw[thinarr,dashed] (topos) -- (games);
\draw[thinarr] (topos) -- (coal);
\draw[thinarr] (games) -- (coal);
\node[align=center,font=\small] at (0,-1.2) {I tried to check the spelling\\of the bridge I want to build.};
\end{tikzpicture}
\end{figure}
\end{columns}
\end{frame}
%AI suggestion: if the audience is already familiar with the project, this can be merged with the next slide.
\begin{frame}{Where this talk comes from}
\small
\begin{itemize}
\item In program semantics, recursive objects are often studied by moving between unfolding \dterm{coalgebras} and folding \iterm{algebras}.
\item In impartial games, the same pattern appears when we define values recursively from immediate options.
\item The familiar \iterm{mex} construction already behaves like an \iterm{integration} operator.
\item The option operator behaves like a \dterm{derivative}: it looks at one-step changes of position.
\end{itemize}
\vspace{0.3cm}
\begin{center}
\safeimage[width=0.95\linewidth,height=0.21\textheight,keepaspectratio]{images/Recursion.png}
\end{center}
\end{frame}
\begin{frame}{The dichotomy in this talk}
\begin{figure}
\centering
\begin{tikzpicture}[node distance=2.0cm and 2.6cm]
\node[dichotomy] (play) at (-3.7,1.5) {play forward};
\node[dichotomy] (analyze) at (3.7,1.5) {analyze backward};
\node[dichotomy,text=dcol] (coalg) at (-3.7,0) {\dterm{coalgebras}};
\node[dichotomy,text=icol] (alg) at (3.7,0) {\iterm{algebras}};
\node[dichotomy,text=dcol] (diff) at (-3.7,-1.5) {\dterm{differentiation}};
\node[dichotomy,text=icol] (intg) at (3.7,-1.5) {\iterm{integration}};
\draw[thinarr] (play) -- node[above,font=\small]{duality of viewpoint} (analyze);
\draw[thinarr] (coalg) -- node[above,font=\small]{fold / unfold} (alg);
\draw[thinarr] (diff) -- node[above,font=\small]{shadow of the same recursion} (intg);
\draw[thinarr,dashed] (play) -- (coalg);
\draw[thinarr,dashed] (coalg) -- (diff);
\draw[thinarr,dashed] (analyze) -- (alg);
\draw[thinarr,dashed] (alg) -- (intg);
\end{tikzpicture}
\end{figure}
\vspace{0.1cm}
\centering
\small The message today: Bouton-type formulas live on the \iterm{integration} side, but they come from a \dterm{Leibniz}-type rule on the \dterm{coalgebraic} side.
\end{frame}
\begin{frame}{Short survey: games $\times$ category theory is not new}
\small
\begin{columns}[T,onlytextwidth]
\column{0.58\textwidth}
\begin{itemize}
\item Conway and many successors organize combinatorial games recursively.
\item Joyal, Blass, Abramsky, Melli\`es, and others relate games to semantics and logic.
\item Category theory already clarifies operations on games, but the \dterm{coalgebra} / \iterm{Rota--Baxter} interface seems less explicit.
\item My point is not to replace classical game theory; it is to repackage one famous identity in a more structural way.
\end{itemize}
\column{0.42\textwidth}
\begin{block}{Slogan}
\centering
impartial games\\[0.25em]
$+$\\[0.25em]
recursion\\[0.25em]
$+$\\[0.25em]
categorical calculus
\end{block}
\end{columns}
\end{frame}
\section[Play vs Analysis]{Play vs Analysis: Generalized Bouton's theorem is Integration!}
\begin{frame}{Preliminaries: Winning Nim!}
\begin{columns}[T,onlytextwidth]
\column{0.5\textwidth}
\begin{itemize}
\item A position of Nim is a finite list $\left( a_1,\dots,a_n \right)$ of heap sizes.
\item A move chooses one heap and decreases it.
\item A $P$-position is losing for the next player; an $N$-position is winning for the next player.
\item The game is impartial, so both players have the same legal moves.
\end{itemize}
\column{0.5\textwidth}
\begin{center}
\safeimage[width=\linewidth,height=0.43\textheight,keepaspectratio]{images/NimSample.jpeg}
\end{center}
\end{columns}
\vspace{0.1cm}
\centering
\small We will reinterpret the classical xor criterion as an \iterm{integral} shadow of a \dterm{differential} rule.
\end{frame}
%AI suggestion: the specific heap sizes here can be changed to match the oral example you plan to talk through at the board.
\begin{frame}{\texorpdfstring{Rule of $n$-heap Nim}{Rule of n-heap Nim}}
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.93]
\node[draw=none,fill=creampanel,rounded corners=11pt,minimum width=10.6cm,minimum height=3.55cm] (bg) at (0,0) {};
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\heapstate{1.3}{1}{0}{1}
\heapstate{5.1}{0}{0}{0}
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\node[font=\large] at (-2.6,-0.05) {$\rightsquigarrow$};
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\node[font=\large] at (0.0,-0.05) {$\rightsquigarrow$};
\node at (3.0,-0.05) {$\cdots$};
\node[text=myred,font=\large] at (4.0,0.44) {$A$};
\node[font=\large] at (4.0,-0.05) {$\rightsquigarrow$};
\end{tikzpicture}
\end{figure}
\begin{itemize}
\item Every move changes exactly one heap.
\item The pattern $A \rightsquigarrow B \rightsquigarrow \cdots \rightsquigarrow A$ suggests that a global invariant is controlling the play.
\end{itemize}
\end{frame}
\begin{frame}{Bouton's winning strategy (1/2): Nim sum}
\small
For natural numbers $a$ and $b$, define the \alert{nim-sum}
\[
a \nimsum b
\]
by bitwise xor. Extend this to finite tuples by associativity.
\vspace{0.25cm}
\begin{columns}[T,onlytextwidth]
\column{0.58\textwidth}
\begin{block}{Example}
\[
13 \nimsum 10 = 7,
\qquad
1101_2 \nimsum 1010_2 = 0111_2.
\]
\end{block}
\begin{itemize}
\item $a \nimsum a = 0$.
\item If $a_1 \nimsum \cdots \nimsum a_n \neq 0$, one can reduce one heap to make the new xor equal to $0$.
\end{itemize}
\column{0.42\textwidth}
\[
\begin{array}{c@{\quad}c@{\quad}c}
a & b & a\nimsum b \\
\hline
0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}
\]
\end{columns}
\end{frame}
\begin{frame}{Bouton's winning strategy (2/2)}
\begin{block}{Theorem (Bouton)}
A Nim position $\left( a_1,\dots,a_n \right)$ is a $P$-position if and only if
\[
a_1 \nimsum \cdots \nimsum a_n = 0.
\]
\end{block}
\begin{columns}[T,onlytextwidth]
\column{0.56\textwidth}
\begin{itemize}
\item From xor $0$, every move creates a nonzero xor.
\item From nonzero xor, there is a unique highest bit witnessing how to move to xor $0$.
\item Hence a winning player alternates between nonzero and zero xor.
\end{itemize}
\column{0.44\textwidth}
\begin{block}{Strategy slogan}
\centering
keep the \alert{xor} at $0$.
\end{block}
\end{columns}
\end{frame}
\begin{frame}{Impartial Games}
\begin{columns}[T,onlytextwidth]
\column{0.54\textwidth}
\begin{itemize}
\item An impartial game is a rooted finite game graph.
\item The only structure we keep is the option relation
\[
x \to x'.
\]
\item Each position is determined by the finite family of its immediate options.
\item This is exactly the kind of data that \dterm{coalgebra} likes.
\end{itemize}
\column{0.46\textwidth}
\begin{figure}
\centering
\begin{tikzpicture}[node distance=1.0cm and 1.0cm]
\node[statebox] (r) {$x$};
\node[statebox,below left=of r] (a) {$x_1$};
\node[statebox,below=of r] (b) {$x_2$};
\node[statebox,below right=of r] (c) {$x_3$};
\node[statebox,below left=of a] (a1) {$\cdot$};
\node[statebox,below right=of c] (c1) {$\cdot$};
\draw[thinarr] (r) -- (a);
\draw[thinarr] (r) -- (b);
\draw[thinarr] (r) -- (c);
\draw[thinarr] (a) -- (a1);
\draw[thinarr] (c) -- (c1);
\end{tikzpicture}
\end{figure}
\end{columns}
\end{frame}
\begin{frame}{Definition of games}
\small
Fix a pointed position $\left( X,x \right)$ in a finite impartial game. Its one-step options form the finite set
\[
\dd\left( X,x \right) := \left\{ \left( X,x' \right) \mid x \to x' \right\}.
\]
\vspace{0.25cm}
\begin{block}{Recursive viewpoint}
Any invariant $F$ of pointed games is determined by a rule describing how to compute
\[
F\left( X,x \right)
\]
from the finite family
\[
\left\{ F\left( X,x' \right) \mid x \to x' \right\}.
\]
\end{block}
\vspace{0.15cm}
\centering
\small This is the game-theoretic origin of both \dterm{differentiation} and \iterm{integration}.
\end{frame}
\begin{frame}{\texorpdfstring{$\times$: Box product of games}{Box product of games}}
\small
For pointed games $\left( X,x \right)$ and $\left( Y,y \right)$, the box product keeps track of a move in exactly one component:
\[
\dd\left( \left( X,x \right) \otimes \left( Y,y \right) \right)
\cong
\dd\left( X,x \right) \otimes \left( Y,y \right)
\sqcup
\left( X,x \right) \otimes \dd\left( Y,y \right).
\]
\begin{figure}
\centering
\begin{tikzpicture}[node distance=1.7cm and 2.5cm]
\node[statebox] (xy) {$X \otimes Y$};
\node[statebox,below left=of xy] (dx) {$\left( \dd X \right) \otimes Y$};
\node[statebox,below right=of xy] (dy) {$X \otimes \left( \dd Y \right)$};
\draw[thinarr] (xy) -- node[above left,font=\small]{move in $X$} (dx);
\draw[thinarr] (xy) -- node[above right,font=\small]{move in $Y$} (dy);
\end{tikzpicture}
\end{figure}
\end{frame}
%AI suggestion: if you plan a longer explanation of mex, consider splitting this into a definition slide and an example slide.
\begin{frame}{\texorpdfstring{\iterm{mex} and Grundy number}{mex and Grundy number}}
\begin{columns}[T,onlytextwidth]
\column{0.58\textwidth}
\begin{block}{Definition}
The Grundy number of a pointed game is defined recursively by
\[
\Gr\left( X,x \right)
:=
\mex\left( \left\{ \Gr\left( X,x' \right) \mid x \to x' \right\} \right).
\]
\end{block}
\begin{itemize}
\item The operation \iterm{mex} picks the least natural number missing from a finite set.
\item This is already an \iterm{integral-like} operation: it adjoins a new value above all immediate options.
\end{itemize}
\column{0.42\textwidth}
\begin{center}
\safeimage[width=0.72\linewidth,height=0.56\textheight,keepaspectratio]{images/Grundy_number_2.jpeg}
\end{center}
\end{columns}
\end{frame}
\begin{frame}{Generalized Bouton's theorem}
\begin{block}{Theorem (Sprague--Grundy in box-product form)}
For impartial pointed games,
\[
\Gr\left( X \otimes Y \right) = \Gr\left( X \right) \nimsum \Gr\left( Y \right).
\]
\end{block}
\begin{itemize}
\item Nim is the universal example where the Grundy number is visibly the heap size.
\item So Bouton's theorem is the special case obtained by evaluating the box product of one-heap Nim positions.
\item Later, we will reinterpret this identity as the \iterm{Rota--Baxter} image of a \dterm{Leibniz} rule.
\end{itemize}
\end{frame}
\section[Coalgebra vs Algebra]{Coalgebra vs Algebra: recursive Pfin-coalgebras}
\begin{frame}{Preliminaries: Coalgebra and recursion}
\small
\begin{columns}[T,onlytextwidth]
\column{0.5\textwidth}
\begin{block}{\dterm{Coalgebra}}
A structure map
\[
c \colon C \to F\left( C \right)
\]
describes how to unfold one step.
\end{block}
\column{0.5\textwidth}
\begin{block}{\iterm{Algebra}}
A structure map
\[
a \colon F\left( A \right) \to A
\]
describes how to fold one step.
\end{block}
\end{columns}
\vspace{0.2cm}
\centering
\small Recursive evaluation is the passage from the unfolding side to the folding side.
\end{frame}
\begin{frame}{Algebra/Coalgebra of an endofunctor}
\begin{center}
\begin{tikzcd}[ampersand replacement=\&,column sep=huge,row sep=large]
C \arrow[r,"c"] \arrow[d,dashed,"h"' ] \& F\left( C \right) \arrow[d,"F\left( h \right)"] \\
A \& F\left( A \right) \arrow[l,"a"']
\end{tikzcd}
\end{center}
\begin{itemize}
\item A map $h$ making this square commute is a coalgebra-to-algebra morphism.
\item Intuitively, $h$ evaluates an unfolding object by repeatedly applying the folding rule $a$.
\item For games, $F = \Pf$ and the unfolding data are exactly finite option sets.
\end{itemize}
\end{frame}
\begin{frame}{Coalgebra-Algebra morphism and Recursive coalgebra}
\small
\begin{block}{Definition}
A coalgebra $c \colon C \to F\left( C \right)$ is \alert{recursive} if for every algebra $a \colon F\left( A \right) \to A$ there exists a unique map
\[
\hylo_{c,a} \colon C \to A
\]
such that
\[
\hylo_{c,a} = a \circ F\left( \hylo_{c,a} \right) \circ c.
\]
\end{block}
\vspace{0.15cm}
\centering
\small This is the abstract form of defining values by structural recursion on immediate options.
\end{frame}
\begin{frame}{Games as recursive coalgebras}
\small
Let $X$ be a finite impartial game graph. The option map
\[
\sigma_X \colon X \to \Pf\left( X \right),
\qquad
\sigma_X\left( x \right) := \left\{ x' \mid x \to x' \right\},
\]
is a \dterm{coalgebra} for $\Pf$.
\vspace{0.25cm}
\begin{block}{Observation}
Any rule of the form
\[
a \colon \Pf\left( A \right) \to A
\]
induces a unique evaluation map
\[
X \to A.
\]
Taking $a = \mex$ yields Grundy values.
\end{block}
\end{frame}
\begin{frame}{\texorpdfstring{Games = Recursive $\Pf$-coalgebras}{Games = Recursive Pfin-coalgebras}}
\begin{columns}[T,onlytextwidth]
\column{0.52\textwidth}
\begin{itemize}
\item Finite well-founded games are built by finitely iterating their option operator.
\item So they are exactly the objects on which recursive evaluation is legitimate.
\item The \dterm{coalgebraic} viewpoint packages this once and for all.
\end{itemize}
\column{0.48\textwidth}
\begin{center}
\safeimage[width=\linewidth,height=0.47\textheight,keepaspectratio]{images/RecursiveIsGames.jpeg}
\end{center}
\end{columns}
\end{frame}
\begin{frame}{Digression: Categorical structure of games}
\small
\begin{itemize}
\item Passing from a single pointed game to a finite family of pointed games gives a category
\[
\Fam\left( \Gsp \right).
\]
\item It carries additive structure by disjoint union of families.
\item It also carries multiplicative structure by gamewise box product.
\item The option operator extends to a \dterm{differential}-like endofunctor on families.
\end{itemize}
\begin{block}{Guiding idea}
The same recursion that yields Grundy values also upgrades to a small piece of categorical calculus.
\end{block}
\end{frame}
\section[Diff vs Integration]{Differentiation vs Integration: Calculus 2-rig of pointed game families}
\begin{frame}{\texorpdfstring{\dterm{Differential} $2$-rig of games}{Differential 2-rig of games}}
\small
On $\Fam\left( \Gsp \right)$ we have:
\[
\begin{array}{rcl}
0 &:=& \text{empty family}, \\
1 &:=& \text{terminal pointed game}, \\
X + Y &:=& X \sqcup Y, \\
X \otimes Y &:=& \text{gamewise box product}, \\
\dd X &:=& \text{family of immediate options of } X.
\end{array}
\]
\vspace{0.15cm}
\begin{block}{Perspective}
Additivity comes from finite coproducts, while the interesting content is the \dterm{Leibniz}-type behavior of $\dd$ with respect to $\otimes$.
\end{block}
\end{frame}
\begin{frame}{Differentiation on families of pointed games}
\small
For a finite family
\[
X = \left\{ \left( X_i,x_i \right) \right\}_{i \in I},
\]
define
\[
\dd X := \bigsqcup_{i \in I} \left\{ \left( X_i,x_i' \right) \mid x_i \to x_i' \right\}.
\]
\vspace{0.2cm}
\begin{center}
\begin{tikzcd}[ampersand replacement=\&,column sep=huge]
X \arrow[r,mapsto] \& \dd X
\end{tikzcd}
\end{center}
\centering
\small The operator $\dd$ forgets the current root and records every possible one-step change.
\end{frame}
\begin{frame}{\texorpdfstring{\dterm{Leibniz rule} for Box product}{Leibniz rule for Box product}}
\small
For the box product, every move changes either the left component or the right component.
\vspace{0.2cm}
\begin{block}{Theorem}
In $\Fam\left( \Gsp \right)$ we have a categorified \dterm{Leibniz} rule:
\[
\dd\left( X \otimes Y \right)
\cong
\left( \dd X \right) \otimes Y
\sqcup
X \otimes \left( \dd Y \right).
\]
\end{block}
\begin{figure}
\centering
\begin{tikzpicture}[scale=0.82,transform shape,node distance=1.75cm and 2.5cm]
\node[statebox] (xy) {$X \otimes Y$};
\node[statebox,below left=of xy] (dx) {$\left( \dd X \right) \otimes Y$};
\node[statebox,below right=of xy] (dy) {$X \otimes \left( \dd Y \right)$};
\draw[thinarr] (xy) -- node[above left,font=\small]{move in $X$} (dx);
\draw[thinarr] (xy) -- node[above right,font=\small]{move in $Y$} (dy);
\end{tikzpicture}
\end{figure}
\end{frame}
\section[RB invariants]{Invariants in Rota-Baxter rig}
\begin{frame}{Invariants in \texorpdfstring{\iterm{Rota--Baxter} rig}{Rota-Baxter rig}}
\small
Suppose $A$ is a rig equipped with an operator $\Int$.
\vspace{0.1cm}
\begin{itemize}
\item The \dterm{coalgebraic} data of a pointed game tell us which values should be added before applying $\Int$.
\item So the natural recursive ansatz is
\[
F\left( X,x \right)
:=
\Int\left( \sum_{x \to x'} F\left( X,x' \right) \right).
\]
\item When $\Int$ satisfies a suitable \iterm{Rota--Baxter} identity, this invariant respects the box product.
\end{itemize}
\end{frame}
\begin{frame}{Decategorification target: \texorpdfstring{\dterm{calculus} / \iterm{Rota--Baxter} rigs}{calculus / Rota-Baxter rigs}}
\scriptsize
\begin{columns}[T,onlytextwidth]
\column{0.5\textwidth}
\begin{block}{Definition (\dterm{Differential} rig)}
A rig $\left( A,0,1,+,\times,\dd \right)$ with
\[
\begin{aligned}
\dd\left( a+b \right) &= \dd a + \dd b,\\
\dd\left( ab \right) &= \left( \dd a \right)b + a\left( \dd b \right).
\end{aligned}
\]
\end{block}
\begin{block}{Definition (Calculus rig)}
A rig carrying both structures, with the fundamental theorem
\[
\dd\left( \Int f \right) = f.
\]
\end{block}
\column{0.5\textwidth}
\begin{block}{Definition (\iterm{Integral} rig)}
A rig with an operator $\Int$ satisfying
\[
\begin{aligned}
1 &= \Int 0,\\
\left( \Int f \right)\left( \Int g \right)
&=
\Int\left( \left( \Int f \right)g + f\left( \Int g \right) \right).
\end{aligned}
\]
\end{block}
\begin{block}{Perspective}
The \dterm{differential} and \iterm{integral} structures live on equal footing, but with distinct semantic roles.
\end{block}
\end{columns}
\end{frame}
\begin{frame}{\texorpdfstring{\iterm{Rota--Baxter}-valued invariants of games}{Rota-Baxter-valued invariants of games}}
\small
Let $A$ be an \iterm{integral rig}. For a pointed game $\left( X,x \right)$ define recursively
\[
F\left( X,x \right)
:=
\Int\left( \sum_{x \to x'} F\left( X,x' \right) \right).
\]
\begin{block}{Theorem}
The assignment $F$ is the unique map induced by the recursive \dterm{coalgebra}
\[
X \to \Pf\left( X \right)
\]
and the \iterm{algebra}
\[
\Pf\left( A \right) \to A,
\qquad
S \mapsto \Int\left( \sum_{s \in S} s \right).
\]
\end{block}
\end{frame}
\section[Examples]{Examples}
\begin{frame}{The universal example: game families themselves}
\small
The category $\Fam\left( \Gsp \right)$ itself already carries all the structures we need:
\begin{center}
\begin{tikzcd}[ampersand replacement=\&,column sep=huge]
X \arrow[r, "{\iterm{\Int}}"] \&
\left\{ X \right\}
\arrow[l, bend left=35, "{\dterm{\dd}}"']
\end{tikzcd}
\end{center}
\begin{columns}[T,onlytextwidth]
\column{0.56\textwidth}
\begin{itemize}
\item $\iterm{\Int}$ packages a family as a single rooted layer.
\item $\dterm{\dd}$ extracts the one-step options.
\item The calculus identities become tautological at this universal level.
\end{itemize}
\column{0.44\textwidth}
\begin{block}{Moral}
The game-valued example is the categorified source of the numerical invariants appearing later.
\end{block}
\end{columns}
\end{frame}
\begin{frame}{Nim-sum gives a concrete \texorpdfstring{\iterm{Rota--Baxter} rig}{Rota-Baxter rig}}
\small
Consider the rig $\Pf\left( \N \right)$ with
\[
S + T := S \cup T,
\qquad
S \times T := \left\{ s \nimsum t \mid s \in S,\ t \in T \right\},
\]
and define
\[
\Int\left( S \right) := \left\{ \mex\left( S \right) \right\}.
\]
\vspace{0.2cm}
\begin{block}{Claim}
This is an \iterm{integral rig}, and the induced invariant on one-heap Nim sends a heap of size $n$ to $\left\{ n \right\}$.
\end{block}
\end{frame}
%AI suggestion: this is the conceptual punchline; if time is short, keep this slide and compress the previous two.
\begin{frame}{The key identity is exactly a \texorpdfstring{\iterm{Rota--Baxter} equation}{Rota-Baxter equation}}
\footnotesize
In the rig $\Pf\left( \N \right)$,
\[
\begin{aligned}
\left( \Int S \right)\left( \Int T \right)
&=
\Int\left(
\left( \left( \Int S \right)T \right)
+
\left( S\left( \Int T \right) \right)
\right),\\
\left\{ \mex\left( S \right) \right\}
\times
\left\{ \mex\left( T \right) \right\}
&=
\Int\left(
\left(
\left\{ \mex\left( S \right) \right\} \times T
\right)
\cup
\left(
S \times \left\{ \mex\left( T \right) \right\}
\right)
\right).
\end{aligned}
\]
\vspace{0.12cm}
Unpacking the product gives
\[
\mex\left( S \right) \nimsum \mex\left( T \right)
=
\mex\left(
\left(
\mex\left( S \right) \nimsum T
\right)
\cup
\left(
S \nimsum \mex\left( T \right)
\right)
\right).
\]
\begin{block}{Interpretation}
The classical xor identity is the \iterm{integral} shadow of the \dterm{Leibniz} rule for the option operator.
\end{block}
\end{frame}
\begin{frame}{Recovering Bouton's theorem}
\small
Let $N_n$ denote the one-heap Nim position of size $n$. Then
\[
F\left( N_n \right) = \left\{ n \right\}
\]
and therefore
\[
\begin{aligned}
F\left( N_{a_1} \otimes \cdots \otimes N_{a_n} \right)
&=
\left\{ a_1 \right\}
\times \cdots \times
\left\{ a_n \right\} \\
&=
\left\{ a_1 \nimsum \cdots \nimsum a_n \right\}.
\end{aligned}
\]
\begin{block}{Conclusion}
The composite position is losing exactly when
\[
a_1 \nimsum \cdots \nimsum a_n = 0.
\]
\end{block}
\end{frame}
\begin{frame}{Take-home messages}
\small
\begin{enumerate}
\setlength{\itemsep}{0.42em}
\item \textbf{Games are recursive \dterm{coalgebras}.}
Their basic numerical invariants come from unique recursive evaluation maps.
\item \textbf{Box product has a \dterm{differential} flavor.}
The option operator satisfies a categorified \dterm{Leibniz} rule.
\item \textbf{Nim-sum is an \iterm{integration} phenomenon.}
Bouton's xor identity is encoded by a concrete \iterm{Rota--Baxter} rig on finite subsets of $\N$.
\end{enumerate}
\vspace{0.15cm}
\begin{block}{Outlook}
Genuine chain rules, partisan variants, and systematic links with differential categories all remain open and interesting.
\end{block}
\end{frame}
\begin{frame}[t]{References}
\footnotesize
\begin{columns}[T,onlytextwidth]
\column{0.49\textwidth}
\begin{itemize}
\setlength{\itemsep}{0.42em}
\item J.~H. Conway, \emph{On Numbers and Games}, Academic Press, 1976.
\item E.~R. Berlekamp, J.~H. Conway, and R.~K. Guy, \emph{Winning Ways for Your Mathematical Plays}, Academic Press, 1982.
\item A. Siegel, \emph{Combinatorial Game Theory}, American Mathematical Society, 2013.
\item S. Abramsky and G. McCusker, ``Game semantics,'' in \emph{Computational Logic}, 1999.
\end{itemize}
\column{0.49\textwidth}
\begin{itemize}
\setlength{\itemsep}{0.42em}
\item J.~J.~M.~M. Rutten, ``Universal coalgebra: a theory of systems,'' \emph{Theoretical Computer Science}, 249(1):3--80, 2000.
\item R. Blute, R. Cockett, and R. Seely, ``Differential categories,'' \emph{Mathematical Structures in Computer Science}, 16(6):1049--1083, 2006.
\item G.-C. Rota, ``Baxter algebras and combinatorial identities I, II,'' \emph{Bulletin of the American Mathematical Society}, 75:325--334, 1969.
\end{itemize}
\end{columns}
\end{frame}
\end{document}