# Affine Normal Play

Urban Larsson<EMAIL_ADDRESS>Richard J. Nowakowski<EMAIL_ADDRESS>Carlos P. Santos<EMAIL_ADDRESS>
###### Abstract

There are many combinatorial games in which a move can terminate the game,
such as a checkmate in chess. These moves give rise to diverse situations that
fall outside the scope of the classical normal play structure. To analyze
these games, an algebraic extension is necessary, including infinities as
elements. In this work, affine normal play, the algebraic structure resulting
from that extension, is analyzed. We prove that it is possible to compare two
affine games using only their forms. Furthermore, affine games can still be
reduced, although the reduced forms are not unique. We establish that the
classical normal play is order-embedded in the extended structure,
constituting its substructure of invertible elements. Additionally, as in
classical theory, affine games born by day $n$ form a lattice with respect to
the partial order of games.

###### keywords:

Combinatorial Game Theory, normal play, terminating moves, infinities,
entailing moves, complimenting moves, carry-on moves.

###### MSC:

[2020] 91A46 ,

###### MSC:

[2020] 06A06

††journal: TO DECIDE

organization=Indian Institute of Technology Bombay, country=India

organization=Dalhousie University, country=Canada

fn1fn1footnotetext: Carlos Santos’ work is funded by national funds through
the FCT - Fundação para a Ciência e a Tecnologia, I.P., under the scope of the
projects UIDB/00297/2020 and UIDP/00297/2020 (Center for Mathematics and
Applications).

organization=Center for Mathematics and Applications (NovaMath), FCT NOVA,
country=Portugal

##  1 Introduction

Combinatorial Game Theory (CGT) is the branch of mathematics that studies
games with perfect information and no chance, where the players take turns
making moves. The required theoretical background for this document pertains
to the normal play convention, where the player who cannot move loses. We
assume the reader has a decent level of knowledge in CGT, while our interest
is solely in non-loopy two-player short games. We refrain from delving into
basic concepts such as game forms, options, followers, numbers, nimbers,
infinitesimals, temperature, and so forth, even though this document is mostly
self-contained. All fundamental concepts are presented and discussed in the
references Albert et al. (2007); Berlekamp et al. (1982); Conway (1976);
Siegel (2013).

Since the classical normal play structure $\mathbbm{N}\mathbbm{p}$ does not
cover the possibility of checkmates, i.e., absorbing _terminating moves_ that
immediately conclude the game, awarding victory to one of the players, it is
imperative to consider an extension. Terminating moves are represented by
$\infty$ and $\overline{\infty}$, where the first is positive (Left111In CGT,
Left is a female and positive player, while Right is a male and negative
player. wins) and the second is negative (Right wins). The sole goal of this
document is to extend $\mathbbm{N}\mathbbm{p}$ with these elements, exploring
the algebraic properties of the resulting structure, which we denote as
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$. The idea is analogous to completing the
real number line with infinities, leading to the affinely extended real number
line. Nevertheless, in the context of games, the analysis of the resulting
algebra is considerably more sophisticated. Inspired by this analogy, we refer
to $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ as _affine normal play_ 222Designation
inspired by geometry, where an affine space includes points at infinity to
represent classes of parallel lines., and its elements are referred to as
_affine normal play games_ (or just affine games).

Still in this introductory part, Subsection 1.1 elaborates on the motivation
for analyzing the mentioned extension. Examples from game practice are shown,
making use of some classical rulesets.333In strict formal terms, in CGT, the
word “game” denotes an abstract algebraic object, and the word “ruleset”
denotes a real-world game. Nevertheless, we maintain a certain flexibility,
occasionally using “game” to also refer to a real-world game. In such cases,
the meaning is evident from the context. We assume the reader knows the rules
of the historical chess and go. In relation to the second, what is required is
familiarity with the rules of atari go, also known as first capture go
(wikiAtari, 2013; wikiChess, 2013). It is also important that the reader is
familiar with rulesets analyzed in the specialized literature, particularly
blue-red-hackenbush, nim, nimstring, and amazons (Berlekamp et al., 1982).
Having pieces or other elements in many colors, Left is bLue and Right is Red.
If the colors are black and white, Left is bLack and Right is White (cleaR).

In Section 2, fundamental definitions related to the structure under analysis
are presented. In this section, Theorem 5 addresses “good” properties of the
disjunctive sum, and Theorem 11 establishes that game equivalence and game
order are indeed an equivalence relation and a partial order relation,
respectively.

A relevant result from Section 3 is Theorem 14 (Fundamental Theorem of Affine
Normal Play), which establishes that, just like in classical theory, a game
$G$ is greater than or equal to zero if and only if Left wins $G$ playing
second. As a consequence of this result, we have Corollary 15 (Order-Outcome
Bijection), indicating that there is a total matching between the order
relation and the outcome classes.

Section 4 contains the first of three main results in this document. To better
understand this result, we recall that the classical structure
$\mathbbm{N}\mathbbm{p}$ is an abelian group in which the inverse of a game
$G$ is its conjugate $-G$, i.e., the game where the players switch roles. It
is known that it is possible to compare two games $G$ and $H$ without
resorting to the universal statement in the definition of order. This is
achieved through an options-only procedure, which consists of playing
$G+(-H)$, as $G\geqslant H$ if and only if Left wins $G+(-H)$ playing second.
The following definition formalizes “Left wins playing second”, and has the
significant advantage of being useful when analyzing structures that are not
groups, as we will see is the case of $\mathbbm{N}\mathbbm{p}^{\infty}\\!$.

######  Definition 1.

Let $G$ and $H$ be two elements of $\mathbbm{N}\mathbbm{p}$. The pair $(G,H)$
_satisfies the maintenance property_ if

  1. 1.

$\forall G^{R}$, $\exists G^{RL}$ such that $G^{RL}\geqslant H$ or $\exists
H^{R}$ such that $G^{R}\geqslant H^{R}$;

  2. 2.

$\forall H^{L}$, $\exists H^{LR}$ such that $G\geqslant H^{LR}$ or $\exists
G^{L}$ such that $G^{L}\geqslant H^{L}$.

In that case, we write $\mathbf{M}(G,H)$.

In other words, in $\mathbbm{N}\mathbbm{p}$, we have $G\geqslant H$ if and
only if $\mathbf{M}(G,H)$. What is new in $\mathbbm{N}\mathbbm{p}^{\infty}\\!$
is that, due to the possibility of forcing sequences of moves, it is possible
to have $G\geqslant H$ without the maintenance property as stated in
Definition 1 being satisfied. Nevertheless, we prove that it is still possible
to have an options-only procedure to compare $G$ with $H$. We have $G\geqslant
H$ if and only if, playing second, Left wins $G\uplus(-H)$, where $\uplus$
refers to inquisitor game that will be formalized later. This is equivalent to
saying that, in $\mathbbm{N}\mathbbm{p}^{\infty}\\!$, $G$ is greater than or
equal to $H$ if and only if a weaker version of the maintenance property is
satisfied, which is the content of Theorem 36. Note that this is a theoretical
fact with little practical relevance, as we estimate that in the vast majority
of cases, the comparison can be made using only the maintenance property as
stated in the previous definition. Moreover, cases where this cannot be done
are immediately recognizable.

In Section 5, it is proven that there are three ways to reduce a game form,
with absorbing reversibility being the only one that does not exist in
$\mathbbm{N}\mathbbm{p}$ (Theorem 40). The most interesting fact is that the
reduced forms cease to be unique.

Section 6 contains the second of three main results in this document. In
addition to the circumstance that, with an appropriate interpretation, one can
prove that $\mathbbm{N}\mathbbm{p}$ is order-embedded in
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$, Theorem 48 states the appealing fact
that the classical structure is effectively the substructure of the invertible
elements of affine normal play. It is not merely a technical proof of a result
with little practical consequences. Alongside a straightforward result
contained in the third section, this theorem allows us to conclude that when
$H$ is a traditional Conway value, comparing $G$ with $H$ can be done by
playing $G+(-H)$, just as in classical theory.

Section 7 contains a classification of affine normal games. One notable
feature is the possibility for a component in which, as long as the current
player is not mortally threatened in another summand, the first thing they
must do is deliver a check there. This possibility led to the formalization of
the important concept of hammerzug, and the definition of the pathetic
infinitesimals
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and
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In a way, $\infty$ and $\overline{\infty}$ close the game line, while
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and
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close the descending scales of infinitesimals.

Section 8 contains our third main result, establishing that the affine games
born by day $n$ form a lattice structure (Theorem 66). Furthermore, it is
noteworthy that, while this does not occur in $\mathbbm{N}\mathbbm{p}$, in
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ a restriction can alter the game
equivalence.

Finally, given that the analysis of combinatorial rulesets is a primary
application of CGT, the document concludes with Section 9, exploring an atari
go endgame by determining the affine game values of its components.

###  1.1 Motivational prelude

The need to use infinities is directly related to the notion of “urgency”, as
measured through the CGT concept of temperature. As an example, consider the
amazons position $G$, illustrated in Figure 1.

setpieces= [label=false,maxfield=f1,showmover=false]$=\\{$ setpieces=
[pgfstyle=cross,color=black,markfields=b1,backfields=b1,label=false,maxfield=f1,showmover=false]
$|$ setpieces=
[pgfstyle=cross,color=black,markfields=e1,backfields=e1,label=false,maxfield=f1,showmover=false]
$\\}$

Figure 1: A hot amazons position.

While the players have other options, the two depicted in Figure 1 dominate
all the others. If Left plays first, she is in time to “conquer” territory (by
moving to $3$); if Right plays first, then it is Right who can play to $-3$.
We have $G=\\{3\,|\,-3\\}=\pm 3$, with a temperature equal to $3$. Often, in a
disjunctive sum $G+X$, both players aim to play in $G$ to gain territory.
However, it is important to note the use of “often” and not “always”. Consider
the disjunctive sum $G+X=\pm 3+\pm 5$, shown in Figure 2. The component $X$
(at the top of the board) is a component where Left has a move to $5$ and
Right has a move to $-5$; therefore, $X$ is hotter than $G$. Of course, in
this position, it is now more urgent to play in $X$ than in $G$.

setpieces= [pgfstyle=cross,color=black,
markfields=a2,markfields=a3,markfields=a4,markfields=a5,markfields=a6,markfields=a7,
markfields=b2,markfields=b3,markfields=b6,markfields=b7,
markfields=c2,markfields=c3,markfields=c4,markfields=c5,markfields=c6,markfields=c7,
markfields=d2,markfields=d3,markfields=d4,markfields=d5,markfields=d6,markfields=d7,
markfields=e2,markfields=e3,markfields=e4,markfields=e5,markfields=e6,markfields=e7,
markfields=f2,markfields=f3,markfields=f4,markfields=f5,markfields=f6,markfields=f7,
markfields=g1,markfields=g2,markfields=g3,markfields=g4,markfields=g6,markfields=g7,
markfields=h1,markfields=h2,markfields=h3,markfields=h5,markfields=h6,markfields=h7,
label=false,maxfield=h8,showmover=false]

Figure 2: First things first.

In the mathematical structure $\mathbbm{N}\mathbbm{p}$, there are no
absolutely mandatory moves. More precisely, given a game $G$, it is always
possible to find a disjunctive sum $G+X$ with a more urgent component $X$.
However, certain rulesets include terminating moves – moves that are
infinitely advantageous for the players making them, leading to an immediate
victory.

The existence of terminating moves enables the existence of threats against
which a player must defend. These are commonly referred to as _checks_. Such
situations have infinite temperature (the urgency is infinitely extreme). It
is worthwhile to re-iterate this point, since it is central to all the
analysis that follows: due to the fact that a terminating move has an
absorbing nature, a player in check must always try to defend himself. In
particular, this allows a player to interpose checks to “re-arrange the
position” before responding to the opponent’s last move with a quiet move.
Even world-class chess players can forget the forcing nature of a
_zwischenzug_ 444“Zwischenzug” (German for “intermediate move”) is a chess
tactic where a player, instead of playing the expected move (commonly a
recapture), first threatens the opponent with a check, and only after their
response plays the expected move. Such a move is also called “in-between
check” or _intermezzo_., as can be seen in the example illustrated in Figure
3.

setpieces=

Figure 3: Carlsen-Anand, World Championship, RUS, 15 Nov 2014, Round 6. An
eventual capture of the white pawn at e5 by the black knight is simply bad
since that knight is captured by the white rook at h5 after two “automatic
moves” resulting from the exchange of a pair of rooks at g8. Due to that,
Carlsen thought he had the position under control and made the very bad move
26.Kd2?. Incredibly, Anand missed the opportunity and replied also with the
bad move 26…a4?. Anand could have explored the opportunity with a pair of
zwischenzugs: 26.Kd2 Ne5! 27.Rg8 Nc4 (zwischenzug) 28.Kd3 Nb2 (one more
zwischenzug) 29.Kd2 Rg8 (only now Black recaptures the white rook at g8, with
a winning position).

Checks and forcing sequences cannot be represented by elements of
$\mathbbm{N}\mathbbm{p}$. Consequently, the
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ framework is needed to analyze some
rulesets. In that framework, $\infty+\overline{\infty}$ is not defined.
However, if $X\neq\overline{\infty}$, then $\infty+X=\infty$, and if
$X\neq\infty$, then $\overline{\infty}+X=\overline{\infty}$. These follow from
the absorbing nature of the terminating moves. There are three important
situations of play that require $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ for their
analysis. Examples are given in the next subsections. In the first, the checks
are overt, but they are hidden in the second and third.

####  1.1.1 Terminating moves and infinite heat components

chess was mentioned because almost all readers are familiar with checks and
checkmates within the context of this ruleset. However, it is a loopy ruleset
that allows ties. Additionally, chess positions do not tend to break down into
disjoint components. In this work, from now on, we will use atari go to
exemplify some concepts, as it is a ruleset for which the use of
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is particularly suitable.

In Figure 4, the black stones marked with triangles are “alive”. This is a
term in go indicating that they cannot be captured; possibly, they are
connected to some group with “eyes”, as indicated by the arrow in the right
corner. Thus, the region bounded by these stones is a disjoint component.

Figure 4: Infinite heat atari go component.

Left, playing first, can capture a white stone, winning the game; Right,
playing first, can capture a black stone, winning the game. Thus,
$H=\\{\infty\,|\,\overline{\infty}\\}=\pm\infty$ has infinite temperature; the
player who has the move wins the game. For $X$ different from $\infty$ and
$\overline{\infty}$, in $H+X$, $X$ is irrelevant, regardless of what $X$ may
be. This is in sharp contrast to the amazons position in Figure 1.

Consider now the atari go position $W$ shown in Figure 5. The Left option
$W^{L}$ corresponding to placing a black stone on point “a” is a check since
it threatens an immediate capture555There is a much better move for Left than
this check, which is left as an exercise for the reader..

Figure 5: The Left option $W^{L}$ is a check.

Right can defend with the option $W^{LR}$, as shown in Figure 6.

Figure 6: $W^{LR}=0$.

Note that in $W^{LR}$, a Left option $W^{LRL}=\overline{\infty}$ is a
_suicidal_ move, that is, an option that Left never chooses unless it is the
only thing she can do. Having this option or not makes no difference.
Sometimes, in certain rulesets, such as go or atari go, these moves are
considered illegal and cannot be made. For the sake of formalization, even in
such cases, $\overline{\infty}$ is still included in the game form as if the
move could be made, since this procedure does not change the nature of the
ruleset. A single Left option $\overline{\infty}$ is the equivalent in
$\mathbbm{N}\mathbbm{p}$ to having an empty set of options; if there are other
Left options, $\overline{\infty}$ is always dominated. On the other hand, the
Right options in $W^{LR}$ are moves that allow Left to immediately play to
$\infty$. These moves take the form $\\{\infty\,|\,\ldots\\}$, and Right never
chooses them unless it is the only thing he can do. In practice, these Right
options can be replaced by $\infty$ since they are also a form of suicide,
just taking one more move. Thus, $W^{LR}=\\{\overline{\infty}\,|\,\infty\\}$
is a game in which neither player has moves except those that self-inflict
defeat. As we will see later, this game is zero, the identity of
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$. It is now easy to understand that the
initial check is $W^{L}=\\{\infty\,|\,0\\}$.

In summary, if a Left option of a game $G$ is $G^{L}=\infty$, then $G^{L}$ is
a checkmate; if a Left option is $G^{L}=\\{\infty\,|\,G^{L\mathcal{R}}\\}$,
then $G^{L}$ is a check; if a Left option is $G^{L}=\overline{\infty}$, then
$G^{L}$ is a suicidal move; if a Left option is
$G^{L}=\\{G^{L\mathcal{L}}\,|\,\overline{\infty}\\}$, then the move allows
being checkmated and can be replaced by $\overline{\infty}$. The same logic
applies to Right options.

####  1.1.2 Entailing moves

_Entailing moves_ 666The designation “entailing move” was proposed in
Berlekamp et al. (1982) precisely to convey the idea that a player is
compelled (“entailed”) to play in a certain way. are moves that _force_ the
opponent to respond locally within a certain range of options. It may be
possible for the opponent to respond locally in more than one way;
nevertheless, the opponent is restricted to those local options by the
entailing move. Of course, this disrupts the usual logic of disjunctive sums.
Consider a disjunctive sum $G_{1}+G_{2}+\ldots+G_{k}$. If Left is not lethally
threatened in any component and makes an entailing move in $G_{1}$ to
$G_{1}^{L}$, then the imposed restriction on Right prevents him from freely
choosing any component from $G_{1}^{L}+G_{2}+\ldots+G_{k}$ for his reply.

As an example, consider nim played on sums of piles of stones
$G_{1}+\ldots+G_{k}$, but with the following change in the rules: whenever a
player leaves three stones in one of the piles, the opponent must play in that
very same pile on the next move. This ruleset does not have mathematical
interest, since a straightforward argument shows that all initial positions
with at least a pile of more than 3 stones are $\mathcal{N}$-positions. The
representation of positions through game forms is what is of interest. A pile
of four stones in traditional nim is described through a game form of
$\mathbbm{N}\mathbbm{p}$, more specifically through
$\\{0,*,*2,*3\,|\,0,*,*2,*3\\}$. In the modified version, a pile of four
stones should be described as
$\\{0,*,*2,*3^{\text{{$\swarrow$}}}\,|\,0,*,*2,*3^{\text{{$\swarrow$}}}\\}$,
where $*3^{\text{{$\swarrow$}}}$ means “next player is compelled to play in
this component, even when the game is not played in isolation”. It is because
there are no game forms of this type in $\mathbbm{N}\mathbbm{p}$ that
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is necessary. The question that arises is
how one can use a form of $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ to represent
the desired situation.

Any chess player knows that a check compels the opponent to defend. In other
words, a check is, by its nature, an entailing move. Since some of elements of
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ are checks, they can be employed as
“gadgets” to describe situations similar to the modified nim. For example,
$\\{0,*,*2,\\{\infty\,|\,0,*,*2\\}\,|\,0,*,*2,\\{\,0,*,*2\,|\,\overline{\infty}\\}\\}$
perfectly represents a pile of four stones in the modified ruleset. As in the
movie Godfather, checks are “offers” (threats) that the opponent cannot refuse
(cannot avoid defending). This explains why these gadgets work. When Left
moves to $\\{\infty\,|\,0,*,*2\\}$, Right has to defend by playing to $0$,
$*$, or $*2$. Since $0$, $*$, or $*2$ are the options Right would have if
forced to play in $*3$, the Left option $\\{\infty\,|\,0,*,*2\\}$ is, in
practice, a move that works as one that compels Right to play in $*3$.

In summary, if a Left entailing option compels Right to respond with an
element from $A$, the way to represent this entailing move through a game form
of $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is to use a check $G^{L}$ where
$G^{L\mathcal{R}}$ is the set $A$, i.e., $G^{L}=\\{\infty\,|\,A\\}$.
Obviously, Right entailing options are represented in a similar manner.

####  1.1.3 Carry-on moves

Rulesets like nimstring allow the possibility of _carry-on_ moves, moves that
compel the player making them to retain the right to play. In this particular
ruleset, when a player closes a box, the player must move again. Figure 7
shows a nimstring position that consists of a disjunctive sum $G+H$, where $G$
is the component on the left and $H$ is the component on the right. For
example, if Left closes a box in the left component, Left _has to play again_
, being able to _choose either of the two components to continue the play_.

(9.5,6.5) rectangle (14.5,11.5); [gray!40!white] (13.,11.) – (11.,11.) –
(11.,9.) – (10.,9.) – (10.,7.) – (13.,7.) – cycle; [line width=1.pt]
(10.,11.)– (10.,10.); [line width=1.pt] (10.,10.)– (10.,9.); [line width=1.pt]
(10.,9.)– (11.,9.); [line width=1.pt] (11.,9.)– (11.,10.); [line width=1.pt]
(11.,10.)– (11.,11.); [line width=1.pt] (12.,11.)– (12.,7.); [line width=1.pt]
(13.,11.)– (13.,7.); [line width=1.pt] (14.,11.)– (14.,7.); [line width=1.pt]
(11.,11.)– (13.,11.); [line width=1.pt] (11.,10.)– (13.,10.); [line
width=1.pt] (11.,9.)– (13.,9.); [line width=1.pt] (10.,9.)– (10.,7.); [line
width=1.pt] (11.,9.)– (11.,7.); [line width=1.pt] (10.,8.)– (13.,8.); [line
width=1.pt] (10.,7.)– (13.,7.);  [fill=black] (10.,11.) circle (2.5pt);
[fill=black] (10.,10.) circle (2.5pt); [fill=black] (10.,9.) circle (2.5pt);
[fill=black] (10.,8.) circle (2.5pt); [fill=black] (10.,7.) circle (2.5pt);
[fill=black] (11.,11.) circle (2.5pt); [fill=black] (11.,10.) circle (2.5pt);
[fill=black] (11.,9.) circle (2.5pt); [fill=black] (11.,8.) circle (2.5pt);
[fill=black] (11.,7.) circle (2.5pt); [fill=black] (12.,7.) circle (2.5pt);
[fill=black] (12.,8.) circle (2.5pt); [fill=black] (12.,9.) circle (2.5pt);
[fill=black] (12.,10.) circle (2.5pt); [fill=black] (12.,11.) circle (2.5pt);
[fill=black] (13.,11.) circle (2.5pt); [fill=black] (13.,10.) circle (2.5pt);
[fill=black] (13.,9.) circle (2.5pt); [fill=black] (13.,8.) circle (2.5pt);
[fill=black] (13.,7.) circle (2.5pt); [fill=black] (14.,7.) circle (2.5pt);
[fill=black] (14.,8.) circle (2.5pt); [fill=black] (14.,9.) circle (2.5pt);
[fill=black] (14.,10.) circle (2.5pt); [fill=black] (14.,11.) circle (2.5pt);

Figure 7:  A disjunctive sum in nimstring.

In Berlekamp et al. (1982), carry-on moves were called _complimenting moves_.
In this work, we prefer the designation carry-on move as we believe it to be
much more descriptive. In Figure 8, we show the component $G$ of the previous
example, along with its two options, one of which is a carry-on move. If the
top bar is drawn, the next player continues, but if the middle bar is drawn,
then the current player has to carry on.

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(4.04,-4.04)– (4.04,-5.04); [line width=1.pt] (4.04,-6.04)– (5.04,-6.04);
[fill=black] (4.04,-4.04) circle (2.5pt); [fill=black] (5.04,-4.02) circle
(2.5pt); [fill=black] (5.04,-5.04) circle (2.5pt); [fill=black] (5.04,-6.04)
circle (2.5pt); [fill=black] (4.04,-5.04) circle (2.5pt); [fill=black]
(4.04,-6.04) circle (2.5pt);  [line width=1.pt] (5.04,-4.02)– (5.04,-5.04);
[line width=1.pt] (5.04,-5.04)– (5.04,-6.04); [line width=1.pt] (4.04,-5.04)–
(4.04,-6.04); [line width=1.pt] (4.04,-4.04)– (4.04,-5.04); [line width=1.pt]
(4.04,-6.04)– (5.04,-6.04); [line width=1.pt] (4.04,-4.02)– (5.04,-4.02);
[fill=black] (4.04,-4.04) circle (2.5pt); [fill=black] (5.04,-4.02) circle
(2.5pt); [fill=black] (5.04,-5.04) circle (2.5pt); [fill=black] (5.04,-6.04)
circle (2.5pt); [fill=black] (4.04,-5.04) circle (2.5pt); [fill=black]
(4.04,-6.04) circle (2.5pt);  [gray!40!white] (4.04,-6.04) rectangle
(5.04,-5.04); [line width=1.pt] (5.04,-4.02)– (5.04,-5.04); [line width=1.pt]
(5.04,-5.04)– (5.04,-6.04); [line width=1.pt] (4.04,-5.04)– (4.04,-6.04);
[line width=1.pt] (4.04,-4.04)– (4.04,-5.04); (5.1,-5.2) node[anchor=north
west] Carry-on; [line width=1.pt] (4.04,-6.04)– (5.04,-6.04); [line
width=1.pt] (4.04,-5.04)– (5.04,-5.04);  [fill=black] (4.04,-4.04) circle
(2.5pt); [fill=black] (5.04,-4.02) circle (2.5pt); [fill=black] (5.04,-5.04)
circle (2.5pt); [fill=black] (5.04,-6.04) circle (2.5pt); [fill=black]
(4.04,-5.04) circle (2.5pt); [fill=black] (4.04,-6.04) circle (2.5pt);

Figure 8:  A nimstring component $G$ with its two options, a “double-box” and
a carry-on.

A carry-on move can be seen as a particular case of an entailing move. As
previously mentioned, even when entailed, a player may have more than one
option to respond. This happens when in a game form like
$G^{L}=\\{\infty\,|\,A\\}$ the set $A$ has more than one element.

When $A$ has a single option, there is only one way for the player to cease
being entailed. Suppose that in $G^{L}=\\{\infty\,|\,A\\}$, Right’s only
option is $G^{LR}$, i.e., $G^{L}=\\{\infty\,|\,G^{LR}\\}$. In this case, when
Left opts for the entailing move, it is as if Left is “jumping” directly from
$G$ to $G^{LR}$, since there will be an opportunity to _play again_ after
replacing $G$ by $G^{LR}$ in the disjunctive sum. After the “jump”, Left can
play again on any component of the sum.

Returning to Figure 8, suppose Left closes a box by drawing the middle bar in
$G$. The remaining position is a $3$-sided box, which is a second player win,
i.e., a $\mathcal{P}$-position (zero). Since Left retains the right to play,
she has to carry on in a position equal to zero. These moves, for Left and
Right, are represented by the options $\\{\infty\,|\,0\\}$ (Left’s carry-on
move) and $\\{0\,|\,\overline{\infty}\\}$ (Right’s carry-on move). If a player
draws the top bar in $G$, then they place a $\mathcal{P}$-position (zero) in
the disjunctive sum, but this time, passing the turn to the opponent. Thus,
the game form that describes $G$ is
$\\{0,\\{\infty\,|\,0\\}\,|\,0,\\{0\,|\,\overline{\infty}\\}\\}$.

The analysis of partizan rulesets with entailing moves, which can be of
various types or only carry-on moves, is absent from the specialized
literature. For example, one can consider whackenbush, which prompts an
interesting mathematical analysis. It is played like the classic blue-red-
hackenbush with an extra rule: whenever a player makes a move that drops one
or more opponent’s edges, they have to play again. Figure 9 illustrates an
intriguing whackenbush position that cannot be adequately described through an
element of $\mathbbm{N}\mathbbm{p}$; thus,
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is invoked. The game value of the
position is $\\{\\{\infty\,|\,0\\}\,|\,0,1\\}=\\{\\{\infty\,|\,0\\}\,|\,0\\}$.
This negative value does not exist in $\mathbbm{N}\mathbbm{p}$.

(1.5,1.5) rectangle (18.9,5.5); [line width=2.pt] (2.,2.)– (4.,2.); [line
width=2.pt] (8.,2.)– (10.,2.); [line width=2.pt] (12.,2.)– (14.,2.); [line
width=2.pt] (16.,2.)– (18.,2.); [line width=2.pt,color=qqqqff] (3.,2.)–
(3.,3.); [line width=2.pt,color=ffqqqq] (3.,3.)– (3.,4.); [line
width=2.pt,color=ffqqqq] (3.,4.)– (3.,5.); [line width=2.pt,color=qqqqff]
(13.,2.)– (13.,3.); [line width=2.pt,color=ffqqqq] (13.,3.)– (13.,4.); [line
width=2.pt,color=qqqqff] (17.,2.)– (17.,3.);  [color=black,fill=white] (3.,2.)
circle (2.5pt); [color=black,fill=white] (3.,3.) circle (2.5pt);
[color=black,fill=white] (3.,4.) circle (2.5pt); [color=black,fill=white]
(3.,5.) circle (2.5pt); [color=black,fill=white] (13.,2.) circle (2.5pt);
[color=black,fill=white] (13.,3.) circle (2.5pt); [color=black,fill=white]
(13.,4.) circle (2.5pt); [color=black,fill=white] (17.,2.) circle (2.5pt);
[color=black,fill=white] (17.,3.) circle (2.5pt);

(4.5,3.75) node[anchor=north west] $=$; (7,5.3) node[anchor=north west]
$\left\\{\begin{array}[]{<EMAIL_ADDRESS>\end{array}\right.$; (18.5,5.3) node[anchor=north west]
$\left\\}\begin{array}[]{<EMAIL_ADDRESS>\end{array}\right.$; (10.9,5.6) node[anchor=north west]
$\makebox[0.0pt][l]{\rule[-99.58464pt]{1.5pt}{99.58464pt}}|$; (14.8,2.1)
node[anchor=north west] ,; (7.82,2.8) node[anchor=north west] Carry-on;

Figure 9:  An intriguing whackenbush position.

In summary, if, with a carry-on move, Left “jumps” directly from $G$ to $H$,
maintaining the right to play, then this option can be described as
$G^{L}=\\{\infty\,|\,H\\}$. This is because it can be seen as a Left option (a
“gadget”) that compels Right to respond with $H$, returning the right to play
again to Left. In game practice, what happens frequently is that Left moves
from $G$ to $H$, playing again without any intervention from Right. With the
gadget, Left moves from $G$ to $G^{L}$, and Right is forced to respond with
$H$, allowing Left to play again. Of course, although Right intervenes, it is
as if he had not, given the forced manner in which he had to do it. The carry-
on option $G^{L}=\\{\infty\,|\,H\\}$ is commonly symbolized by
$\circlearrowright^{H}$. Analogously, Right carry-on options
$\\{H\,|\,\overline{\infty}\\}$ are denoted as $\circlearrowleft^{H}$. For
example, the nimstring position shown in Figure 7 can be represented as
$\\{0,\circlearrowright^{0}\,|\,0,\circlearrowleft^{0}\\}$. The whackenbush
position shown in Figure 8 can be represented as
$\\{\circlearrowright^{0}\,|\,0\\}$.

##  2 Mathematical structure $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ of affine
game forms

In this section, we present the fundamental definitions and some initial
results related to affine normal play. The first step, before delving into the
structure, is to define rigorously the objects under study, which are,
naturally, games. Like many others in CGT, it is a recursive definition.

######  Definition 2 (Affine Games).

Let $\mathbb{A}_{-1}=\\{\infty,\overline{\infty}\\}$ be the base case, the set
of _affine_ games born on day $-1$; these are the only games in which players
have no options. For $i>-1$, $\mathbb{A}_{i}$ is the set of games
$\\{G^{\mathcal{L}}\,|\,G^{\mathcal{R}}\\}$, where $G^{\mathcal{L}}$ and
$G^{\mathcal{R}}$ are non-empty finite sets of games contained in
$\cup_{j<i}\mathbb{A}_{j}$; these are the affine games born by day $i>-1$.
Moreover, if $i>-1$, $G\in\mathbb{A}_{i}$, and $G\not\in\mathbb{A}_{i-1}$,
then $G$ is born on day $i$; it is said to have a formal birthday of $i$,
denoted by $\tilde{b}(G)=i$. The set of affine games is
$\mathbb{A}=\cup_{i\geqslant-1}\mathbb{A}_{i}$.

The definition of affine games deserves some considerations. We observe that,
except for games of day $-1$, both players have options. In other words,
affine games are dicotic, meaning either both players have options, or neither
does – this latter scenario pertains to cases where the game is a terminating
game, i.e., it is either $\infty$ or $\overline{\infty}$. One might ask if the
structure we are about to propose is limited by not being able to encompass
rulesets with positions where one player has moves and the other does not. The
answer to the question is easy: _the structure also covers those rulesets_.
The only thing that needs to be done is a modification by choosing
$G^{\mathcal{L}}=\\{\overline{\infty}\\}$ or $G^{\mathcal{R}}=\\{\infty\\}$
for the player without moves. In practice, if the empty set is replaced with a
single suicidal move, the same effect is achieved, as the existence of that
move is equivalent to having no moves at all. It should also be mentioned that
the dicotic nature of the structure streamlines the thinking process in many
proofs, as one can assume for a non-terminating game that both players have
moves. Note also that all games end with a move to $\infty$ or to
$\overline{\infty}$, i.e., all games end with a “checkmate”, which can be
self-inflicted or not. In the case where the move is a suicidal move, it may
seem that the name “normal play” is poorly chosen, as in that case the last
player loses. However, this is not the case, as it is again a matter of
interpretation. Before making the suicidal move, in practice, the player was
deprived of satisfactory moves, and that is the fundamental reason for their
loss. In other words, the structure preserves the logic of the normal play
convention.

The reason why the recursion begins on day $-1$ instead of day $0$ is purely
instrumental. In Section 6, we will prove that the classical structure
$\mathbbm{N}\mathbbm{p}$ is order-embedded in
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$. Starting the recursion on day $-1$
allows a game’s formal birthday to match in both structures if the game
belongs to both. In a way, the infinities play a role analogous to the empty
set in the classical structure, with a difference: in the classical structure,
the empty set is always “horrible” for a player, whereas in the structure we
are proposing, an infinity can be “horrible” or “wonderful”, depending on its
sign. Somehow, in the classical structure, the empty set “exists” on day $-1$,
before day zero, as it is used to make the only game of day zero. Similarly,
infinities exist on day $-1$. Since there are two infinities and not just one
empty set, there is more than one game born by day zero. There is another
difference, but it is not very crucial. In the classical structure, the empty
set is a useful _atom_ for constructing game forms, but it is not a game
itself. The infinities are games. However, in a certain sense, the infinities
also serve as atoms for building game forms. This subject will be revisited in
Section 8.

Moving on to the definition of disjunctive sum, it makes sense to establish an
initial parallel with what happens in the extended real number line. The
disjunctive sum is well-defined except in the case where there are infinities
of different signs in the sum. Regarding the game practice, this poses no
issue because there is never more than one terminating move in a game, given
its absorbing nature; the game ends as soon as one occurs.

######  Definition 3 (Disjunctive Sum).

The disjunctive sum of two affine games $G$ and $H$ is given by

$G+H=\begin{cases}\textrm{not defined},\quad\textrm{ if }(G=\infty\textrm{ and
}H=\overline{\infty})\textrm{ or }(G=\overline{\infty}\textrm{ and
}H=\infty);\\\ \infty,\quad\quad\quad\quad\,\,\,\,\textrm{ if
}(G=\infty\textrm{ and }H\neq\overline{\infty})\textrm{ or
}(G\neq\overline{\infty}\textrm{ and }H=\infty);\\\
\overline{\infty},\quad\quad\quad\quad\,\,\,\,\textrm{ if
}(G=\overline{\infty}\textrm{ and }H\neq\infty)\textrm{ or
}(G\neq\infty\textrm{ and }H=\overline{\infty});\\\
\\{G^{\mathcal{L}}+H,G+H^{\mathcal{L}}\mid
G^{\mathcal{R}}+H,G+H^{\mathcal{R}}\\},\quad\textrm{ otherwise.}\end{cases}$

######  Observation 4.

The definition of disjunctive sum is recursive and is presented with some
abuse of language: $G^{\mathcal{L}}+H,G+H^{\mathcal{L}}$ means
$\\{G^{L}+H\,|\,G^{L}\in G^{\mathcal{L}}\\}\cup\\{G+H^{L}\,|\,H^{L}\in
H^{\mathcal{L}}\\}$ (and the same for Right’s options).

A second parallel concerns the following theorem. The extended real number
line is “almost” an abelian group, with only one issue when infinities are
involved. Similarly, the affine normal play structure is “almost” an abelian
monoid.

######  Theorem 5.

Let $G$, $H$, and $J$ be affine games. We have the following:

  * 1.

$G+(H+J)$ and $(G+H)+J$ are either equal or both undefined;

  * 2.

$G+H$ and $H+G$ are either equal or both undefined.

###### Proof.

The items are established as follows.

  * 1.

If one of the summands is $\infty$ and one of the others is
$\overline{\infty}$, it is easy to verify that both $G+(H+J)$ and $(G+H)+J$
are undefined. If one of the summands is $\infty$ without either of the others
being $\overline{\infty}$, then both $G+(H+J)$ and $(G+H)+J$ are equal to
$\infty$. If one of the summands is $\overline{\infty}$ without either of the
others being $\infty$, then both $G+(H+J)$ and $(G+H)+J$ are equal to
$\overline{\infty}$. Thus, assume that none of $G$, $H$, and $J$ are
infinities. In that case,

$\displaystyle\left(G+(H+J)\right)^{\mathcal{L}}$
$\displaystyle=\left(G^{\mathcal{L}}+(H+J)\right)\cup\left(G+(H+J)^{\mathcal{L}}\right)\quad$
$\displaystyle=\left(G^{\mathcal{L}}+(H+J)\right)\cup\left(G+(H^{\mathcal{L}}+J)\right)\cup\left(G+(H+J^{\mathcal{L}})\right)\quad$
(Induction)
$\displaystyle=\left((G^{\mathcal{L}}+H)+J\right)\cup\left((G+H^{\mathcal{L}})+J\right)\cup\left((G+H)+J^{\mathcal{L}}\right)$
$\displaystyle=\left((G+H)^{\mathcal{L}}+J\right)\cup\left((G+H)+J^{\mathcal{L}}\right)$
$\displaystyle=\left((G+H)+J\right)^{\mathcal{L}}.\quad$

Note that the induction works well, given that the sums are defined because
none of $G$, $H$, and $J$ are infinities. The fact that
$\left(G+(H+J)\right)^{\mathcal{R}}=\left((G+H)+J\right)^{\mathcal{R}}$ is
proven in a similar way. Hence,
$G+(H+J)=\\{(G+(H+J))^{\mathcal{L}}\,|\,(G+(H+J))^{\mathcal{R}}\\}=\\{((G+H)+J)^{\mathcal{L}}\,|\,((G+H)+J)^{\mathcal{R}}\\}=(G+H)+J$.

  * 2.

If one of the summands is $\infty$ and the other is $\overline{\infty}$, then
both $G+H$ and $H+G$ are undefined. If one of the summands is $\infty$ without
the other being $\overline{\infty}$, then both $G+H$ and $H+G$ are equal to
$\infty$. If one of the summands is $\overline{\infty}$ without the other
being $\infty$, then both $G+H$ and $H+G$ are equal to $\overline{\infty}$.
Thus, assume that neither $G$ nor $H$ are infinities. In that case,

$\displaystyle G+H$ $\displaystyle=\\{G^{\mathcal{L}}+H,G+H^{\mathcal{L}}\mid
G^{\mathcal{R}}+H,G+H^{\mathcal{R}}\\}\quad$
$\displaystyle=\\{H^{\mathcal{L}}+G,H+G^{\mathcal{L}}\mid
H^{\mathcal{R}}+G,H+G^{\mathcal{R}}\\}\quad$ (Induction)
$\displaystyle=H+G.\quad$

Note that the induction works well, given that the sums are defined because
neither $G$ nor $H$ are infinities.

∎

In order to define an equivalence relation and a partial order of games, it is
necessary to first define the concept of outcomes. This is done in the usual
manner.

######  Definition 6 (Individualized Outcomes).

The _individualized outcomes_ of $G$ when Left plays first and when Right
plays first are recursively defined by

$o^{L}(G)=\begin{cases}\mathbf{L},\quad\textrm{ if }G=\infty\textrm{ or
}\exists G^{L}\in G^{\mathcal{L}}\textrm{ such that
}o^{R}(G^{L})=\mathbf{L};\\\ \mathbf{R},\quad\textrm{ otherwise.}\end{cases}$
$o^{R}(G)=\begin{cases}\mathbf{R},\quad\textrm{ if
}G=\overline{\infty}\textrm{ or }\exists G^{R}\in G^{\mathcal{R}}\textrm{ such
that }o^{L}(G^{R})=\mathbf{R};\\\ \mathbf{L},\quad\textrm{
otherwise.}\end{cases}$

######  Observation 7.

$\mathbf{L}$ and $\mathbf{R}$ mean “Left wins” and “Right wins”, respectively.
Note that Definition 6 formalizes the idea of _optimal play_ in that, having
that move at their disposal, a player chooses a winning move. In CGT, the
number of moves in which a player wins a game is not relevant; what matters is
winning. In other words, speed is irrelevant.

######  Definition 8 (Outcomes).

The combined outcome, or simply the _outcome_ of a game $G$, is the pair
$(o^{L}(G),o^{R}(G))$ which can be one of the four possibilities
$(\mathbf{L},\mathbf{L})$, $(\mathbf{L},\mathbf{R})$,
$(\mathbf{R},\mathbf{L})$, and $(\mathbf{R},\mathbf{R})$. Traditionally, these
four possibilities are denoted as $\mathscr{L}$, $\mathscr{N}$, $\mathscr{P}$,
and $\mathscr{R}$.

From Left’s perspective777In CGT, Left is the “positive player”, and Right is
the “negative player”: Left always aims for the highest possible, while Right
seeks the lowest possible. Given that in two-player games, each pulls in their
direction, this convention aligns with game practice., $\mathscr{L}$ is the
best outcome (she wins, regardless of whether playing first or second) and
$\mathscr{R}$ is the worst (she loses, regardless of whether playing first or
second). On the other hand, regarding $\mathscr{N}$ and $\mathscr{P}$, the
victory depends on playing first or second, so these outcomes are not
comparable. These considerations explain how the partial order of outcomes is
conventionally established:

[node distance=1cm] (L) at (0,2) $\mathscr{L}$; (P) at (-1,1) $\mathscr{P}$;
(N) at (1,1) $\mathscr{N}$; (R) at (0,0) $\mathscr{R}$;

(L) – (P); (L) – (N); (P) – (R); (N) – (R);

The outcome function $o(G)$ is used to denote the outcome of $G$. The outcome
classes $\mathcal{L},\mathcal{N},\mathcal{R},\mathcal{P}$ are the sets of all
games with the indicated outcome, so that we can write $G\in\mathcal{L}$ when
$o(G)=\mathscr{L}$.

In accordance with what has already been mentioned, when Left wins the game,
the last move is a move to $\infty$, made by one of the two players. Of
course, Right only makes this move when he has no other option. Analogously,
when Right wins the game, the last move is a move to $\overline{\infty}$.

With a partial order of outcomes defined, it makes sense to write $o(G)=o(H)$,
$o(G)\geqslant o(H)$ or $o(G)\,\|\,o(H)$. Up to this point, the symbol $=$ has
been used exclusively to mean “by definition”; from now on, it will also be
used to indicate the equality of outcomes. The distinction will be made by the
context.

With the concept of outcome established, it is now possible to define the
equivalence and partial order of games.

######  Definition 9 (Equivalence and Order).

Let $G$ and $H$ be affine games. We have the following:

  * 1.

$G\sim H\textrm{ if and only if }o(G+X)=o(H+X)\textrm{ for all games
}X\in\mathbb{A}\setminus\\{\infty,\overline{\infty}\\}.$

  * 2.

$G\gtrsim H\textrm{ if and only if }o(G+X)\geqslant o(H+X)\textrm{ for all
games }X\in\mathbb{A}\setminus\\{\infty,\overline{\infty}\\};$

######  Observation 10.

The relation $G\gtrsim H$ means that replacing $H$ by $G$ can never hurt Left,
no matter what the disjunctive sum is; the relation $G\sim H$ means that
replacing $H$ by $G$ can never hurt Left or Right in any disjunctive sum.

######  Theorem 11.

The relations $\sim$ and $\gtrsim$ are, respectively, an equivalence relation
and a partial order relation.

###### Proof.

The facts that $\sim$ is reflexive and symmetric are direct consequences of
Definition 9. Suppose now that $G\sim H$ and $H\sim J$. For every
$X\in\mathbb{A}\setminus\\{\infty,\overline{\infty}\\}$, it is a consequence
of $G\sim H$ that $o(G+X)=o(H+X)$. On the other hand, it is a consequence of
$H\sim J$ that $o(H+X)=o(J+X)$. Consequently, as we have
$o(G+X)=o(H+X)=o(J+X)$, it follows that $G\sim J$ and $\sim$ is also
transitive.

The fact that $\gtrsim$ is reflexive is a direct consequence of Definition 9.
Suppose now that $G\gtrsim H$ and $H\gtrsim G$. For every
$X\in\mathbb{A}\setminus\\{\infty,\overline{\infty}\\}$, it is a consequence
of $G\gtrsim H$ that $o(G+X)\geqslant o(H+X)$. On the other hand, it is a
consequence of $H\gtrsim G$ that $o(H+X)\geqslant o(G+X)$. Consequently, as we
must have $o(G+X)=o(H+X)$, it follows that $G\sim H$ and $\gtrsim$ is
antisymmetric. Finally, suppose that $G\gtrsim H$ and $H\gtrsim J$. For every
$X\in\mathbb{A}\setminus\\{\infty,\overline{\infty}\\}$, it is a consequence
of $G\gtrsim H$ that $o(G+X)\geqslant o(H+X)$. On the other hand, it is a
consequence of $H\gtrsim J$ that $o(H+X)\geqslant o(J+X)$. Thus, as we have
$o(G+X)\geqslant o(H+X)\geqslant o(J+X)$, it follows that $G\gtrsim J$ and
$\gtrsim$ is transitive. ∎

Notation: From now on, the structure $(\mathbb{A},+,\sim,\gtrsim)$ is simply
designated as $\mathbbm{N}\mathbbm{p}^{\infty}\\!$. For ease, we will use the
symbol $\geqslant$ instead of $\gtrsim$, and $=$ instead of $\sim$; different
situations determine if the symbols are being used for definitions, outcomes
or games. Furthermore, $G>H$ means that $G\geqslant H$ and $G\neq H$. When two
games have exactly the same game form, they are _isomorphic_ , denoted as
$G\cong H$. Naturally, if $G$ is isomorphic to $H$, $G$ is equal to $H$ in
terms of game equivalence, but the opposite direction may not be true.

##  3 Fundamental Theorem of Normal Play and the partial order of games

We begin this section dedicated to some initial results on the partial order
of games by clarifying what is the identity of
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ with respect to the disjunctive sum.

######  Theorem 12.

The game $\\{\overline{\infty}\,|\,\infty\\}$, denotated as $0$, is the
identity of $\mathbbm{N}\mathbbm{p}^{\infty}\\!$.

###### Proof.

According to Theorem 3, $G+\\{\overline{\infty}\,|\,\infty\\}$ commutes, thus
it suffices to prove that $G+\\{\overline{\infty}\,|\,\infty\\}=G$.

If $G=\infty$, according to Definition 3,
$G+\\{\overline{\infty}\,|\,\infty\\}=G=\infty$. If $G=\overline{\infty}$,
according to Definition 3,
$G+\\{\overline{\infty}\,|\,\infty\\}=G=\overline{\infty}$. Hence, assume that
$G\neq\infty$ and $G\neq\overline{\infty}$. Let
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.

Suppose that, playing first, a player wins $G+X$. Without loss of generality,
assume that the player is Left, and she wins with some option $(G+X)^{L}$.
Then, she also wins $G+X+\\{\overline{\infty}\,|\,\infty\\}$ with the option
$(G+X)^{L}+\\{\overline{\infty}\,|\,\infty\\}$. Essentially, she mimics the
strategy used when $G+X$ is played alone, since neither player can make a move
in $\\{\overline{\infty}\,|\,\infty\\}$ without immediately leading to defeat.
On the other hand, if, playing first, Left wins
$G+X+\\{\overline{\infty}\,|\,\infty\\}$, this must be done with an option
$(G+X)^{L}+\\{\overline{\infty}\,|\,\infty\\}$. She also wins $G+X$ with the
option $(G+X)^{L}$, since, once more, one can never touch the component
$\\{\overline{\infty}\,|\,\infty\\}$, and the mimicry can be employed again.
The argument is entirely analogous from Right’s perspective, thus
$o(G+X+\\{\overline{\infty}\,|\,\infty\\})=o(G+X)$, implying
$G+\\{\overline{\infty}\,|\,\infty\\}=G$. ∎

The next theorem follows directly from the absorbing nature of the infinities.

######  Theorem 13.

If $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$, then $\infty\geqslant G$ and
$G\geqslant\overline{\infty}$.

###### Proof.

If
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
then, according to Definition 3, $\infty+X=\infty$. Hence, by Definition 6, we
have $o(\infty+X)=o(\infty)=\mathscr{L}$. Thus, for every
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$,
we have $o(\infty+X)\geqslant o(G+X)$, implying $\infty\geqslant G$. Proving
that $G\geqslant\overline{\infty}$ is analogous. ∎

Regarding the classical structure $\mathbbm{N}\mathbbm{p}$, it is a well-known
fact that $G\geqslant 0$ if and only if $G\in\mathcal{L}\cup\mathcal{P}$ –
here, we refer to this result as the Fundamental Theorem of Normal Play. The
ultimate reason for this theorem is that if Left has a winning strategy in a
game $X$, she also has it in $G+X$. She can use a _local response strategy_ ,
meaning that in $G+X$, Left responds to Right’s moves in a component with a
move in that same component, as if she were playing it in isolation. Under the
normal play convention, Left makes the last move in both components and,
consequently, in the disjunctive sum as a whole. The following theorem states
that this result remains valid in $\mathbbm{N}\mathbbm{p}^{\infty}\\!$,
providing yet another reason to support the choice of the name “affine normal
play”.

######  Theorem 14 (Fundamental Theorem of Affine Normal Play).

Let $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. We have that $G\geqslant 0$ if
and only if $G\in\mathcal{L}\cup\mathcal{P}$.

###### Proof.

Assume that $G\geqslant 0$. We have $0\in\mathcal{P}$, and so, by order of
outcomes, $G\in\mathcal{L}\cup\mathcal{P}$.

Suppose now that $G\in\mathcal{L}\cup\mathcal{P}$. If $G=\infty$, according to
Theorem 13, $G\geqslant 0$; hence, assume $G\neq\infty$ and let
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
If, playing first, Left wins $X$ with the option $X^{L}$, then she also wins
$G+X$ with the option $G+X^{L}$. Essentially, she mimics the strategy used
when $X$ is played alone, answering locally when Right plays in $G$. Due to
the assumption $G\in\mathcal{L}\cup\mathcal{P}$, this is a winning strategy
for Left in $G+X$. If Left, playing second, wins $X$. Then, on $G+X$, she can
respond to each of Right’s moves locally, with a winning move on the same
component, because $G\in\mathcal{L}\cup\mathcal{P}$. Thus Left can win $G+X$
playing second. Consequently, $o(G+X)\geqslant o(X)$, implying $G\geqslant 0$.
∎

Once again, as is the case in $\mathbbm{N}\mathbbm{p}$, Theorem 14 is
sufficient to prove that there is a total matching between the order relation
and the outcome classes.

######  Corollary 15 (Order-Outcome Bijection).

If $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ then

  * 1.

$G>0$ if and only if $G\in\mathcal{L}$;

  * 2.

$G=0$ if and only if $G\in\mathcal{P}$;

  * 3.

$G\mathrel{\|}0$ if and only if $G\in\mathcal{N}$;

  * 4.

$G<0$ if and only if $G\in\mathcal{R}$.

###### Proof.

In this proof, we also use the dual version of Theorem 14: “$G\leqslant 0$ if
and only if $G\in\mathcal{R}\cup\mathcal{P}$”.

Suppose that $G>0$. By Theorem 14, $G\in\mathcal{L}\cup\mathcal{P}$. But, we
cannot have $G\in\mathcal{P}$, for otherwise $G\in\mathcal{R}\cup\mathcal{P}$
and $G\leqslant 0$. Therefore, $G\in\mathcal{L}$. Conversely, suppose that
$G\in\mathcal{L}$. By Theorem 14, we have $G\geqslant 0$. But, we cannot have
$G=0$, for otherwise $G\leqslant 0$, and $G\in\mathcal{R}\cup\mathcal{P}$.
Hence, $G>0$. Thus, the first equivalence holds.

The proof of the fourth equivalence is analogous.

For the second equivalence, we have that if $G=0$, then $G\geqslant
0\;\wedge\;G\leqslant 0$. Therefore,
$G\in(\mathcal{L}\cup\mathcal{P})\cap(\mathcal{R}\cup\mathcal{P})=\mathcal{P}$.

The third equivalence is a consequence of eliminating all other possibilities.
∎

The subsequent theorems offer a set of valuable results that relate the
partial order of games to the disjunctive sum.

######  Theorem 16.

If $G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ and
$J\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$,
then

$G\geqslant H\implies G+J\geqslant H+J.$

###### Proof.

Consider any
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
and let $X^{\prime}=J+X$. Since $X$ and $J$ are neither $\infty$ nor
$\overline{\infty}$, we have that $X^{\prime}$ is neither $\infty$ nor
$\overline{\infty}$. Definition 9 implies $o(G+X^{\prime})\geqslant
o(H+X^{\prime})$, that is $o(G+J+X)\geqslant o(H+J+X)$. Thus, the
arbitrariness of $X$ implies $G+J\geqslant H+J$. ∎

######  Theorem 17.

If $G,H,J\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ and $J$ is an invertible
element, then

$G\geqslant H\iff G+J\geqslant H+J.$

###### Proof.

Firstly, note that the fact that $J$ is invertible implies that $J$ is neither
$\infty$ nor $\overline{\infty}$, since, given the absorbing nature of
infinities, there cannot be any element that, when added to these elements,
equals $0$. Thus, the implication $G\geqslant H\implies G+J\geqslant H+J$
directly follows from Theorem 16.

Regarding the reciprocal implication, consider any
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
and let$X^{\prime}=-J+X$ ($J$ is invertible, i.e. $-J$ exists and $J+(-J)=0$).
Since $-J$ is invertible, $-J$ is neither $\infty$ nor $\overline{\infty}$,
and so, $X^{\prime}$ is neither $\infty$ nor $\overline{\infty}$. By
Definition 9, $o(G+J+X^{\prime})\geqslant o(H+J+X^{\prime})$, that is
$o(G+J-J+X)\geqslant o(H+J-J+X)$. Hence, $o(G+X)\geqslant o(H+X)$, and so,
given the arbitrariness of $X$, we have $G\geqslant H$. ∎

The following theorem is highly relevant as it establishes that the comparison
between two games, one of which is invertible, can be conducted by playing,
exactly in the same manner as in the classical structure. In Section 6, an
elegant characterization of invertible elements will be presented, reinforcing
the significance of the theorem.

######  Theorem 18.

If $G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ and $H$ is an invertible
element, then

$G\geqslant H\iff G+(-H)\in\mathcal{L}\cup\mathcal{P}\text{ and }G=H\iff
G+(-H)\in\mathcal{P}.$

###### Proof.

By Theorem 17, $G\geqslant H\iff G+(-H)\geqslant H+(-H)$. Therefore, we have
$G\geqslant H\iff G+(-H)\geqslant 0$. By Theorem 14, this leads to $G\geqslant
H\iff G+(-H)\in\mathcal{L}\cup\mathcal{P}$. Finally, $G=H\iff
G+(-H)\in\mathcal{P}$, since $G\geqslant H$ and $H\leqslant G$. ∎

The two following theorems are straightforward consequences of the fact that
the order and equivalence of affine games constitute a order relation and an
equivalence relation.

######  Theorem 19.

If
$G,H,J,W\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$,
then we have the following:

  * 1.

$G\geqslant H\text{ and }J\geqslant W\implies G+J\geqslant H+W$;

  * 2.

$G=H\text{ and }J=W\implies G+J=H+W$.

###### Proof.

Starting with the first implication, it follows from Theorem 16 that
$G+J\geqslant H+J$ and $H+J\geqslant H+W$. Next, it follows from the fact that
we have an order relation that $G+J\geqslant H+W$. The second implication is a
trivial consequence of the first one. ∎

######  Theorem 20.

Let $G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. If $G>0$ and $H\geqslant 0$
then $G+H>0$.

###### Proof.

If either $G$ or $H$ is $\infty$, the theorem holds trivially. Assuming that
this is not the case, by Theorem 19, we already know that $G+H\geqslant 0$.
So, it is enough to show that $G+H\neq 0$. Since $G>0$ then, without loss of
generality, we may assume that $o^{L}(G+X)=\mathbf{L}$ and
$o^{L}(X)=\mathbf{R}$ for some $X$ different from $\infty$ and
$\overline{\infty}$. Because $H\geqslant 0$, we have $o^{L}(G+X+H)\geqslant
o^{L}(G+X+0)=o^{L}(G+X)=\mathbf{L}$. Since $o^{L}(G+H+X)=\mathbf{L}$ and
$o^{L}(X)=\mathbf{R}$, we have $G+H\neq 0$. ∎

We conclude this section with a very simple theorem that encapsulates the idea
that having extra options can never be a disadvantage.

######  Theorem 21 (Monotone Principle).

If
$G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$,
then, for any $H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$, we have
$\\{{G^{\mathcal{L}}}\cup\\{H\\}\mid{G^{\mathcal{R}}}\\}\geqslant G$.

###### Proof.

Let $G^{\prime}=\\{{G^{\mathcal{L}}}\cup\\{H\\}\mid{G^{\mathcal{R}}}\\}$ and
consider any
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
If Left has a winning strategy in $G+X$ (playing first or second), then Left
also has a winning strategy in $G^{\prime}+X$ since she never has to use the
extra option, simply sticking to the winning strategy that exists in $G+X$. ∎

##  4 Options-only comparison

As mentioned earlier, it is possible to compare two elements $G$ and $H$ of
the classical structure $\mathbbm{N}\mathbbm{p}$ using only their followers.
This is done through an options-only procedure, which consists of playing
$G+(-H)$ to check if $G\geqslant H$. It is known that $G\geqslant H$ if and
only if, playing second, Left wins $G+(-H)$.

In this section, we will see that the options-only comparison in
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is not as straightforward. In fact,
unlike in the majority of CGT monoids, it is possible to have
$G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ such that $G\geqslant H$, without
the maintenance property, as established in Definition 1, being satisfied.
This fact is due to the forcing nature of checks and sequences of checks.

Nevertheless, it is possible to obtain an options-only procedure using a
weaker version of the maintenance property, in order to adress the effect of
the forcing sequences. Thus, we begin this section by presenting an example of
two games $G$ and $H$ such that $G\geqslant H$ without the traditional
maintenance property being satisfied. Subsequently, still at an intuitive
level, we propose the inquisitor game, a compound game that allows comparing
$G$ with $H$ by playing, using only their followers, as done in the classical
structure. We conclude the section with a suitable formalization and rigorous
proofs.

###  4.1 A challenging example

Let
$G=\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}$
and
$H=\left\\{\\{1\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2,*3\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}$.
In these game forms, $0$ is the game $\\{\overline{\infty}\,|\,\infty\\}$, $1$
is the game $\\{0\,|\,\infty\\}$, and the nimbers are the games defined
recursively from $0$ in the traditional way.

Claim 1: We have $G\geqslant H$.

_Proof of Claim 1_ : Let
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
If Right wins $G+X$ by moving to $G+X^{R}$, then, by induction, he wins $H+X$
by moving to $H+X^{R}$. On the other hand, if Right wins $G+X$ by moving to
$0+X$, then the exact same move is available in $H+X$. Therefore, for every
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$,
if Right wins $G+X$ when playing first, then Right also wins $H+X$.

If Left wins $H+X$ by moving to $H+X^{L}$, then, by induction, she wins $G+X$
by moving to $G+X^{L}$. The sensible option happens when Left wins $H+X$ by
moving to $H^{L}+X$. In that case, she must have a winning move against both
Right-checks $\\{0\,|\,\overline{\infty}\\}+X$ and
$\\{*2,*3\,|\,\overline{\infty}\\}+X$. The first implies that $0+X$ is a
winning move for Left, while the second implies that either $*2+X$ or $*3+X$
is a winning move for Left. Hence, we conclude that either both $0+X$ and
$*2+X$ are winning moves for Left or both $0+X$ and $*3+X$ are winning moves
for Left. In the latter case, Left, playing first, secures a win in $G+X$ by
moving to $*3+X$. In the former case, Left, playing first, wins $G+X$ by
moving to
$\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}+X$.
Thus, for every
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$,
if Left wins $H+X$ when playing first, then she also wins $G+X$.

Claim 2: The maintenance property (Definition 1) is not satisfied.

_Proof of Claim 2_ : Given the games $G$ and $H$, we have

$G^{L_{1}}=*3$, and
$G^{L_{2}}=\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}$,

$H^{LR_{1}}=\\{0\,|\,\overline{\infty}\\}$, and
$H^{LR_{2}}=\\{*2,*3\,|\,\overline{\infty}\\}$.

We will show that $G^{L_{1}}\not\geqslant H^{L}$, $G^{L_{2}}\not\geqslant
H^{L}$, $G\not\geqslant H^{LR_{1}}$, and $G\not\geqslant H^{LR_{2}}$.

$G^{L_{1}}\not\geqslant H^{L}$ because, playing first, Left loses
$G^{L_{1}}-1=*3-1$,

and wins
$H^{L}-1=\\{1\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2,*3\,|\,\overline{\infty}\\}\\}-1.$

$G^{L_{2}}\not\geqslant H^{L}$ because, playing second, Left loses

$G^{L_{2}}+\\{0\,|\,*2\\}=\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}+\\{0\,|\,*2\\},$

and wins
$H^{L}+\\{0\,|\,*2\\}=\\{1\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2,*3\,|\,\overline{\infty}\\}\\}+\\{0\,|\,*2\\}.$

$G\not\geqslant H^{LR_{1}}$ because, playing first, Left loses

$G=\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\},$

and wins $H^{LR_{1}}=\\{0\,|\,\overline{\infty}\\}.$

$G\not\geqslant H^{LR_{2}}$ because, playing first, Left loses

$G+*2=\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}+*2,$

and wins $H^{LR_{2}}+*2=\\{*2,*3\,|\,\overline{\infty}\\}+*2.$

This finishes the proof of Claim 2.

###  4.2 inquisitor game

The results that will be proved next establish that in affine normal play, to
check if $G\geqslant H$ ($G$ and $H$ not infinities), it is enough to verify
that Left, playing second, wins $G\uplus(-H)$, where $\uplus$ is “a kind of”
disjunctive sum, but with two extra rules. Just like in a disjunctive sum,
players choose one of the two components to play, leaving the other untouched.
The two extra rules are related to two specificities of affine normal play,
namely the possibility of forcing sequences, and will now be discussed at an
intuitive level. Note that $\uplus$ is not the disjunctive sum as defined in
Definition 3; it is a compound operator whose sole purpose is to compare
games.

The _conjugate_ of a given game switches the roles of the players. In the game
that will be proposed, $-H$ is the conjugate of $H$. We will see in Section 6
that $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ does not deviate from the usual
rule: when a game is invertible, its inverse is its conjugate.

######  Definition 22 (Conjugate).

The conjugate of $H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is

$-H\,=\begin{cases}\overline{\infty},&\mbox{if $H=\infty$}\\\ \infty,&\mbox{if
$H=\overline{\infty}$}\\\
\\{-H^{\mathcal{R}}\,|\,-H^{\mathcal{L}}\\}&\mbox{otherwise},\end{cases}$

where $-H^{\mathcal{R}}$ denotes the set of games $-H^{R}$, for $H^{R}\in
H^{\mathcal{R}}$, and similarly for $-H^{\mathcal{L}}$.

Recall that in order to have $G\geqslant H$, it is necessary to have
$o^{L}(G+X)\geqslant o^{L}(H+X)$ and $o^{R}(G+X)\geqslant o^{R}(H+X)$ for
every game $X$ different from $\infty$ and $\overline{\infty}$. In other
words, if Left, playing first, wins $H+X$, she must also win $G+X$; if Right,
playing first, wins $G+X$, he must also win $H+X$. Often, moves in $X$ are
covered by simple induction. Therefore, the sensible moves are in $G$ or in
$H$.

Now, suppose we are playing $G\uplus(-H)$, with Right playing first. The first
additional rule stipulates that whenever Right moves to $\overline{\infty}$ in
one of the components, Left still has _one last opportunity_ to respond with
$\infty$ in the other component. If Left can still do so on the immediately
next move, she wins the game; otherwise, she loses the game. To understand why
this rule is imperative, let us assume that Right has an option
$G^{R}=\overline{\infty}$. Of course, playing first, Right wins $G+X$. In this
case, for Right to also win $H+X$ for any $X$, it is absolutely necessary that
there exists a Right option $H^{R}=\overline{\infty}$ as well. A symmetrical
situation occurs when there is a Left option $H^{L}=\infty$; it is absolutely
necessary that there also exists a Left option $G^{L}=\infty$. As an example,
consider again the infinitely hot game $G=\\{\infty\,|\,\overline{\infty}\\}$.
It is clear that if $H=\\{\infty\,|\,\overline{\infty}\\}$, then $G\geqslant
H$; in fact, in this case, $G$ and $H$ are equal and isomorphic! In the game
$G\uplus(-H)=\\{\infty\,|\,\overline{\infty}\\}\uplus\\{\infty\,|\,\overline{\infty}\\}$,
Left only wins against Right’s move to
$\overline{\infty}\uplus\\{\infty\,|\,\overline{\infty}\\}$ due to the extra
rule and to the final move to $\overline{\infty}\uplus\infty$.

Understanding the second extra rule at an intuitive level is a bit more
delicate and requires some preliminary definitions.

######  Definition 23 (Check Games).

If $\infty$ is a Left option of a game, then that game is a _Left-check_. A
Left option of a game $G$ that is a Left-check is denoted by
$G^{\overrightarrow{L}}$. Moreover, a _Right-check_ and the notation
$G^{\overleftarrow{R}}$ are defined symmetrically. If a game is either a Left-
check or a Right-check, then it may be simply referred to as a _check_.

######  Definition 24 (Quiet Games).

Let $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. If $G\neq\infty$ and $G$ is not
a Left-check, then $G$ is _Left-quiet_. The definition of a _Right-quiet_ game
is similar. If $G$ is Left-quiet and Right-quiet then $G$ is _quiet_.

######  Definition 25 (Silent Games).

Let $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. If $G$ is not a Left-check, then
$G$ is _Left-silent_. The definition of a _Right-silent_ game is done in a
similar manner. If $G$ is Left-silent and Right-silent then $G$ is _silent_.

######  Observation 26.

A quiet-game cannot be $\infty$ or $\overline{\infty}$. A silent-game may be
$\infty$ or $\overline{\infty}$. A Left option $G^{L}$ can be either a check
or a silent game; a Left option $G^{\overrightarrow{L}}$ is a Left-check.

Let us suppose that Left, playing first, wins $H+X$ with a move $H^{L}+X$. It
may happen that Right has forcing sequences with checks in the first
component, against which Left has to defend. Left’s responses at each moment
take the form $H^{L\overrightarrow{R}\ldots\overrightarrow{R}L}+X$ and _must
be winning responses_ , otherwise, the initial move $H^{L}+X$ would not be a
winning move. Right may have many forcing sequences
$H^{L\overrightarrow{R_{1}}\ldots}+X$,
$H^{L\overrightarrow{R_{2}}\ldots}+X$,$\,\ldots$, and Left always must be able
to respond victoriously. It is crucial to observe that both players may have a
choice. Against each check, Left must chose a winning response; she has a
victorious behavior against checks, a strategy of responses. However, _it is
Right who may choose the moment to stop giving checks_ ; he may do so
immediately with a silent move $H^{LR}+X$, but he eventually may choose to do
so after some checks with a silent move
$H^{L\overrightarrow{R}\ldots\overrightarrow{R}LR}+X$.

Now, in order to have $G\geqslant H$, there must also be a winning move for
Left in $G+X$. Once again, it can be directly a silent move to $G^{L}+X$, but
it can also be a silent move
$G^{\overrightarrow{L}R\ldots\overrightarrow{L}RL}+X$ after a sequence of Left
checks and the responses that Right gives, always avoiding a Left winning move
on the second component (if Right is unable to respond in that way, no issue
is raised). We are now ready to mention the fundamental difference between
classical normal play and affine normal play. In the first case, all moves are
silent moves. Thus, for each $H^{L}$, either there is a silent move $H^{LR}$
such that $G\geqslant H^{LR}$ or there is a silent move $G^{L}$ such that
$G^{L}\geqslant H^{L}$. In affine normal play, it is Left who interrupts the
forcing sequences in $G$, and it is Right who interrupts the forcing sequences
in $H^{L}$. It is sufficient that there is always a pair of games
$G^{\overrightarrow{L}R\ldots\overrightarrow{L}R}$ and
$H^{L\overrightarrow{R}\ldots\overrightarrow{R}L}$ where the forcing sequences
end with a silent move
$G^{\overrightarrow{L}R\ldots\overrightarrow{L}R}\geqslant
H^{L\overrightarrow{R}\ldots\overrightarrow{R}LR}$ or with a silent move
$G^{\overrightarrow{L}R\ldots\overrightarrow{L}RL}\geqslant
H^{L\overrightarrow{R}\ldots\overrightarrow{R}L}$, for $G\geqslant H$ to hold.
This is a weaker version of the classical maintenance property: _it is not
necessary to have immediately $G\geqslant H^{LR}$ or $G^{L}\geqslant H^{L}$.
Achieving the effect after sequences of checks is sufficient_.

Notice that saying it is Right who chooses the moment to stop giving checks in
$H^{L}$ is the same as saying it is Left who chooses the moment to stop giving
checks in $-H^{L}$. This explains the second extra rule. Let $G^{\prime}$ and
$-H^{L^{\prime}}$ be followers of $G$ and $-H^{L}$, respectively. Whenever, in
a position $G^{\prime}\uplus(-H^{L^{\prime}})$, on her turn, Left has the
opportunity to initiate sequences of consecutive checks in either of the two
components, _she can inquire about Right’s responses_. More precisely, Right
must commit to a strategy of responses
$(G^{\prime},\ldots,G^{\prime\overrightarrow{L}R\ldots\overrightarrow{L}R},\ldots)$
to all forcing sequences from $G^{\prime}$ and to a strategy of responses
$(-H^{L^{\prime}},\ldots,-H^{L^{\prime}\overrightarrow{R}L\ldots\overrightarrow{R}L},\ldots)$
to all forcing sequences from $-H^{L^{\prime}}$. This is done in advance, and,
after that, Left may jump directly from $G^{\prime}\uplus(-H^{L^{\prime}})$ to
some
$G^{\prime\overrightarrow{L}R\ldots\overrightarrow{L}R}\uplus(-H^{L^{\prime}\overrightarrow{R}L\ldots\overrightarrow{R}L})$
continuing the play in that position with a silent move, which can be
$G^{\prime\overrightarrow{L}R\ldots\overrightarrow{L}RL}\uplus(-H^{L^{\prime}\overrightarrow{R}L\ldots\overrightarrow{R}L})$
or
$G^{\prime\overrightarrow{L}R\ldots\overrightarrow{L}R}\uplus(-H^{L^{\prime}\overrightarrow{R}L\ldots\overrightarrow{R}LR})$.
Hence, the inquisitor game is a compound game $G_{1}\uplus G_{2}$ where

  1. (Disjunctive sum rule) Players make their moves as if it were a disjunctive sum;

  2. (Last opportunity rule) After a move by Right to $\overline{\infty}$ in one of the components, Left has one last opportunity to win by playing to $\infty$ in the other component;

  3. (Inquiry rule) On her turn, and being able to initiate forcing sequences _in both components_ , Left can inquire in advance about Right’s behaviour in each of them. After that, Left can jump directly to a pair of responses, continuing the game from there.

Let
$G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
What we will prove is that $G\geqslant H$ if and only if Left, playing second,
wins the inquisitor game $G\uplus(-H)$. Of course, there is a dual
formalization $G_{1}\,$ $\uplus$ $\,G_{2}$ from Right’s perspective.

As an example, let us revisit the games $G$ and $H$ from the previous
subsection. To check if $G\geqslant H$, we have to play $G\uplus(-H)$.
Consider the following Right’s move:

(10.,8.3) rectangle (31.,12.8-2); [line width=1.pt] (12.,12.-2) – (15.,9.);
(10.36,12.7-2) node[anchor=north west] $G\uplus(-H)$; (12,8.86)
node[anchor=north west]
$\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}\uplus\\{\\{\infty\,|\,0\\},\\{\infty\,|\,*2,*3\\}\,|\,-1\\}$;

Now it is Left’s turn, and there are Left-checks in both components. Regarding
the first one, suppose that, after inquiring Right, she finds out that the
response to the check is to $\\{0\,|\,\overline{\infty}\\}$. Regarding the
second component, against the first check, Right commits to the answer $0$,
and against the second check, Right commits to the answer $*2$. We have the
following situation:

(10.,7.5-0.5) rectangle (31.,12.8-2); [line width=1.pt] (12.,12.-2) –
(15.,9.); (10.36,12.7-2) node[anchor=north west] $G\uplus(-H)$; (12,8.86)
node[anchor=north west]
$\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}\uplus\\{\\{\infty\,|\,0\\},\\{\infty\,|\,*2,*3\\}\,|\,-1\\}$;
(13.8,8.1) node[anchor=north west]
$\left(G,\\{0\,|\,\overline{\infty}\\}\right)$; (19,8.1) node[anchor=north
west] $\left(-H^{L},0,*2\right)$;

Given these strategies of responses, Left chooses
$\\{0\,|\,\overline{\infty}\\}$ and $0$, securing the victory by playing from
$\\{0\,|\,\overline{\infty}\\}\uplus 0$ to $0\uplus 0$.

(10.,1+3) rectangle (31.,12.8-2); [line width=1.pt] (12.,12.-2) – (15.,9.);
(10.36,12.7-2) node[anchor=north west] $G\uplus(-H)$; (12,8.86)
node[anchor=north west]
$\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}\uplus\\{\\{\infty\,|\,0\\},\\{\infty\,|\,*2,*3\\}\,|\,-1\\}$;
(13.8,8.1) node[anchor=north west]
$\left(G,\bm{\mathsf{\\{0\,|\,\overline{\infty}\\}}}\right)$; (19,8.1)
node[anchor=north west] $\left(-H^{L},\bm{\mathsf{0}},*2\right)$; (15,6+1)
node[anchor=north west] $\\{0\,|\,\overline{\infty}\\}\uplus 0$; [line
width=1.pt] (15.,5.+1) – (12.,2.+3); (10.9,1.85+3) node[anchor=north west]
$0\uplus 0$;

Regardless of other strategies of responses, Left still wins. Suppose that,
regarding the check in the first component, Right’s commitment is the response
to $\\{*2\,|\,\overline{\infty}\\}$. Regarding the second component, against
the first check, Right commits to the answer $0$, and against the second
check, Right commits to the answer $*3$. Given this strategy of responses,
Left chooses $G$ and $*3$, securing victory by playing from $G\uplus*3$ to
$*3\uplus*3$.

(10.,1+3) rectangle (31.,12.8-2); [line width=1.pt] (12.,12.-2) – (15.,9.);
(10.36,12.7-2) node[anchor=north west] $G\uplus(-H)$; (12,8.86)
node[anchor=north west]
$\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}\uplus\\{\\{\infty\,|\,0\\},\\{\infty\,|\,*2,*3\\}\,|\,-1\\}$;
(13.8,8.1) node[anchor=north west]
$\left(\bm{G},\\{*2\,|\,\overline{\infty}\\}\right)$; (19,8.1)
node[anchor=north west] $\left(-H^{L},0,\bm{\mathsf{*3}}\right)$; (15,6+1)
node[anchor=north west] $G\uplus*3$; [line width=1.pt] (15.,5.+1) –
(12.,2.+3); (10.9,1.85+3) node[anchor=north west] $*3\uplus*3$;

Once it is possible to verify that in all cases, playing second, Left wins
$G\uplus(-H)$, it follows that $G\geqslant H$.

###  4.3 Strategies of responses and game trees

In the same way that in classic normal play, the statement “$G\geqslant H$ if
and only if $o^{R}(G+(-H))=\mathbf{L}$” can be transformed into “$G\geqslant
H$ if and only if $\mathbf{M}(G,H)$”, in affine normal play, “$G\geqslant H$
if and only if $o^{R}(G\uplus(-H))=\mathbf{L}$” can be transformed into
“$G\geqslant H$ if and only if $\mathbf{M}^{\infty}(G,H)$”. In this last
sentence, $\mathbf{M}^{\infty}(G,H)$ is a weaker form of maintenance property
that will be formalized next. For now, so that the proofs can be done, it is
necessary to define the meaning of “strategy of responses” to forcing
sequences.

######  Definition 27 (Strategy of Responses).

Let $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. A _strategy of Right responses_
in $G$ is a network of decisions in response to forcing sequences carried out
by Left, constructed as follows:

  * 1.

For $i=0$, $\mathbb{R}^{0}(G)=\\{G\\}$ is the singular set that contains $G$.

  * 2.

For $i>0$, $\mathbb{R}^{i}(G)$ contains a choice of exactly one
$K^{\overrightarrow{L}R}\neq\infty$ for each Left-check
$K^{\overrightarrow{L}}$, where $K\in\mathbb{R}^{i-1}(G)$ (whenever such a
choice can be made). $\mathbb{R}^{i}(G)$ may be empty; when this happens, for
every $k>i$, $\mathbb{R}^{k}(G)$ is also empty, indicating the end of the
decision-making process.

The set $\mathbb{R}(G)=\cup_{i\geqslant 0}\mathbb{R}^{i}(G)$ is the strategy
of Right responses. A _strategy of Left responses_ $\mathbb{L}(G)$ is defined
analogously.

The set of all strategies of Right responses in $G$ is denoted by
$\mathfrak{R}(G)$. Analogously, the set of all strategies of Left responses in
$G$ is denoted by $\mathfrak{L}(G)$.

Given a game $G$, a strategy of Right responses is a set of Right choices in
response to all forcing sequences by Left (a strategy of Left responses is the
symmetrical concept). When comparing games, there is the possibility for Left
to play in $G$, so $G$ must belong to the strategy of Right responses. That
explains the base case $\mathbb{R}^{0}(G)=\\{G\\}$. While Left is forcing with
checks, Right must answer, and the answers must belong to the strategy. That
explains the recursion. What is new in $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is
that the Left-silent options in $G$ are not the only Left-silent options that
matter. We can have relevant Left-silent options after a sequence of checks,
and these options depend on the strategy chosen by Right.

######  Example 28.

Consider once again $G$ and $H^{L}$ as given in Subsection 4.1. Starting with
$G=\left\\{*3,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2\,|\,\overline{\infty}\\}\\}\,\|\,0\right\\}$,
there are two strategies of Right responses:

  * 1.

$\mathbb{R}_{1}(G)=\\{G,\\{0\,|\,\overline{\infty}\\}\\}$

  * 2.

$\mathbb{R}_{2}(G)=\\{G,\\{*2\,|\,\overline{\infty}\\}\\}$

In
$H^{L}=\\{1\,|\,\\{0\,|\,\overline{\infty}\\},\\{*2,*3\,|\,\overline{\infty}\\}\\}$,
there are two strategies of Left responses:

  * 1.

$\mathbb{L}_{1}(H^{L})=\\{H^{L},0,*2\\}$

  * 2.

$\mathbb{L}_{2}(H^{L})=\\{H^{L},0,*3\\}$

Notation: $\overrightarrow{G}$ denotes a typical element of $\mathbb{R}(G)$
and $\overleftarrow{G}$ denotes a typical element of $\mathbb{L}(G)$. Hence,
in $\mathbb{R}_{1}(G)$ in the previous example, $\overrightarrow{G}_{1}=G$ and
$\overrightarrow{G}_{2}=\\{0\,|\,\overline{\infty}\\}$.

The choice of the term “strategy” was not accidental, as this same term is
used in Economic Game Theory (Stengel, 2021). In fact, the concept we just
described is an interesting case in which research in Combinatorial Game
Theory borrows a procedure used in Economic Game Theory.

For a better understanding of this observation, we need to resort to game
trees. A game tree of an affine game is drawn exactly the same way as in
classical theory, with two nuances. Whenever a player’s only option is a
suicidal move, no edge is placed in the tree (as having or not having that
move is equivalent); this is the first nuance. Whenever a player has at their
disposal a winning terminating move, a small arrow is placed in the game tree;
this is the second nuance. That said, as in classical theory, the followers of
a game are identified with the nodes of its game tree, as these nodes are the
roots of the subtrees that represent them. The games $\infty$ and
$\overline{\infty}$ do not have a game tree, as in these cases, the play has
already ended. Figure 10 shows some examples.

Figure 10: On top, the game trees of the four games
$0=\\{\overline{\infty}\,|\,\infty\\}$,
$\pm\infty=\\{\infty\,|\,\overline{\infty}\\}$,
$\infty_{1}=\\{\infty\,|\,\infty\\}$, and
$\overline{\infty}_{1}=\\{\overline{\infty}\,|\,\overline{\infty}\\}$ are
displayed. In the center, the game trees of the games $-1$,
$\\{\infty\,|\,0\\}$, and $\\{\\{\infty\,|\,0\\}\,|\,0\\}$ are easy to
understand. Regarding the game tree below, we have $J=\\{\infty\,|\,0,*\\}$,
$P=\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},-1\\}$,
$T=\\{1,P,J\,|\,\overline{\infty}\\}$, and $W=\\{\infty\,|\,T,0\\}$. Since
$G=\\{W,J,0\,|\,0,\\{0\,|\,\overline{\infty}\\}\\}$, the game tree of $G$
pertains to the game form
$\\{\\{\infty\,|||\,\\{1,\\{\infty\,|\,\\{0\,|\,\overline{\infty}\\},-1\\},\\{\infty\,|\,0,*\\}\,||\,\overline{\infty}\\},0\\},\\{\infty\,|\,0,*\\},0\,||||\,0,\\{0\,|\,\overline{\infty}\\}\\}$.

Returning to the procedure used in Economic Game Theory, given an affine game
$K$, it is possible to consider the _Left-forcing-tree_ of $K$ with all the
information about the forcing sequences that can be carried out by Left. To do
this, the following four steps are taken: (1) the root, identified with the
game $K$ itself, is not removed from the game tree; (2) all nodes related to
the elements of $K^{\mathcal{R}}$ and their followers are removed from the
game tree; (3) regarding the Left moves, only the nodes identified with
followers of the type
$K^{\overrightarrow{L}R\overrightarrow{L}R\ldots\overrightarrow{L}}$ are
retained; (4) regarding the Right moves, only the nodes identified with the
responses to the Left-checks from the previous item are retained.

We observe that a Left-forcing-tree _is not_ a game tree. It is an object
constructed from a game tree, keeping games as labels for some chosen nodes.
Figure 11 illustrates the Left-forcing-tree of the game $G$ shown in Figure
10.

Figure 11: Left-forcing-tree of the game $G$ shown in Figure 10.

It is possible to think of $\mathfrak{R}(G)$ by considering the Left-forcing-
tree of $G$ as if it were the extensive form of an economic game without
payoffs. Note that in Economic Game Theory, node labels are typically payoffs,
and edge labels help identifying strategies. Since, in our case, there are no
payoffs, the node labels remain the games identified with those nodes. The
fundamental idea is to not use labels on the edges, identifying the strategies
directly with the games (node labels). The larger ones (in black) are decision
nodes, nodes where Right can be faced with a decision. The smaller nodes are
possible choices and a strategy of Right responses is a set of choices.

The dashed line, also used in Economic Game Theory, indicates that Right’s
decision can be considered the same in both nodes (the check is identical).
Making different decisions against the same check, only because it occurs at
different places of the game tree, merely increases the number of elements in
the strategy, benefiting Left. Therefore, assuming that Right always responds
in the same way against the same check reduces the work in determining
strategies. Having that in mind, there are $6$ strategies of Right responses
in $G$:

  * 1.

$\mathbb{R}_{1}(G)=\\{G,0\\}$ – Right moves to $0$ against both checks.

  * 2.

$\mathbb{R}_{2}(G)=\\{G,0,*\\}$ – Right moves to $0$ against the first check
and to $*$ against the second.

  * 3.

$\mathbb{R}_{3}(G)=\\{G,T,\\{0\,|\,\overline{\infty}\\},0\\}$ – Against the
first check, Right moves to $T$ with the intention of continuing with
$\\{0\,|\,\overline{\infty}\\}$ or with $0$. The response to the second check
does not need to be specified since that check has already been analyzed.

  * 4.

$\mathbb{R}_{4}(G)=\\{G,T,\\{0\,|\,\overline{\infty}\\},*\\}$ – Against the
first check, Right moves to $T$ with the intention of continuing with
$\\{0\,|\,\overline{\infty}\\}$ or with $*$. The response to the second check
does not need to be specified.

  * 5.

$\mathbb{R}_{5}(G)=\\{G,T,-1,0\\}$ – Against the first check, Right moves to
$T$ with the intention of continuing with $-1$ or with $0$. The response to
the second check does not need to be specified.

  * 6.

$\mathbb{R}_{6}(G)=\\{G,T,-1,*\\}$ – Against the first check, Right moves to
$T$ with the intention of continuing with $-1$ or with $*$. The response to
the second check does not need to be specified.

The set of strategies of Right responses in $G$ is
$\mathfrak{R}(G)=\\{\mathbb{R}_{1},\mathbb{R}_{2},\mathbb{R}_{3},\mathbb{R}_{4},\mathbb{R}_{5},\mathbb{R}_{6}\\}$.

###  4.4 Options-only comparison in $\mathbbm{N}\mathbbm{p}^{\infty}\\!$

To formalize the maintenance property appropriate to affine normal play, it is
necessary to combine two strategies of responses, one for each player.

######  Definition 29 (Locked Pairs).

Let $G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. The pair of games $(G,H)$ is
_locked_ if for all
$(\mathbb{R},\mathbb{L})\in\mathfrak{R}(G)\times\mathfrak{L}(H)$ there are two
games $G^{\prime}\in\mathbb{R}$ and $H^{\prime}\in\mathbb{L}$ for which

  * 1.

there exists a Right-silent game $H^{\prime R}$ such that $G^{\prime}\geqslant
H^{\prime R}$ or

  * 2.

there exists a Left-silent $G^{\prime L}$ such that $G^{\prime L}\geqslant
H^{\prime}$.

######  Definition 30 ($\infty$-Maintenance Property).

Let $G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ be two affine games. We say
that $(G,H)$ _satisfies the $\infty$-maintenance property_, denoted by
$\mathbf{M}^{\infty}(G,H)$, if $(G^{R},H)$ is locked for all
$G^{R}\in{G^{\mathcal{R}}}$, and $(G,H^{L})$ is locked for all
$H^{L}\in{H^{\mathcal{L}}}$.

######  Observation 31.

If $(G,H)$ satisfies the $\infty$-maintenance property, then, for all $G^{R}$
and for each pair $(\mathbb{R}(G^{R}),\mathbb{L}(H))$, there are two games
$\overrightarrow{G^{R}}_{a}\in\mathbb{R}(G^{R})$ and
$\overleftarrow{H}_{b}\in\mathbb{L}(H)$ “compatible with the traditional
maintenance property”. That means that there exists a Right-silent game
$(\overleftarrow{H}_{b})^{R}$ for which
$\overrightarrow{G^{R}}_{a}\geqslant(\overleftarrow{H}_{b})^{R}$ or there
exists a Left-silent game $(\overrightarrow{G^{R}}_{a})^{L}$ for which
$(\overrightarrow{G^{R}}_{a})^{L}\geqslant\overleftarrow{H}_{b}$. The same
happens with any $H^{L}$. In that sense, we have a weaker version of the
traditional maintenance property, as will be stated in Theorem 37.

######  Example 32.

Consider again $G$ and $H$ from Subsection 4.1, to check that $H^{L}$ poses no
issues regarding the $\infty$-maintenance property. We have 2 strategies of
Right responses, $\mathbb{R}_{1}(G)=\\{G,\\{0\,|\,\overline{\infty}\\}\\}$ and
$\mathbb{R}_{2}(G)=\\{G,\\{*2\,|\,\overline{\infty}\\}\\}$, and 2 strategies
of Left responses, $\mathbb{L}_{1}(H^{L})=\\{H^{L},0,*2\\}$ and
$\mathbb{L}_{2}(H^{L})=\\{H^{L},0,*3\\}$. There are $4$ pairs to consider.

  * 1.

For $\left(\mathbb{R}_{1}(G),\mathbb{L}_{1}(H^{L})\right)$, we have the pair
$(\\{0\,|\,\overline{\infty}\\},0)$ and the Left-silent game
$0\in\\{0\,|\,\overline{\infty}\\}^{\mathcal{L}}$ such that $0\geqslant 0$.

  * 2.

For $\left(\mathbb{R}_{1}(G),\mathbb{L}_{2}(H^{L})\right)$, we have the pair
$(\\{0\,|\,\overline{\infty}\\},0)$ and the Left-silent game
$0\in\\{0\,|\,\overline{\infty}\\}^{\mathcal{L}}$ such that $0\geqslant 0$.

  * 3.

For $\left(\mathbb{R}_{2}(G),\mathbb{L}_{1}(H^{L})\right)$, we have the pair
$(\\{*2\,|\,\overline{\infty}\\},*2)$ and the Left-silent game
$*2\in\\{*2\,|\,\overline{\infty}\\}^{\mathcal{L}}$ such that $*2\geqslant*2$.

  * 4.

For $\left(\mathbb{R}_{2}(G),\mathbb{L}_{2}(H^{L})\right)$, we have the pair
$(G,*3)$ and the Left-silent game $*3\in G^{\mathcal{L}}$ such that
$*3\geqslant*3$.

It is time to prove the two implications that underpin the main result of this
section.

######  Theorem 33.

Let
$G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
If $G\geqslant H$ then $\mathbf{M}^{\infty}(G,H)$.

For the proof of Theorem 33, a preliminary lemma is required.

######  Lemma 34.

Let $G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$.

  * 1.

If $G\not\geqslant H$ and $G$ is Left-silent, then there is some
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
such that $o^{L}(G+X)=\mathbf{R}<o^{L}(H+X)=\mathbf{L}$.

  * 2.

If $G\not\geqslant H$ and $H$ is Right-silent, then there is some
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
such that $o^{R}(G+X)=\mathbf{R}<o^{R}(H+X)=\mathbf{L}$.

###### Proof.

It is enough to prove the first item, as the proof for the second is entirely
analogous. By Definition 9, $G\not\geqslant H$ implies that

  * 1.

$\exists
X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
such that $o^{L}(G+X)=\mathbf{R}<o^{L}(H+X)=\mathbf{L}$ or

  * 2.

$\exists
Y\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
such that $o^{R}(G+Y)=\mathbf{R}<o^{R}(H+Y)=\mathbf{L}$.

If the first case is satisfied, then the proof is complete. If the second case
is satisfied, then let $X$ be the game $\\{Y\,|\,\overline{\infty}\\}$. Cases
where $G=\overline{\infty}$ or $H=\infty$ are very simple to handle.
Otherwise, we have $o^{L}(G+X)=\mathbf{R}<o^{L}(H+X)=\mathbf{L}$. These
individualized outcomes arise from the fact that the moves to $G+Y$ and $H+Y$
are mandatory, and $o^{R}(G+Y)=\mathbf{R}<o^{R}(H+Y)=\mathbf{L}$.
∎

###### Proof.

(Theorem 33) For the sake of contradiction, suppose that we have $G\geqslant
H$ and that $(G,H)$ does not satisfy the $\infty$-maintenance property as
stated in Definition 30.

Without loss of generality, suppose that the $\infty$-maintenance property
fails due to the existence of some $H^{L}\in H^{\mathcal{L}}$ opposing it. If
the failure were due to the existence of some $G^{R}\in G^{\mathcal{R}}$
opposing it, the proof would be similar.

Saying that is the same as saying that there is a particular _pernicious pair_
$(\mathbb{R}(G),\mathbb{L}(H^{L}))=(\\{\overrightarrow{G}_{1},\ldots,\overrightarrow{G}_{a},\ldots\\},\\{\overleftarrow{H^{L}}_{1},\ldots,\overleftarrow{H^{L}}_{b},\ldots\\})$
where the following holds for each pair
$(\overrightarrow{G}_{a},\overleftarrow{H^{L}}_{b})$:

there are no Left-silent games $(\overrightarrow{G}_{a})^{L_{i}}$ or, for all
$i$, $\underbrace{(\overrightarrow{G}_{a})^{L_{i}}}_{Left-
silent}\not\geqslant\overleftarrow{H^{L}}_{b}$

and

there are no Right-silent games $(\overleftarrow{H^{L}}_{b})^{R_{j}}$ or, for
all $j$,
$\overrightarrow{G}_{a}\not\geqslant\underbrace{(\overleftarrow{H^{L}}_{b})^{R_{j}}}_{Right-
silent}$.

Note that there is a slight abuse of language in the previous expressions. The
games $(\overrightarrow{G}_{a})^{L_{i}}$ should ideally be expressed as
$(\overrightarrow{G}_{a})^{L_{i}}$, where $i\in\\{1,\ldots,k_{a}\\}$, given
that we are enumerating the Left-silent options of $\overrightarrow{G}_{a}$.
However, to prevent an overwhelming use of subscripts and superscripts, we
have chosen a simplified writing style.

Fact 1: Regarding the Left-silent games, Lemma 34 ensures that, for all
$a,b,i$, there exist distinguishing games $X_{a,b}^{i}$ such that

$o^{L}(\underbrace{(\overrightarrow{G}_{a})^{Li}}_{Left-
silent}+X_{a,b}^{i})=\mathbf{R}<o^{L}(\overleftarrow{H^{L}}_{b}+X_{a,b}^{i})=\mathbf{L}.$

Fact 2: Regarding the Right-silent games, Lemma 34 ensures that, for all
$a,b,j$, there exist distinguishing games $Y_{a,b}^{j}$ such that

$o^{R}(\overrightarrow{G}_{a}+Y_{a,b}^{j})=\mathbf{R}<o^{R}(\underbrace{(\overleftarrow{H^{L}}_{b})^{R_{j}}}_{Right-
silent}+Y_{a,b}^{j})=\mathbf{L}.$

With the aim of constructing a distinguishing game leading to the desired
contradiction, for each $b$, let
$\overline{X}_{b}=\bigcup_{a,i}\\{X_{a,b}^{i}\\}$ , and for each $a$, let
$\overline{Y}_{a}=\bigcup_{b,j}\\{Y_{a,b}^{j}\\}$. The slight abuse of
language is repeated. The set $\bigcup_{a,i}\\{X_{a,b}^{i}\\}$ is
$\bigcup_{a}\bigcup_{i\in\\{1,\ldots,k_{a}\\}}\\{X_{a,b}^{i}\\}$, and the same
for $\bigcup_{b,j}\\{Y_{a,b}^{j}\\}$. Supported by these sets, let

$X^{L}=\left\\{\begin{array}[]{ll}\overline{\infty}&\text{if there are no
Right-silent }(\overleftarrow{H^{L}}_{b})^{R_{j}}\\\
\\{\infty\,|\,\\{\overline{Y}_{1}\,|\,\overline{\infty}\\},\\{\overline{Y}_{2}\,|\,\overline{\infty}\\},\ldots\\}&\text{if
there are Right-silent }(\overleftarrow{H^{L}}_{b})^{R_{j}}\end{array}\right.$

and

$X^{R}=\left\\{\begin{array}[]{ll}\infty&\text{if there are no Left-silent
}(\overrightarrow{G}_{a})^{Li}\\\
\\{\\{\infty\,|\,\overline{X}_{1}\\},\\{\infty\,|\,\overline{X}_{2}\\},\ldots\,|\,\overline{\infty}\\}&\text{if
there are Left-silent }(\overrightarrow{G}_{a})^{Li}\end{array}\right.$

and consider the distinguishing game $X=\\{X^{L}\,|\,X^{R}\\}$.

Claim 1: Left, playing first, loses $G+X$. Right follows his strategy of the
pernicious pair, described in the following picture. The dashed line means
“forcing sequence”. After that, Left either chooses a Left-silent game on the
first component or makes a move on the second component. The “Right wins” are
supported by Facts 1 and 2. If there are no Left-silent games
$(\overrightarrow{G}_{a})^{Li}$, then only the branch on the right matters.
According to Definition 27, Right never moves to $\infty$, and, if Right moves
to $\overline{\infty}$, then Left loses anyway. If $X^{L}=\overline{\infty}$,
then Left loses when playing to $(\overrightarrow{G}_{a})+X^{L}$.

Claim 2: Left, playing first, wins $H+X$ by moving to $H^{L}+X$. Left follows
her strategy of the pernicious pair, described in the following picture. The
dashed line means “forcing sequence”. After that, Right either makes a move on
the second component or chooses a Right-silent game on the first component.
The “Left wins” are supported by Facts 1 and 2. If there are no Right-silent
games $(\overleftarrow{H^{L}}_{b})^{R_{j}}$ then only the branch on the left
matters. According to Definition 27, Left never moves to $\overline{\infty}$,
and, if Left moves to $\infty$, then Left wins anyway. If $X^{R}=\infty$ then
Right loses when playing to $(\overleftarrow{H^{L}}_{b})+X^{R}$.

It follows from Claims 1 and 2 that $G\not\geqslant H$, resulting in a
contradiction. ∎

The opposite direction is also true, as the following Theorem shows.

######  Theorem 35.

Let
$G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
If $\mathbf{M}^{\infty}(G,H)$ then $G\geqslant H$.

###### Proof.

For the sake of contradiction, suppose that $\mathbf{M}^{\infty}(G,H)$ and
$G\not\geqslant H$. According to Definition 9, there exists a distinguishing
game
$X\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$
such that either $o^{L}(G+X)=\mathbf{R}<o^{L}(H+X)=\mathbf{L}$ or
$o^{R}(G+X)=\mathbf{R}<o^{R}(H+X)=\mathbf{L}$.

Without loss of generality, assume that $X$ has the smallest formal birthday,
and such that $o^{L}(G+X)=\mathbf{R}<o^{L}(H+X)=\mathbf{L}$. We can establish
the following three facts:

  1. (a)

$X$ is Left-silent since $o^{L}(G+X)=\mathbf{R}$.

  2. (b)

If $o^{L}(H+X)=o^{R}(H^{L}+X)=\mathbf{L}$ for some $H^{L}\in H^{\mathcal{L}}$,
then there exists a strategy $\mathbb{L}(H^{L})$ such that
$o^{R}(\overleftarrow{H^{L}}+X)=\mathbf{L}$ for all
$\overleftarrow{H^{L}}\in\mathbb{L}(H^{L})$. If this were not the case, Right
could maneuver with checks, and we would not have $o^{R}(H^{L}+X)=\mathbf{L}$.
We can also conclude that, for all
$\overleftarrow{H^{L}}\in\mathbb{L}(H^{L})$, we have
$o^{L}((\overleftarrow{H^{L}})^{R}+X)=\mathbf{L}$, regardless of the Right
option $(\overleftarrow{H^{L}})^{R}$. On the other hand, there exists a
strategy $\mathbb{R}(G)$ such that $o^{L}(\overrightarrow{G}+X)=\mathbf{R}$
for all $\overrightarrow{G}\in\mathbb{R}(G)$. If this were not the case, Left
could manoeuver with checks and we would not have $o^{L}(G+X)=\mathbf{R}$. We
can also conclude that, for all $\overrightarrow{G}\in\mathbb{R}(G)$, we have
$o^{R}(\overrightarrow{G}^{L}+X)=\mathbf{R}$, regardless of the Left option
$\overrightarrow{G}^{L}$. Thus, if $\overrightarrow{G}\in\mathbb{R}(G)$ and
$\overleftarrow{H^{L}}\in\mathbb{L}(H^{L})$, there are neither
$\overrightarrow{G}^{L}\geqslant\overleftarrow{H^{L}}$ nor
$\overrightarrow{G}\geqslant(\overleftarrow{H^{L}})^{R}$. This is equivalent
to saying that the pair $(\mathbb{R}(G),\mathbb{L}(H^{L}))$ breaks the
$\infty$-maintenance property, which constitutes a contradiction.

  3. (c)

According to the initial smallest birthday assumption,
$o^{R}(G+X^{L})\geqslant o^{R}(H+X^{L})$.If
$o^{L}(H+X)=o^{R}(H+X^{L})=\mathbf{L}$ for some game $X^{L}\in
X^{\mathcal{L}}$, then we conclude that $o^{L}(G+X)\geqslant
o^{R}(G+X^{L})\geqslant o^{R}(H+X^{L})=o^{L}(H+X)$. Consequently, in that
case, $o^{L}(G+X)=\mathbf{L}$, which is once again a contradiction.

A contradiction is unavoidable given the assumptions. Hence, we have shown
that $G\geqslant H$. ∎

Finally, the main result of this section follows.

######  Theorem 36.

Let
$G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
We have that $G\geqslant H$ if and only if $\mathbf{M}^{\infty}(G,H)$.

###### Proof.

Combine Theorem 33 and Theorem 35. ∎

Situations with many checks are relatively rare in game practice. In most
cases, the traditional maintenance property, as stated in Definition 1, is
enough to verify if $G\geqslant H$.

######  Theorem 37.

Let
$G,H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!\setminus\\{\infty,\overline{\infty}\\}$.
If $\mathbf{M}(G,H)$ then $G\geqslant H$.

###### Proof.

If $\mathbf{M}(G,H)$, then $\mathbf{M}^{\infty}(G,H)$, since $G$ belongs to
every strategy $\mathbb{R}(G)$, $H^{L}$ belongs to every strategy
$\mathbb{L}(H^{L})$, $G^{R}$ belongs to every strategy $\mathbb{R}(G^{R})$,
and $H$ belongs to every strategy $\mathbb{L}(H)$.

As $\mathbf{M}^{\infty}(G,H)$, Theorem 36 ensures that $G\geqslant H$. ∎

##  5 Reductions and reduced forms

In $\mathbbm{N}\mathbbm{p}^{\infty}\\!$, the game reductions are essentially
the same as in $\mathbbm{N}\mathbbm{p}$. The following three theorems are
stated from Left’s perspective, but, of course, there are dual statements from
Right’s perspective.

_Domination_ captures the idea that if a player has an option at least as good
as another, that other option becomes superfluous.

######  Theorem 38 (Domination).

If $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is a game such that
$G^{L_{1}},G^{L_{2}}\in{G^{\mathcal{L}}}$ and$G^{L_{2}}\geqslant G^{L_{1}}$,
then $G=\\{{G^{\mathcal{L}}}\setminus\\{G^{L_{1}}\\}\mid{G^{\mathcal{R}}}\\}$.

###### Proof.

Let
$G^{\prime}=\\{{G^{\mathcal{L}}}\setminus\\{G^{L_{1}}\\}\mid{G^{\mathcal{R}}}\\}$.
We want to verify that $o^{L}(G+X)=o^{L}(G^{\prime}+X)$ and
$o^{R}(G+X)=o^{R}(G^{\prime}+X)$ for every game $X$ different from $\infty$
and $\overline{\infty}$.

If Left wins $G^{\prime}+X$ with a move $G^{\prime}+X^{L}$, then, by
induction, she also wins $G+X$ with the move $G+X^{L}$. Analogously, if Left
wins $G+X$ with a move $G+X^{L}$, then, by induction, she also wins
$G^{\prime}+X$ with the move $G^{\prime}+X^{L}$. The argument for the case
where Right plays first in the component $X$ is similar.

Regarding moves other than in the component $X$, the only situation that
escapes direct mimicry pertains to a possible winning move by Left in $G+X$ to
$G^{L_{1}}+X$. In that case, playing first, she also wins $G^{\prime}+X$ by
moving to $G^{L_{2}}+X$, since $G^{L_{2}}\geqslant G^{L_{1}}$. ∎

_Open reversibility_ captures the idea that if a player has an option in a
component against which the opponent, in a worst-case scenario, can respond in
a way to improve their situation, the player chooses that option only if they
intend to continue playing locally. If the intention is to play in another
component, it is better to play in that component right from the start, thus
avoiding the opportunity for the mentioned opponent’s improvement.

######  Theorem 39 (Open Reversibility).

If $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is a game with
$G^{L}\in{G^{\mathcal{L}}}$ and $G^{LR}\neq\overline{\infty}$ such that
$G\geqslant G^{LR}$, then
$G=\\{{G^{\mathcal{L}}}\setminus\\{G^{L}\\},G^{LR\mathcal{L}}\mid{G^{\mathcal{R}}}\\}$.

###### Proof.

We note that $G^{LR\mathcal{L}}$ is not empty. This happens because $G^{LR}$
cannot be $\infty$ due to the condition $G\geqslant G^{LR}$ and it cannot be
$\overline{\infty}$ due to the condition $G^{LR}\neq\overline{\infty}$. Let
$H=\\{{G^{\mathcal{L}}}\setminus\\{G^{L}\\},G^{LR\mathcal{L}}\mid{G^{\mathcal{R}}}\\}$.
We want to prove that $H=G$, i.e., $G\geqslant H$ and $H\geqslant G$.

First, we prove that $H\geqslant G^{LR}$. Let $X$ be a game different from
$\infty$ and $\overline{\infty}$. If Left wins $G^{LR}+X$ with $G^{LR}+X^{L}$,
then, by induction, she also wins $H+X$ with the move $H+X^{L}$. If Left wins
$G^{LR}+X$ with a move $G^{LRL}+X$, she also wins $H+X$ with the move
$G^{LRL}+X$, since $G^{LRL}\in H^{\mathcal{L}}$. If Right, playing first, wins
$H+X$ with $H+X^{R}$, then, by induction, she also wins $G^{LR}+X$ with the
move $G^{LR}+X^{R}$. Suppose now that Right, playing first, wins $H+X$ with
$H^{R}+X$. Since $H^{R}\in{G^{\mathcal{R}}}$, we have that
$H^{R}+X\in(G+X)^{\mathcal{R}}$. Consequently, Right has a winning move in
$G+X$, and due to the fact that $G\geqslant G^{LR}$, he also has a winning
move in $G^{LR}+X$.

Second, we prove that $H\geqslant G$. The only case whose analysis is not
straightforward is a possible winning move for Left from $G+X$ to $G^{L}+X$.
In that case, it is mandatory for Left, playing first, to have a winning move
in $G^{LR}+X$. As we already know that $H\geqslant G^{LR}$, Left must have a
winning move in $H+X$ as well.

Third, we prove that $G\geqslant H$. The only cases whose analysis is not
straightforward are possible winning moves for Left from $H+X$ to $G^{LRL}+X$.
Since these winning moves are also available in $G^{LR}+X$, the fact that
$G\geqslant G^{LR}$ implies that Left must also have a winning move in $G+X$.
∎

_Absorbing reversibility_ captures the idea that if a player has an option
that allows the opponent to make a winning terminating move, in practice, that
option works as if the player were directly committing suicide.

######  Theorem 40 (Absorbing Reversibility).

If $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is a game with a Left option
$G^{L}=\\{G^{L\mathcal{L}}\,|\,\overline{\infty}\\}$, then
$G=\\{{G^{\mathcal{L}}}\setminus\\{G^{L}\\},\overline{\infty}\mid{G^{\mathcal{R}}}\\}$.

###### Proof.

Let
$H=\\{{G^{\mathcal{L}}}\setminus\\{G^{L}\\},\overline{\infty}\mid{G^{\mathcal{R}}}\\}$.
We want to prove that $H=G$, i.e., $G\geqslant H$ and $H\geqslant G$.

First, we prove that $G\geqslant H$. Let $X$ be a game different from $\infty$
and $\overline{\infty}$. If Left wins $H+X$ with $H+X^{L}$, then, by
induction, she also wins $G+X$ with the move $G+X^{L}$. Regarding moves other
than in the component $X$, the only Left option in $H+X$ that escapes direct
mimicry is $\overline{\infty}+X$. However, there is no need to consider that
possibility, as it is a suicidal move for Left. To prove that the existence of
a winning move for Right in $G+X$ implies the existence of a winning move for
Right in $H+X$, it is sufficient to use induction and mimicry.

Second, we prove that $H\geqslant G$. Once again, winning moves in component
$X$ are addressed by induction. Regarding moves other than in the component
$X$, the only Left option in $G+X$ that escapes direct mimicry is
$\\{G^{L\mathcal{L}}\,|\,\overline{\infty}\\}+X$. However, there is no need to
consider that possibility, as Right can win by responding to
$\overline{\infty}+X$. To prove that the existence of a winning move for Right
in $H+X$ implies the existence of a winning move for Right in $G+X$, it is
sufficient to use induction and mimicry. ∎

Similar to what is formalized in classical theory, a game form $G$ is reduced
if none of its followers admits any of the three reductions stated in Theorems
38, 39, and 40 (also in terms of dual formulations according to the Right’s
perspective). A very interesting fact is that there can be two reduced forms,
$G$ and $G^{\prime}$, such that $G=G^{\prime}$ and $G\not\cong G^{\prime}$,
meaning the uniqueness of reduced forms is lost. Once again, it is a
consequence of checks and forcing sequences. Still, this fact is not a serious
issue, since reduced forms can still be used in proofs and for the purpose of
simplifications. The following games constitute an example of this occurrence.

$G=\\{\\{\infty\,|\,\\{0,*2\,|\,\overline{\infty}\\},\\{*,*3\,|\,\overline{\infty}\\}\\},\\{\infty\,|\,\\{0,*\,|\,\overline{\infty}\\},\\{*2,*3\,|\,\overline{\infty}\\}\\}\,\|\,\overline{\infty}\\}\text{,
and}$
$H=\\{\\{\infty\,|\,\\{0,*2\,|\,\overline{\infty}\\},\\{*,*3\,|\,\overline{\infty}\\}\\},\\{\infty\,|\,\\{0,*3\,|\,\overline{\infty}\\},\\{*,*2\,|\,\overline{\infty}\\}\\}\,\|\,\overline{\infty}\\}.$

We have that $G\not\cong H$ because, although the game forms share a Left-
check, there is a pair of distinct Left-checks. On the other hand, a careful
application of Theorem 36 allows us to verify that $G\geqslant H$ and
$H\geqslant G$, i.e., $G=H$. The same theorem allows us to check that there
are no reducible followers in either of the two game forms.

In practice, what happens is that, although there are different checks, the
control provided by its forcing nature is essentially the same. The following
definition formalizes the concept of reduced form.

######  Definition 41 (Reduced Forms).

An affine game form $G$ is a _reduced form_ if the following two conditions
are satisfied:

  * 1.

None of the followers of $G$ admits any of the three reductions stated in
Theorems 38, 39, and 40;

  * 2.

There is no $G^{\prime}$ such that $G^{\prime}=G$ and
$\tilde{b}(G^{\prime})<\tilde{b}(G)$.

##  6 Conway-embedding and invertibility

It has been emphasized that the effect of the absence of moves (empty set) can
be achieved in $\mathbbm{N}\mathbbm{p}^{\infty}\\!$ by choosing
$G^{\mathcal{L}}=\\{\overline{\infty}\\}$ or $G^{\mathcal{R}}=\\{\infty\\}$
for the player without moves, as a single suicidal move is the same as having
no moves at all. In this section, we formalize this idea, proving that the
traditional Conway values are order embedded in the affine structure. In fact,
we will prove that the group structure of Conway values is effectively the
substructure of the invertible elements of
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$.

######  Definition 42 (Conway Forms and Games).

A game form $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is a _Conway form_ if $G$
is distinct from $\infty$ and $\overline{\infty}$, and has no checks as
followers. Let
$\mathbbm{N}\mathbbm{p}^{\emph{C}}\\!\subseteq\mathbbm{N}\mathbbm{p}^{\infty}\\!$
denote the substructure of Conway forms. A game form
$G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is a _Conway game_ if it is equal to
a Conway form in terms of equivalence of games. A game form
$G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ is a _pressing game_ if it is not a
Conway game.

######  Example 43.

The game
$G=\\{\\{\overline{\infty}\,|\,\infty\\}\,|\,\\{\overline{\infty}\,|\,\infty\\}\\}=\\{0\,|\,0\\}=*$
is a Conway form (the game form has no checks as followers). The game
$G^{\prime}=\\{\\{\infty\,|\,*\\}\,|\,\\{*\,|\,\overline{\infty}\\}\\}$ is not
a Conway form because it has checks as followers. However, since
$G^{\prime}=G$, it is a Conway game.

######  Definition 44.

The _Conway map_
$c:\mathbbm{N}\mathbbm{p}^{\emph{C}}\\!\rightarrow\mathbbm{N}\mathbbm{p}$ is
recursively defined in the following way:

$c(G)=\begin{cases}0=\\{\emptyset\,|\,\emptyset\\}\quad\quad\quad\quad\quad\textrm{
if }G=0=\\{\overline{\infty}\,|\,\infty\\};\\\
\\{\emptyset\,|\,c({G^{\mathcal{R}}})\\}\quad\quad\quad\quad\,\,\,\,\textrm{
if }G=\\{\overline{\infty}\,|\,{G^{\mathcal{R}}}\\}\textrm{ and
}{G^{\mathcal{R}}}\neq\\{\infty\\};\\\
\\{c({G^{\mathcal{L}}})\,|\,\emptyset\\}\quad\quad\quad\quad\,\,\,\,\,\textrm{
if }G=\\{{G^{\mathcal{L}}}\,|\,\infty\\}\textrm{ and
}{G^{\mathcal{L}}}\neq\\{\overline{\infty}\\};\\\
\\{c({G^{\mathcal{L}}})\,|\,c({G^{\mathcal{R}}})\\}\quad\quad\,\,\,\,\textrm{
otherwise.}\end{cases}$

Here, $c({G^{\mathcal{L}}})$ means $\\{c(G^{L})|G^{L}\in{G^{\mathcal{L}}}\\}$,
and the same for $c({G^{\mathcal{R}}})$.

######  Lemma 45.

If $G\in\mathbbm{N}\mathbbm{p}^{\emph{C}}\\!$, then $c(-G)=-c(G)$.

###### Proof.

It is an immediate consequence of the Definitions 22 and 44. ∎

######  Theorem 46.

The Conway map is a one-to-one map.

###### Proof.

We define first the following map from $\mathbbm{N}\mathbbm{p}$ to
$\mathbbm{N}\mathbbm{p}^{\text{C}}$:

$c^{-1}(G)=\begin{cases}0=\\{\overline{\infty}\,|\,\infty\\}\quad\quad\quad\quad\quad\,\,\,\;\textrm{
\emph{if} }G=0=\\{\emptyset\,|\,\emptyset\\};\\\
\\{\overline{\infty}\,|\,c^{-1}({G^{\mathcal{R}}})\\}\quad\quad\quad\quad\,\,\,\,\textrm{
\emph{if} }G=\\{\emptyset\,|\,{G^{\mathcal{R}}}\\}\textrm{ \emph{and}
}{G^{\mathcal{R}}}\neq\emptyset;\\\
\\{c^{-1}({G^{\mathcal{L}}})\,|\,\infty\\}\quad\quad\quad\quad\,\,\,\,\,\textrm{
\emph{if} }G=\\{{G^{\mathcal{L}}}\,|\,\emptyset\\}\textrm{ \emph{and}
}{G^{\mathcal{L}}}\neq\emptyset;\\\
\\{c^{-1}({G^{\mathcal{L}}})\,|\,c^{-1}({G^{\mathcal{R}}})\\}\quad\quad\textrm{
\emph{otherwise}.}\end{cases}$

Now, we prove that $c^{-1}c=\mathbf{1}$, where $\mathbf{1}$ is the identity
map. Proving that $cc^{-1}$ is also the identity is entirely analogous, so we
refrain from doing it. The proof is carried out for each of the four cases of
the definitions of the maps, using induction:

$\bullet\,\,cc^{-1}(0)=cc^{-1}(\\{\emptyset\,|\,\emptyset\\})=c(\\{\overline{\infty}\,|\,\infty\\})=\\{\emptyset\,|\,\emptyset\\}=0$;

$\bullet\,\,cc^{-1}(\\{\emptyset\,|\,{G^{\mathcal{R}}}\\})=c(\\{\overline{\infty}\,|\,c^{-1}({G^{\mathcal{R}}})\\})=\\{\emptyset\,|\,cc^{-1}({G^{\mathcal{R}}})\\}\underbrace{=}_{\text{ind.}}\\{\emptyset\,|\,{G^{\mathcal{R}}}\\}$;

$\bullet\,\,cc^{-1}(\\{{G^{\mathcal{L}}}\,|\,\emptyset\\})=c(\\{c^{-1}({G^{\mathcal{R}}})\,|\,\infty\\})=\\{cc^{-1}({G^{\mathcal{L}}})\,|\,\emptyset\\}\underbrace{=}_{\text{ind.}}\\{{G^{\mathcal{L}}}\,|\,\emptyset\\}$;

$\bullet\,\,cc^{-1}(\\{{G^{\mathcal{L}}}\,|\,{G^{\mathcal{R}}}\\})=c(\\{c^{-1}({G^{\mathcal{L}}})\,|\,c^{-1}({G^{\mathcal{R}}})\\})=\\{cc^{-1}({G^{\mathcal{L}}})\,|\,cc^{-1}({G^{\mathcal{R}}})\\}\underbrace{=}_{\text{ind.}}\\{{G^{\mathcal{L}}}\,|\,{G^{\mathcal{R}}}\\}$.
∎

######  Observation 47.

The game tree of a Conway form has no small arrows and is identical to the
game tree of the corresponding form in $\mathbbm{N}\mathbbm{p}$.

The following theorem constitutes a fundamental result, directly related to
Theorem 18. Since Conway games are invertible, with their inverses being their
conjugates, the comparison involving these elements can be done by playing, in
the traditional way. From now on, for ease, we write $G-H$ instead of
$G+(-H)$, where $-H$ is the conjugate of $H$.

One can also establish a third parallel with what happens on the extended real
number line. In that structure, infinities are the only non-invertible
elements. In $\mathbbm{N}\mathbbm{p}^{\infty}\\!$, non-invertible elements are
those where there is some pressure exerted by an infinity in one of their
followers (this also explains the choice of the term “pressing game”).

######  Theorem 48.

Let $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$. We have that $G$ is invertible
if and only if $G$ is a Conway game. Moreover, if $G$ is a Conway game, then
$G-G=0$.

###### Proof.

($\Leftarrow:$) Suppose first that $G$ is a Conway form. If $G=0$, then $-G=0$
and $G-G=0$. Otherwise, according to Corollary 15, it is sufficient to verify
that $G-G$ is a $\mathcal{P}$-position. If Left, playing first, chooses
$G^{L}-G$, because this game is not $\infty$ ($G$ is not a check), Right can
answer with $G^{L}-G^{L}$ and, by induction, because $G^{L}$ is a Conway form
with no checks as followers, that option is equal to zero. Because of that, by
Corollary 15, that option is a $\mathcal{P}$-position, and Right wins.
Analogous arguments work for the other options of the first player, and so,
$G-G$ is a $\mathcal{P}$-position. Again, by Corollary 15, $G-G=0$.

Suppose now that $G$ is a Conway game, but not a Conway form. Because it is a
Conway game, by definition, it is equal to some
$G^{\prime}\in\mathbbm{N}\mathbbm{p}^{\text{C}}$. The first paragraph proved
that $G^{\prime}-G^{\prime}=0$. Also, by symmetry, $-G$ is equal to
$-G^{\prime}$. Therefore, $G^{\prime}-G^{\prime}=0$ implies $G-G=0$.

($\Rightarrow:$) Let $G\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ be a reduced
form of a pressing game. Suppose that there is
$H\in\mathbbm{N}\mathbbm{p}^{\infty}\\!$ such that $G+H=0$, i.e.,
$G+H\in\mathcal{P}$. Assume that $H$ is also a reduced form and that
$\tilde{b}(G)+\tilde{b}(H)$ is the smallest possible for $G$ and $H$
satisfying these conditions.

If $G$ has no options, $G$ is either $\infty$ or $\overline{\infty}$, and
$G+H\not\in\mathcal{P}$, which is a contradiction. If all options of $G$ are
Conway games, then $G$ itself must be a Conway game, leading to a
contradiction as well. If $G=\\{\overline{\infty}\,|\,\infty\\}$, then $G=0$,
leading to another contradiction. So, without loss of generality, let
$G^{L_{1}}+H$ be a move such that $G^{L_{1}}\neq\overline{\infty}$ is a
pressing game.

Against $G^{L_{1}}+H$, Right must have a winning response less than or equal
to zero. Suppose that Right’s winning answer is some $G^{L_{1}R}+H\leqslant
0$. If so, by adding $G$ to both sides, $G^{L_{1}R}\leqslant G$ is obtained.
This contradicts the assumption (reduced forms), as if $G^{L_{1}R}$ is
$\overline{\infty}$, then $G^{L_{1}}$ is absorbing reversible, and if
$G^{L_{1}R}$ is not $\overline{\infty}$, then $G^{L_{1}}$ is open reversible.

Suppose now Right’s winning response is some $G^{L_{1}}+H^{R_{1}}\leqslant 0$.
If $G^{L_{1}}+H^{R_{1}}=0$, then it contradicts the assumption once more
(smallest possible $\tilde{b}(G)+\tilde{b}(H)$), since $G^{L_{1}}$ is a
pressing game. Therefore, we have $G^{L_{1}}+H^{R_{1}}<0$.

In order to initiate a “carousel argument”, often used in CGT, consider the
first move by Right in $G+H$ to $G+H^{R_{1}}$. By similar reasons, Left must
have an option $G^{L_{2}}+H^{R_{1}}\geqslant 0$. If $H^{R_{1}}$ is a Conway
game, it is invertible, and we have $G^{L_{1}}<-H^{R_{1}}\leqslant G^{L_{2}}$.
In this case, $G^{L_{1}}$ is a dominated option, contradicting the assumption
(reduced forms). On the other hand, due to the fact that $H^{R_{1}}$ is not a
Conway game, we cannot have $G^{L_{2}}+H^{R_{1}}=0$ as it once again
contradicts the assumption (smallest possible $\tilde{b}(G)+\tilde{b}(H)$).
Thus, we have $G^{L_{2}}+H^{R_{1}}>0$. Turning the carousel, a first move by
Left in $G+H$ to $G^{L_{2}}+H$ must be answered with some
$G^{L_{2}}+H^{R_{2}}<0$. And turning again, a first move by Right in $G+H$ to
$G+H^{R_{2}}$ must be answered with some $G^{L_{3}}+H^{R_{2}}>0$. Note that,
in this case, it is not possible to have $G^{L_{1}}+H^{R_{2}}>0$, since we
would have

  * 1.

$G^{L_{1}}+H^{R_{1}}<0$;

  * 2.

$-G^{L_{2}}-H^{R_{1}}<0$ (since $G^{L_{2}}+H^{R_{1}}>0$);

  * 3.

$G^{L_{2}}+H^{R_{2}}<0$;

  * 4.

$-G^{L_{1}}-H^{R_{2}}<0$ (since $G^{L_{1}}+H^{R_{2}}>0$).

Combining these inequalities by making use of Theorem 20 would lead to the
inequality
$(G^{L_{1}}+H^{R_{1}}+G^{L_{2}}+H^{R_{2}})+(-G^{L_{1}}-H^{R_{1}}-G^{L_{2}}-H^{R_{2}})<0$.
That cannot occur, given that the specified game, due to its symmetry, can
belong to $\mathcal{P}$ or $\mathcal{N}$, but never to $\mathcal{R}$. Entirely
analogous arguments are sufficient to show that the winning responses must
always be obtained with options different from the previous ones. Since the
carousel cannot rotate indefinitely, it is not possible to have $G$ and $H$
such that $G+H\in\mathcal{P}$, i.e., such that $G+H=0$. ∎

We conclude this section by proving the Conway-embedding.

######  Theorem 49 (Conway-Embedding).

For any $G,H\in\mathbbm{N}\mathbbm{p}^{\emph{C}}\\!$, if $G\geqslant H$ then
$c(G)\geqslant c(H)$, where the first is the order relation of
$\mathbbm{N}\mathbbm{p}^{\infty}\\!$ and the second is the order relation of
$\mathbbm{N}\mathbbm{p}$. Similarly, for any $G,H\in\mathbbm{N}\mathbbm{p}$,
if $G\geqslant H$ then $c^{-1}(G)\geqslant c^{-1}(H)$, where the first is the
order relation of $\mathbbm{N}\mathbbm{p}$ and the second is the order
relation of $\mathbbm{N}\mathbbm{p}^{\infty}\\!$.

###### Proof.

Let $G$ and $H$ be two Conway forms such that $G\geqslant H$. According to
Theorem 48 $H$ is invertible and, by Theorem 18, we have
$G-H\in\mathcal{L}\cup\mathcal{P}$. Let us verify that, in
$\mathbbm{N}\mathbbm{p}$, we also have