Documentation

CombinatorialGames.Misere.LeftDeadEnd

Left Dead Ends #

The basic theory #

Core predicate, ambient universe, notation #

Comparison of Left dead ends is multiversal, so we fix the dead-ending universe $\mathcal{E}$ as the ambient set, writing $G \ge H$ for $G \ge_{\mathcal{E}} H$ and $G = H$ for $G =_{\mathcal{E}} H$.

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      theorem GameForm.LeftDeadEnd.IsLde.add {g h : GameForm} (hg : IsLde g) (hh : IsLde h) :
      IsLde (g + h)
      theorem GameForm.LeftDeadEnd.isLde_sum {s : Multiset GameForm} (hs : ∀ a ∈ s, IsLde a) :

      If $\mathcal{U}$ is a universe, and $G$ and $H$ are Left dead ends, then $G \ge_{\mathcal{U}} H$ if and only if the following hold:

      1. if $G \cong 0$, then $H \cong 0$; otherwise,

      2. for every $G^R \in G^{\mathcal{R}}$, there exists some $H^R \in H^{\mathcal{R}}$ such that $G^R \ge_{\mathcal{U}} H^R$.

      This is Davies (Theorem 2.4 on p. 7).

      Runs #

      theorem GameForm.LeftDeadEnd.IsLde.of_mem_isRun {g : GameForm} (hg : IsLde g) {l : List GameForm} (hl : IsRun (g :: l)) (x : GameForm) :
      x ∈ g :: l → IsLde x
      theorem GameForm.LeftDeadEnd.exists_run_dominated {g h : GameForm} (hg : IsLde g) (hh : IsLde h) (hge : g ≥m Form.DeadEnding.ShortDeadEnding h) {l : List GameForm} (hl : IsRun (g :: l)) :
      ∃ (m : List GameForm), IsRun (h :: m) ∧ List.Forall₂ (fun (a b : GameForm) => a ≥m Form.DeadEnding.ShortDeadEnding b) (g :: l) (h :: m)

      If $G$ and $H$ are Left dead ends with $G \ge H$, and $k$ is an integer satisfying $0 \le k \le \tilde{b}(G)$, then, for any run $(G_0, G_1, \ldots, G_k)$ of $G$ of length $k$, there exists a run $(H_0, H_1, \ldots, H_k)$ of $H$ of length $k$ such that $G_i \ge H_i$ for every $0 \le i \le k$.

      This is Davies (Corollary 3.2 on p. 10).

      Good options #

      If $G$ is a combinatorial game, then we will say that a (Left, respectively Right) option $G'$ of $G$ is good if it is not strictly dominated by another (Left, respectively Right) option of $G$.

      This is Davies (Definition 3.3 on p. 11).

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        If $G \ge H$ and $G^R$ is a good option of $G$, then some (good) option $H^R$ of $H$ has $G^R \ge H^R$.

        This is the $\ge$ case of Davies (Corollary 3.4 on p. 11).

        If $G = H$ and $G^R$ is a good option of $G$, then some (good) option $H^R$ of $H$ has $G^R = H^R$.

        This is the $=$ case of Davies (Corollary 3.4 on p. 11).

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          The monoid of Left dead ends $\mathcal{L}$ is reduced.

          This is Davies (Theorem 3.1 on p. 10).

          Terminal lengths, race and birthday #

          @[irreducible]

          If $G$ is a short game, then we define the set of terminal lengths of $G$, denoted $T(G)$, as the set of all $n$ such that there exists a terminal run of $G$ of length $n$.

          This is Davies (Definition 3.12 on p. 14), taken here in the equivalent recursive form of Lemma 3.13.

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            If $G$ is a non-zero Left dead end, then $T(G) = \bigcup (T(G') + 1)$, where $G'$ ranges over all options of $G$.

            This is Davies (Lemma 3.13 on p. 14).

            If $G$ is a short game, define the race of $G$, denoted $r(G)$, as $r(G) := \min(T(G))$.

            This is Davies (Definition 3.14 on p. 15).

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              A Left dead end is an integer if it is isomorphic either to $0 := \{\cdot \mid \cdot\}$ or to $\overline{n} := \{\cdot \mid \overline{n - 1}\}$ for some $n \ge 1$ (where $\overline{0} := 0$).

              This is Davies (p. 15).

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                @[irreducible]

                If $G$ is a Left dead end, then we define the flexibility of $G$, written $f(G)$, recursively as $f(G) = 1 + \max(f(G'))$, where $G'$ ranges over all options of $G$, when $G$ is not an integer, and as $f(G) := 0$ otherwise.

                This is Davies (Definition 5.1 on p. 22).

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                  @[irreducible]

                  A Left dead end $G$ is an integer if and only if $|T(G)| = 1$. In particular, $G = \overline{n}$ if and only if $T(G) = \{n\}$.

                  This is Davies (Lemma 3.15 on p. 15).

                  @[irreducible]

                  If $G$ and $H$ are Left dead ends with $G \ge H$, then $T(G) \subseteq T(H)$.

                  This is Davies (Theorem 3.17 on p. 16).

                  If $G$ is a combinatorial game and $(G_0, G_1, \ldots, G_k)$ is a run of $G$ such that $G_k \cong 0$, then we say the run is terminal.

                  This is Davies (Definition 3.11 on p. 14).

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                    @[irreducible]

                    $n \in T(G)$ if and only if there exists a terminal run of $G$ of length $n$.

                    This is Davies (Definition 3.12 on p. 14), recovered as a theorem from the recursive definition of terminalLengths.

                    theorem GameForm.LeftDeadEnd.exists_run_terminalLengths_subset {g h : GameForm} (hg : IsLde g) (hh : IsLde h) (hge : g ≥m Form.DeadEnding.ShortDeadEnding h) {l : List GameForm} (hl : IsRun (g :: l)) :
                    ∃ (m : List GameForm), IsRun (h :: m) ∧ List.Forall₂ (fun (a b : GameForm) => terminalLengths a ⊆ terminalLengths b) (g :: l) (h :: m)

                    If $G$ and $H$ are Left dead ends with $G \ge H$, and $k$ is an integer satisfying $0 \le k \le \tilde{b}(G)$, then, for any run $(G_0, G_1, \ldots, G_k)$ of $G$ of length $k$, there exists a run $(H_0, H_1, \ldots, H_k)$ of $H$ of length $k$ such that $T(G_i) \subseteq T(H_i)$ for every $0 \le i \le k$.

                    This is Davies (Corollary 3.19 on p. 17).

                    theorem GameForm.LeftDeadEnd.race_ge_of_ge {g h : GameForm} (hg : IsLde g) (hh : IsLde h) (hge : g ≥m Form.DeadEnding.ShortDeadEnding h) :

                    If $G$ and $H$ are Left dead ends with $G \ge H$, then $r(G) \ge r(H)$.

                    This is the race half of Davies (Corollary 3.18 on p. 16).

                    @[irreducible]

                    On a Left dead end, $b(G) = \max(T(G))$.

                    If $G$ and $H$ are Left dead ends with $G \ge H$, then $b(G) \le b(H)$.

                    This is the birthday half of Davies (Corollary 3.18 on p. 16).

                    If $G$ is a Left dead end, then the following are equivalent: $G = 0$; $G \cong 0$; $0 \in T(G)$; $r(G) = 0$; $b(G) = 0$.

                    This is Davies (Lemma 3.16 on p. 15).

                    @[irreducible]
                    theorem GameForm.LeftDeadEnd.terminalLengths_add {g h : GameForm} (hg : IsLde g) (hh : IsLde h) :
                    terminalLengths (g + h) = {x : ℕ | ∃ a ∈ terminalLengths g, ∃ b ∈ terminalLengths h, a + b = x}

                    If $G$ and $H$ are Left dead ends, then $T(G + H) = T(G) + T(H)$, a Minkowski sum.

                    This is Davies (Proposition 3.20 on p. 18).

                    @[irreducible]
                    theorem GameForm.LeftDeadEnd.race_add {g h : GameForm} (hg : IsLde g) (hh : IsLde h) :
                    race (g + h) = race g + race h

                    If $G$ and $H$ are Left dead ends, then $r(G + H) = r(G) + r(H)$.

                    This is the race half of Davies (Proposition 3.20 on p. 18).

                    If $G$ and $H$ are Left dead ends, then $|T(G + H)| \ge \max(|T(G)|, |T(H)|)$.

                    This is the inequality half of Davies (Corollary 3.21 on p. 18).

                    If $G$ and $H$ are Left dead ends, then $|T(G + H)| \ge \max(|T(G)|, |T(H)|)$, with equality if and only if at least one of $G$ and $H$ is an integer.

                    This is Davies (Corollary 3.21 on p. 18).

                    Pocancellativity and cancellativity #

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                        theorem GameForm.LeftDeadEnd.lde_ge_of_add_pocancellative {A : GameForm → Prop} {g h j : GameForm} (hgA : A g) (hhA : A h) (hj : IsPocancellativeIn A j) (hge : (g + j) ≥m A h + j) :
                        g ≥m A h

                        If $\mathcal{U}$ is a universe, $G, H \in \mathcal{U}$ are Left dead ends, and $J$ is a pocancellative element of $\mathcal{U}$ satisfying $G + J \ge H + J$, then $G \ge H$.

                        This is Davies (Lemma 3.5 on p. 12).

                        theorem GameForm.LeftDeadEnd.lde_misereEQ_of_add_cancellative {A : GameForm → Prop} {g h j : GameForm} (hgA : A g) (hhA : A h) (hj : IsCancellativeIn A j) (heq : (g + j) =m A h + j) :
                        g =m A h

                        If $\mathcal{U}$ is a universe, $G, H \in \mathcal{U}$ are Left dead ends, and $J$ is a cancellative element of $\mathcal{U}$ satisfying $G + J = H + J$, then $G = H$.

                        This is Davies (Lemma 3.6 on p. 12).

                        @[irreducible]

                        If $G$ is a dead Right end and $H$ is a dead Left end, then $o^-(G + H)$ is $\mathcal{N}^-$ when $l(G) = r(H)$, $\mathcal{L}^-$ when $l(G) < r(H)$, and $\mathcal{R}^-$ when $l(G) > r(H)$.

                        This is Milley, Renault (Lemma 5 on p. 7).

                        @[irreducible]

                        If $G$ is a dead end then $G + \overline{G} \equiv 0 \pmod{\mathcal{E}}$.

                        This is Milley, Renault (Theorem 6 on p. 8).

                        theorem GameForm.LeftDeadEnd.lde_pocancellative {g h k : GameForm} (hg : IsLde g) (hh : IsLde h) (hk : IsLde k) (hge : (g + k) ≥m Form.DeadEnding.ShortDeadEnding h + k) :

                        The monoid of Left dead ends $\mathcal{L}$ is pocancellative.

                        This is Davies (Theorem 3.7 on p. 12).

                        The monoid of Left dead ends $\mathcal{L}$ is cancellative.

                        This is Davies (Theorem 3.7 on p. 12).

                        theorem GameForm.LeftDeadEnd.lde_nsmul_cancel {g h : GameForm} (hg : IsLde g) (hh : IsLde h) {n : ℕ} (hn : 0 < n) (heq : (n • g) =m Form.DeadEnding.ShortDeadEnding n • h) :

                        If $G$ and $H$ are Left dead ends with $n \cdot G = n \cdot H$ for some $n > 0$, then $G = H$.

                        This is Davies (Proposition 3.8 on p. 13).

                        Divisibility, atoms, primes and factorisations #

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                          In a monoid $M$, we call a non-invertible element $m \in M$ a molecule if $m$ is atomic but not an atom.

                          This is Davies (Definition 2.6 on p. 9).

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                                The monoid of Left dead ends $\mathcal{L}$ is atomic.

                                This is Davies (Lemma 3.23 on p. 19).

                                theorem GameForm.LeftDeadEnd.birthday_lt_of_add {g h k : GameForm} (hg : IsLde g) (hh : IsLde h) (hk : IsLde k) (hk0 : k ≠ 0) (heq : g =m Form.DeadEnding.ShortDeadEnding h + k) :

                                If $G$, $H$, and $K$ are non-zero Left dead ends with $G = H + K$, then $b(H) < b(G)$.

                                This is Davies (Lemma 3.22 on p. 19).

                                The monoid of Left dead ends $\mathcal{L}$ is a BF-monoid. In particular, if $G$ is a Left dead end, then the length of each factorisation of $G$ is at most $r(G)$; that is, $\overline{\mathsf{L}(G)} \le r(G)$.

                                This is Davies (Lemma 3.24 on p. 19).

                                If $G$ is a Left dead end with $r(G) = 1$, then $G$ is an atom.

                                This is Davies (Corollary 3.25 on p. 20).

                                The monoid of Left dead ends $\mathcal{L}$ is an FF-monoid.

                                This is Davies (Theorem 3.26 on p. 20).

                                The prime $\overline{1}$ #

                                @[irreducible]

                                If $G$ is a Left dead end, then $\overline{1} + G' \ge G$ for every option $G'$ of $G$.

                                This is Davies (Lemma 4.1 on p. 20).

                                If $G$ is a Left dead end, then $\overline{1} + G = \{\cdot \mid G\}$.

                                This is Davies (Proposition 4.2 on p. 21).

                                If $G$ and $G'$ are Left dead ends, then $G$ has exactly one good option (equal to) $G'$ if and only if $G = \overline{1} + G'$.

                                This is Davies (Proposition 4.3 on p. 21).

                                The game $\overline{1}$ is prime in the monoid of Left dead ends $\mathcal{L}$.

                                This is Davies (Theorem 4.4 on p. 21).

                                theorem GameForm.LeftDeadEnd.neg_intCast_add (a b : ℕ) :
                                -↑(a + b) = -↑a + -↑b

                                Flexibility and the atom criteria #

                                If $G$ and $H$ are Left dead ends with $f(G) \le 1$ and $T(G) \subseteq T(H)$, then $G \ge H$.

                                This is Davies (Proposition 5.2 on p. 23).

                                If $G$ is a Left dead end, then we say an option $G'$ of $G$ is versatile if $f(G') \ge f(G) - 1$.

                                This is Davies (Definition 5.3 on p. 23).

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                                  If $G$ is a Left dead end, then $f(G) \le b(G)$, with equality if and only if $G = 0$.

                                  This is Davies (Lemma 5.4 on p. 24).

                                  @[irreducible]

                                  If $G$ and $H$ are Left dead ends with $G \ge H$, then $f(G) \le f(H)$.

                                  This is Davies (Proposition 5.5 on p. 24).

                                  If $G$ is a Left dead end, then we say a run $(G_0, G_1, \ldots, G_n)$ of $G$ of length $n > 0$ is flexible if $G_i$ is not an integer for $i < n$. We define all runs of length $0$ to be flexible.

                                  This is Davies (Definition 5.6 on p. 25).

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                                    A flexible run is maximal if it ends at an integer; that is, if it cannot be extended to a longer flexible run.

                                    This is Davies (Definition 5.6 on p. 25).

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                                      If $G$ is a Left dead end, then $f(G)$ is equal to the length of a longest flexible run of $G$.

                                      This is Davies (Lemma 5.7 on p. 25).

                                      If $G$ is a Left dead end, then we define a racing option of $G$ to be an option $G'$ such that $r(G') = r(G) - 1$.

                                      This is Davies (Definition 5.8 on p. 25).

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                                        If $G$ is a Left dead end, then we define a stalling option of $G$ to be an option $G'$ such that $b(G') = b(G) - 1$.

                                        This is Davies (Definition 5.9 on p. 25).

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                                          theorem GameForm.LeftDeadEnd.exists_isRacingOption {g : GameForm} (hg : IsLde g) (hg0 : g ≠ 0) :
                                          ∃ (gr : GameForm), IsRacingOption gr g

                                          If $G$ is a non-integer Left dead end, then $f(G + \overline{n}) = f(G) + n$ for all $n \ge 0$.

                                          This is Davies (Proposition 5.10(1) on p. 26).

                                          @[irreducible]
                                          theorem GameForm.LeftDeadEnd.flexibility_add {g h : GameForm} (hg : IsLde g) (hh : IsLde h) (hgi : ¬IsInteger g) (hhi : ¬IsInteger h) :

                                          If $G$ and $H$ are non-integer Left dead ends, then $f(G + H) = \max(b(G) + f(H), f(G) + b(H))$.

                                          This is Davies (Proposition 5.10(2) on p. 26).

                                          theorem GameForm.LeftDeadEnd.birthday_lt_flexibility_of_ge {g h k : GameForm} (hg : IsLde g) (hh : IsLde h) (hk : IsLde k) (hh0 : h ≠ 0) (hk_ni : ¬IsInteger k) (hge : (h + k) ≥m Form.DeadEnding.ShortDeadEnding g) :

                                          If $G$, $H$, and $K$ are Left dead ends such that $H$ is non-zero, $K$ is not an integer, and $H + K \ge G$, then $b(H) < f(G)$.

                                          This is Davies (Proposition 5.11 on p. 27).

                                          theorem GameForm.LeftDeadEnd.not_isInteger_of_multi_good {g h k : GameForm} (hg : IsLde g) (hh : IsLde h) (hk : IsLde k) (hk0 : k ≠ 0) (hmulti : ∃ (a : GameForm) (b : GameForm), IsGoodOption a g ∧ IsGoodOption b g ∧ ¬a =m Form.DeadEnding.ShortDeadEnding b) (heq : g =m Form.DeadEnding.ShortDeadEnding h + k) :
                                          theorem GameForm.LeftDeadEnd.birthday_lt_flexibility_of_multi_good {g h k : GameForm} (hg : IsLde g) (hh : IsLde h) (hk : IsLde k) (hh0 : h ≠ 0) (hk0 : k ≠ 0) (hmulti : ∃ (a : GameForm) (b : GameForm), IsGoodOption a g ∧ IsGoodOption b g ∧ ¬a =m Form.DeadEnding.ShortDeadEnding b) (heq : g =m Form.DeadEnding.ShortDeadEnding h + k) :

                                          If $G$ is a Left dead end with more than one good option, then, for any non-zero Left dead ends $H$ and $K$ with $H + K = G$, it holds that $b(H), b(K) < f(G)$.

                                          This is Davies (Corollary 5.12 on p. 28).

                                          If $G$ is a Left dead end with more than one good option, and $b(G) > 2 \cdot f(G) - 2$, then $G$ is an atom.

                                          This is Davies (Corollary 5.13 on p. 28).

                                          If $G$ is a Left dead end with $f(G) = 1$, then $G$ is an atom.

                                          This is Davies (Proposition 5.14 on p. 28).

                                          If $G$ is a Left dead end with $f(G) = 2$, then either $G$ has exactly one good option (i.e. $\overline{1} \mid G$), or $G$ is an atom.

                                          This is Davies (Proposition 5.14 on p. 28).

                                          If $G$ is a Left dead end with $f(G) = 3$, then either $G$ has exactly one good option, $G = 2 \cdot W_2$, or $G$ is an atom.

                                          This is Davies (Proposition 5.14 on p. 28).

                                          Lengths of factorisations #

                                          $\overline{\mathsf{L}(G)}$, the length of a longest factorisation of $G$, with $\overline{\mathsf{L}(0)} = 0$.

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                                            If $G$ is a Left dead end, then $\overline{\mathsf{L}(G)} \le \overline{\mathsf{L}(G')} + 1$ for every good option $G'$ of $G$.

                                            This is Davies (Lemma 6.1 on p. 30).

                                            If $G$ is a Left dead end with exactly one good option $G'$, then $\overline{\mathsf{L}(G)} = \overline{\mathsf{L}(G')} + 1$.

                                            This is Davies (Theorem 6.2 on p. 30).

                                            If $G$ is a Left dead end with more than one good option, then $$\overline{\mathsf{L}(G)} \le \min\left\{r(G),\ \overline{\mathsf{L}(G')} + 1,\ |T(G)| - 1,\ \frac{f(G) + 1}{2}\right\},$$ where $G'$ ranges over all good options of $G$.

                                            This is Davies (Theorem 6.3 on p. 30).

                                            Uniqueness of factorisations #

                                            theorem GameForm.LeftDeadEnd.exists_mem_of_isLdePrime_dvd_sum {p : GameForm} (hp : IsLdePrime p) {s : Multiset GameForm} (hs : ∀ a ∈ s, IsLde a) (hdvd : LdeDvd p s.sum) :
                                            ∃ a ∈ s, LdeDvd p a
                                            theorem GameForm.LeftDeadEnd.isGoodOption_of_add_isGoodOption {g h gr : GameForm} (hg : IsLde g) (hh : IsLde h) (hgr : gr ∈ Moves.moves Player.right g) (hgood : IsGoodOption (gr + h) (g + h)) :

                                            If $G + H$ is a sum of Left dead ends, and $G' + H$ is a good option of $G + H$, then $G'$ is a good option of $G$.

                                            This is Davies (Lemma 7.7 on p. 35).

                                            theorem GameForm.LeftDeadEnd.isGoodOption_add_nsmul_iff {g gr : GameForm} (hg : IsLde g) (hgr : gr ∈ Moves.moves Player.right g) (n : ℕ) :
                                            IsGoodOption (gr + n • g) ((n + 1) • g) ↔ IsGoodOption gr g

                                            If $G$ is a Left dead end, then, for every $n \in \mathbb{N}_1$, $G'$ is a good option of $G$ if and only if $G' + (n - 1) \cdot G$ is a good option of $n \cdot G$.

                                            This is Davies (Lemma 7.13 on p. 38).

                                            If $G = J + p_1 + \cdots + p_m$ with every $p_i$ prime, then every factorisation of $G$ splits as atoms matching the $p_i$ one-for-one together with a factorisation of $J$.

                                            Every longest factorisation of $G$ is of the form $H_1 + \cdots + H_k + K$, where $H_1 + \cdots + H_k$ is a longest factorisation of $H$.

                                            This is Davies (Proposition 7.1(3) on p. 32).

                                            There exists an atom $K$ such that $G = H + K$, and $K$ is terminable.

                                            This is the existence half of Davies (Proposition 7.1(2) on p. 32).

                                            If $G$ is a Left dead end with a good option $H$ such that $\overline{\mathsf{L}(G)} = \overline{\mathsf{L}(H)} + 1$, then

                                            1. $H$ is a racing option of $G$; and

                                            2. there exists a unique atom $K$ such that $G = H + K$, and $K$ is terminable.

                                            This is Davies (Proposition 7.1 on p. 32).

                                            If $G$ is a Left dead end with a good option to an atom $H$, then either $G$ is an atom, or $H$ is a racing option of $G$ and there exists a unique (terminable) atom $K$ such that $G = H + K$; in particular, $G$ is uniquely factorisable.

                                            This is the "in particular" of Davies (Corollary 7.2 on p. 33).

                                            If $G$ is a Left dead end with good options to distinct atoms $H$ and $K$, then either $G$ is an atom, or $H + K$ is the unique factorisation of $G$.

                                            This is Davies (Corollary 7.3 on p. 33).

                                            theorem GameForm.LeftDeadEnd.isLdeAtom_of_good_atoms_of_race_ne_two {g h k : GameForm} (hg : IsLde g) (hh : IsLdeAtom h) (hk : IsLdeAtom k) (hgh : IsGoodOption h g) (hgk : IsGoodOption k g) (hne : ¬h =m Form.DeadEnding.ShortDeadEnding k) (hrace : race g ≠ 2) :

                                            If either of $H$ or $K$ is not terminable (i.e. if $r(G) \ne 2$), then $G$ is an atom.

                                            This is the "in particular" of Davies (Corollary 7.3 on p. 33).

                                            If $G$ is a Left dead end with at least three good options to pairwise distinct atoms, then $G$ is an atom.

                                            This is Davies (Corollary 7.4 on p. 33).

                                            theorem GameForm.LeftDeadEnd.factor_eq_of_good_pf {a : GameForm} {s' t : Multiset GameForm} (hs : IsFactorisation (a ::ₘ s') (a + s'.sum)) (ht : IsFactorisation t (a + s'.sum)) {ar : GameForm} (har : ar ∈ Moves.moves Player.right a) (hgood : IsGoodOption (ar + s'.sum) (a + s'.sum)) (hpf : IsPrimeFactorisable ar) :

                                            If $G := G_1 + \cdots + G_n = H_1 + \cdots + H_m =: H$ are factorisations of a Left dead end $G$, and there is a good option of $G$ playing on the component $G_1$ to some prime-factorisable $G'_1$, then there exists an index $j$ such that $G_1 = H_j$.

                                            This is Davies (Lemma 7.5 on p. 34).

                                            Reading a factorisation as $G_1 + \cdots + G_k$: for every $1 \le i \le k$, there exists a good option of $G_i + \cdots + G_k$ on the component $G_i$ to some prime-factorisable $G'_i$.

                                            This is the hypothesis of Davies (Theorem 7.6 on p. 35).

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                                              If $G := G_1 + \cdots + G_k$ is a factorisation of a Left dead end such that, for every $1 \le i \le k$, there exists a good option of $G_i + \cdots + G_k$ on the component $G_i$ to some prime-factorisable $G'_i$, then $G$ is uniquely factorisable.

                                              This is Davies (Theorem 7.6 on p. 35).

                                              theorem GameForm.LeftDeadEnd.uniquelyFactorisable_S_racing {S : GameForm → Prop} (hS : ∀ (a : GameForm), S a → IsLdeAtom a) (hpf : ∀ (a : GameForm), S a → ∀ (ar : GameForm), IsGoodOption ar a → IsRacingOption ar a → IsPrimeFactorisable ar) {s : Multiset GameForm} (hs : ∀ a ∈ s, S a) :

                                              If $S \subseteq \mathcal{L}$ is a set of atoms such that every good racing option of every $G \in S$ is prime-factorisable, then every $S$-factorisation is uniquely factorisable.

                                              This is Davies (Theorem 7.8 on p. 36).

                                              theorem GameForm.LeftDeadEnd.uniquelyFactorisable_S_stalling {S : GameForm → Prop} (hS : ∀ (a : GameForm), S a → IsLdeAtom a) (hpf : ∀ (a : GameForm), S a → ∀ (ar : GameForm), IsGoodOption ar a → IsStallingOption ar a → IsPrimeFactorisable ar) {s : Multiset GameForm} (hs : ∀ a ∈ s, S a) :

                                              If $S \subseteq \mathcal{L}$ is a set of atoms such that every good stalling option of every $G \in S$ is prime-factorisable, then every $S$-factorisation is uniquely factorisable.

                                              This is Davies (Theorem 7.9 on p. 36).

                                              If $G$ and $H$ are Left dead ends, where $G$ is uniquely factorisable, and $H$ is prime-factorisable, then $G + H$ is uniquely factorisable.

                                              This is Davies (Lemma 7.14 on p. 38).

                                              If $G$ is a Left dead end with a good prime-factorisable option $H$, then $G$ is uniquely factorisable.

                                              This is Davies (Proposition 7.10 on p. 37).

                                              If $G$ is a Left dead end with a good option $\overline{n}$, then either $G = \overline{n + 1}$, or else $G$ is an atom.

                                              This is Davies (Proposition 7.11 on p. 37).

                                              This is Davies (Question 7.12 on p. 38).

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                                              Instances For

                                                If $G$ is a Left dead end that is an atom with a good prime-factorisable option, then $G$ is a strong atom.

                                                This is Davies (Lemma 7.15 on p. 38).

                                                If $G$ is a Left dead end with a good prime-factorisable option, then $n \cdot G$ is uniquely factorisable for all $n$.

                                                This is Davies (Theorem 7.16 on p. 39).

                                                The central conjecture #

                                                A minimal counterexample to unique factorisation has more than one good option.

                                                This is part of the Section 8 discussion in Davies (p. 39).

                                                Every good option of a minimal counterexample to unique factorisation is a molecule.

                                                This is part of the Section 8 discussion in Davies (p. 39).

                                                theorem GameForm.LeftDeadEnd.counterexample_factorisations_disjoint {g : GameForm} (hg : IsLde g) (hmin : ∀ (g' : GameForm), IsLde g' → Form.birthday g' < Form.birthday g → UniquelyFactorisable g') {s t : Multiset GameForm} (hs : IsFactorisation s g) (ht : IsFactorisation t g) (hst : ¬Multiset.Rel (fun (x1 x2 : GameForm) => x1 =m Form.DeadEnding.ShortDeadEnding x2) s t) {x y : GameForm} (hx : x ∈ s) (hy : y ∈ t) :

                                                Two factorisations of a minimal counterexample to unique factorisation share no atom.

                                                This is part of the Section 8 discussion in Davies (p. 39).

                                                A minimal counterexample to unique factorisation has at least three atoms that are factors.

                                                This is part of the Section 8 discussion in Davies (p. 39).