2. A proof of Theorem 1.2

In this section we assume that $x,y,z\in Z_{\geqq0}$ and we write them in base 2, so $x=\displaystyle\sum_{i=0}^{n} x_i2^i$ , $y=\displaystyle\sum_{i=0}^{n} y_i2^i$ and $z=\displaystyle\sum_{i=0}^{n} z_i2^i$ with $x_{i},y_{i},z_{i}\in \{0,1\}$ , where $n = log_2 max(x,y,z)$ .

In this section we prove Theorem 1.2, and we need several facts about the relations between numbers in base $2$ , the nim-sum of numbers and the floor function.

Lemma 2.1   $(1)$ $y=\left\lfloor \frac{z}{2} \right\rfloor$ if and only if $y_{n} = 0$ and $y_{i} = z_{i+1}$ for $i = 0,1,2,...,n-1$ .
$(2)$ $y < \left\lfloor \frac{z}{2} \right\rfloor$ if and only if $y_{n} = 0$ , $y_{i} = z_{i+1}$ for $i = m+1,m+2,...,n-1$ and $y_{m} = 0 < z_{m+1} = 1$ for some $m\in Z_{\geq 0}$ .

Proof    When we write $y,z$ in base 2, $(1)$ and $(2)$ of this lemma is clear from the definition of floor function $\left\lfloor \ \right\rfloor$ .

Lemma 2.2   We suppose that

$$x\oplus y\oplus z=0.$$ (2.1)

$(1)$ Then

$$y= \left\lfloor \frac{z}{2} \right\rfloor$$ (2.2)

if and only if $x_{n} = z_{n} = 1, y_{n} = 0$ and for each $i = 0,1,2,...,n-1$

$$y_{i}=z_{i+1}$$ (2.3)

and

$$z_{i}=x_{i}+y_{i} \ (mod \ 2).$$ (2.4)

$(2)$

$$y< \left\lfloor \frac{z}{2} \right\rfloor$$ (2.5)

if and only if there exists $m\in Z_{\geqq0}$ such that $x_{n} = z_{n} = 1, y_{n} = 0$ ,(2.4) is valid for each $i = 0,1,2,...,n-1$ , (2.3) is valid for each $i = m+1,m+2,...,n-1$ and $y_{m} = 0 < z_{m+1} = 1$ .

Proof    $(1)$ and $(2)$ are direct from Lemma 2.1 and the Definition 1.3 of nim-sum.

Remark 2.1   Suppose that $x\oplus y\oplus z= 0$ and $y \leq \left\lfloor \frac{z}{2} \right\rfloor$ , then by using (2.3) and (2.4) repeatedly we get $z_{i}=\displaystyle\sum_{j=i}^{n} x_j \ (mod\ 2)$ and $y_{i}=\displaystyle\sum_{j=i+1}^{n} x_j \ (mod\ 2)$ for $i = k,k+1,k+2,...n-1$ with some $k \in Z_{\geqq0}$ .
Clearly $y=\left\lfloor \frac{z}{2} \right\rfloor$ if and only if $k = 0$ , and $y < \left\lfloor \frac{z}{2} \right\rfloor$ if and only if $k > 0$ and $y_{k-1} < z_k$ . In particular, for any $x$ there exist unique $y,z$ that satisfy (2.1) and (2.2).

Lemma 2.3   We suppose that

$$x\oplus y\oplus z=0 y = \left\lfloor \frac{z}{2} \right\rfloor.$$ (2.6)

If there exist $v,w\in Z_{\geq 0}$ such that $x\oplus v\oplus w=0$ and $v<\left\lfloor \frac{w}{2} \right\rfloor$ ,then $v<y$ .

Proof    By Lemma 2.2 we have ${x_n} = {z_{_n}} = 1,{y_n} = 0$ and for each $i = 0,1,2,...,n-1$

$${y_i} = {z_{i + 1}},{z_i} = {x_i} + {y_i}\ (mod\ 2).$$ (2.7)

Suppose that there exist $v,w\in Z_{\geq 0}$ such that

$$x\oplus v\oplus w=0\ and\ v < \left\lfloor {w/2} \right\rfloor.$$ (2.8)

Clearly $v_{n}=0$ and $w_n=1$ . By Lemma 2.2 and Remark 2.1 there exists $m \in {Z_{ \ge 0}}$ such that
for each $i=m+1,m+2,...n-1$

$${w_i} = \sum\limits_{j = i}^n {{x_i}}\ (mod\ 2),{v_i} = \sum\limits_{j = i + 1}^n {{x_i}}\ (mod\ 2)$$ (2.9)

and $v_m<w_{m+1}$ .
By using Lemma 2.2 and Remark 2.1 for $x,y,z$ we have for each $i = 0,1,2,...,n-1$

$${z_i} = \sum\limits_{j = i}^n {{x_i}}\ (mod\ 2),{y_i} = \sum\limits_{j = i + 1}^n {{x_i}}\ (mod\ 2).$$ (2.10)

Then by (2.9) and (2.10) we have
$v_i=y_i$ for each $i=m+1,m+2,...n-3,n-2,n-1$ and $v_m<y_m$ , and hence we have $v<y$ .

Theorem 2.1   If $x\oplus y\oplus z= 0$ and $y \le \left\lfloor {z/2} \right\rfloor$ ,then
$(1)$ $u \oplus y \oplus z \ne 0$ for any $u\in Z_{\geq 0}$ with $u<x$ .
$(2)$ $x \oplus v \oplus z \ne 0$ for any $v\in Z_{\geq 0}$ with $v<y$ .
$(3)$ $x \oplus y \oplus w \ne 0$ for any $w\in Z_{\geq 0}$ with $w<z$ .
$(4)$ $x \oplus v \oplus w \ne 0$ for any $v,w\in Z_{\geq 0}$ with $v<y,w<z$ and $v = \left\lfloor {w/2} \right\rfloor$ .

Proof (1)   , (2) and (3) are direct from Theorem 1.1. We suppose that $x\oplus v\oplus w=0$ and $v = \left\lfloor {w/2} \right\rfloor$ for some $v,w \in {Z_{ \ge 0}}$ with $v<y,w<z$ . If $y < \left\lfloor {z/2} \right\rfloor$ , then by Lemma 2.3 we have $y<v$ , and this contradicts the fact $v<y$ . If $y = \left\lfloor {z/2} \right\rfloor$ , then by Remark 2.1 we have $y=v$ , and this contradicts the fact $v<y$ . Therefore $x \oplus v \oplus w \ne 0$ .

Lemma 2.4   For any $x,y \in {Z_{ \ge 0}}$ there exists $z$ that satisfies one of the following two conditions.
$[i]$ $x\oplus y\oplus z= 0$ and $y \le \lfloor {z/2} \rfloor$ .
$[ii]$ Let $y^\prime = \lfloor z/2 \rfloor$ , then $y^\prime < y$ and $x\oplus y^\prime \oplus z=0$ .

Proof   Let

$$u_i = x_i+y_i \ (mod \ 2)$$ (2.11)

for $i = 0,1,2,...,n$ .
$[a]$   We assume that $y_n = 0$ .
$[a-1]$ Suppose that the following $(\ref{condition1})$ is valid for $i = 0,1,...,n-1$ .

$$y_i = u_{i+1}.$$ (2.12)

Then let $z =u = \sum\limits_{i = 0}^n {{u_i}} {2^i}$ . By $(1)$ of Lemma 2.1 we have $y = \lfloor {z/2} \rfloor$ , and hence by $(\ref{condition12a})$ we have $[i]$ of this lemma.
$[a-2]$ Suppose that there exists $m$ such that

$$y_m=0 < 1 = u_{m+1},$$ (2.13)

and $(\ref{condition1})$ is valid for $i = m+1,m+2,...,n-1$ . Then let $z =u = \sum\limits_{i = 0}^n {{u_i}} {2^i}$ . By $(2)$ of Lemma 2.1 we have $y < \lfloor {z/2} \rfloor$ , and hence we have $[i]$ of this lemma.
$[a-3]$ Suppose that there exists $m$ such that $y_i$ $=u_{i+1}$ for $i = m+1,m+2,...,n-1$ and $y_m=1>$ $0 = u_{m+1}$ .
We define $y^\prime _i = y_i$ for $i = m+1,m+2,...,n$ , $y^\prime _m = 0$ , $z_i = u_i$ for $i = m+1,m+2,...,n$ , $z_m = y^\prime _m + x_m$ $(mod 2)$ and

$$y^\prime _i = z_{i+1}$$ (2.14)

$$z_i = x_i + y^\prime _i \ (mod \ 2)$$ (2.15)

for $i = 0,1,2,...,m-1$ .
Let $z = \sum\limits_{i = 0}^n {{z_i}} {2^i}$ and $y^\prime = \sum\limits_{i = 0}^n {{y^\prime _i}} {2^i}$ . Clearly $y^\prime < y$ . By Lemma 2.1 we have $y^\prime = \lfloor z/2 \rfloor$ , and hence we have $[ii]$ of this lemma.
$[a-4]$ Suppose that $y_{n-1} = 1 > 0 = u_n$ . Then by the same method used in $[a-3]$ we get $[ii]$ of this lemma.
$[b]$ Suppose that $y_n=1$ . Then let $y^\prime _n = 0$ , $z_n = x_n + y^\prime _n \ (mod \ 2 )$ , and for $i = 0,1,2,...,n-1$ $y^\prime _i =z_{i+1}$ and $z_i = y^\prime _i + x_i$ $(mod 2)$ .
Let $z = \sum\limits_{i = 0}^n {{z_i}} {2^i}$ and $y^\prime = \sum\limits_{i = 0}^n {{z^\prime _i}} {2^i}$ . Then we have $[ii]$ .

Theorem 2.2   Suppose that $x \oplus y \oplus z \ne 0$ and $y \le \left\lfloor {z/2} \right\rfloor$ .
Then at least one of the following (1), (2), (3) and (4) is true.
$(1)$ $u \oplus y \oplus z = 0$ for some $u \in {Z_{ \ge 0}}$ with $u<x$ .
$(2)$ $x \oplus v \oplus z = 0$ for some $v \in {Z_{ \ge 0}}$ with $v<y$ .
$(3)$ $x \oplus y \oplus w = 0$ for some $w \in {Z_{ \ge 0}}$ with $w<z$ .
$(4)$ $x\oplus v\oplus w=0$ for some $v,w \in {Z_{ \ge 0}}$ with $v<y,w<z$ and $v = \left\lfloor {w/2} \right\rfloor$ .

Proof   Suppose that $x_i +y_i +z_i =0\ (mod\ 2)$ for $i = m+1,m+2,...,n$ and $x_m +y_m +z_m \neq 0\ (mod\ 2)$ .
$(a)$ If $x_m=1$ , we define $u= \sum_{i=0}^{n} u_i 2^i$ by $u_i=x_i$ for $i=m+1,m+2,...n,u_m=0<x_m$ and $u_i=y_i+z_i$ for $i=0,1,...,m-1$ . Then we have $u \oplus y \oplus z \oplus =0$ and $u<x$ . Therefore we have (1).
$(b)$ If $y_m=1$ , we define $v= \sum_{i=0}^{n} v_i 2^i$ by $v_i=y_i$ for $i=m+1,m+2,...n,v_m=0<y_m$ and $v_i=x_i+z_i$ for $i=0,1,...,m-1$ . Then we have $x \oplus v \oplus z \oplus =0$ and $v<y$ . Therefore we have (2).
$(c)$ Next we suppose that $x_m=y_m=0$ and $z_m=1$ .

By Lemma 2.4 For $x,y$ there exists $w= \sum_{i=0}^{n} w_i 2^i$ that satisfies one of the following two conditions $[i]$ and $[ii]$ .
$[i]$ $\{x,y,w\}$ satisfies $x \oplus y \oplus w = 0$ and $y \le \lfloor {w/2} \rfloor$ , then by the fact that $x_i +y_i +z_i =0\ (mod\ 2)$ for $i = m+1,m+2,...,n$ we have $w_i = z_i$ for $i = m+1,m+2,...,n$ , and hence by the fact that $w_m = x_m + y_m = 0 < 1 = z_m$ we have $w<z$ . Therefore we have $(3)$
$[ii]$ Let $y^\prime = \lfloor w/2 \rfloor$ , then $y^\prime < y$ and $x\oplus y^\prime \oplus w=0$ . By the fact that $x_i +y_i +z_i =0\ (mod\ 2)$ for $i = m+1,m+2,...,n$ we have $w_i = z_i$ for $i = m+1,m+2,...,n$ , and hence by the fact that $w_m = x_m + y_m = 0 < 1 = z_m$ we have $w<z$ . Let $v = y^\prime$ , then we have $(4)$ .

Next we are going to define the function $move (\{x, y, z\})$ for a state $\{x, y, z\}$ of chocolates.
$move (\{x, y, z\})$ is the set of all states that can be reached from the state $\{x, y, z\}$ in one step (directly).

Definition 2.1   For $x,y,z \in Z_{\ge 0}$ we define $move(\{x,y,z\})=\{\{u,y,z \};u<x \} \cup \{\{x,v,z \};v<y \} \cup \{ \{x,y,w \};w<z \} \cup \{ \{x,min(y, \lfloor w/2 \rfloor ),w \};w<z \}$ , where $u,v,w \in Z_{\ge 0}$ .

Example 2.1   We study the function $move (\{x, y, z\})$ using examples of states in Example 1.2 . If we start with the state $\{x,y,z\}=\{2,2,5\}$ and reduce $z=5$ to $w=3$ , then the second coordinate will be $min(2, \lfloor 3/2 \rfloor )=min(2,1)=1$ .
Therefore we have $\{2,1,3\} \in move(\{ 2,2,5 \})$ . It is easy to see that
$\{0,2,5\},\{2,0,5\} \in move(\{2,2,5\})$ . It is clear that $\{0,2,5\} \notin move(\{2,1,3\})$ , since we cannot move to $\{0, 2, 5\}$ from $\{2, 1, 3\}$ . Note that $\{0,1,5\} \notin move(\{2,2,5\})$ , since it will take 2 steps to reach $\{0, 1, 5\}$ from $\{2,2,5\}$ .

Next we prove that if you start with an element of $A_2$ , then any move leads to an element of $B_2$ .

Theorem 2.3   For any $\{x,y,z\} \in A_2$ we have $move(\{x,y,z\}) \subset B_2$ .

Proof   Let $\{x,y,z\} \in A_2$ , then we have

$$x \oplus y \oplus z \oplus =0$$ (2.16)

and

$$y \leq \left\lfloor z/2 \right\rfloor.$$ (2.17)

Suppose that we move from $\{x, y, z\}$ to $\{p,q,r\}$ ,i.e., $\{p,q,r\} \in move(\{x,y,z \})$ . Next we prove that $\{p,q,r\} \in B_2$ .
Since $move(\{x,y,z\})=\{\{u,y,z\};u<x\} \cup \{\{x,y,z\};v<y\} \cup \{ \{x,y,w \};w<z \} \cup \{\{x,min(y,\lfloor w/2 \rfloor ),w\};w<z\}$ ,by Theorem 2.1 $p \oplus q \oplus r \oplus \ne 0$ .

Next we prove that if you start with an element of $B_2$ , then there is a proper move that leads to an element of $A_2$ .

Theorem 2.4   Let $\{x,y,z\} \in B_2$ , then $move(\{x,y,z\})\cap A_2 \ne \phi$ .

Proof   Let $\{x,y,z\} \in B_2$ . Then have

$$x \oplus y \oplus z \oplus \ne 0$$ (2.18)

and

$$y \leq \left\lfloor z/2 \right\rfloor.$$ (2.19)

Since $move(\{x,y,z\})= \{\{u,y,z\};u < x\} \cup \{\{x,v,z\};v<y\}$ $\cup \{ \{x,y,w \};w<z \} \cup \{\{x,min\left( y,\lfloor \frac{w}{2} \rfloor\right),w\};w<z\}$ , by Theorem 2.2 there exists $\{p,q,r\} \in move\left(\{x,y,z\}\right)$ such that
$p\oplus q\oplus r= 0$ . Therefore $\{p,q,r\} \in move(\{x,y,z\})\cap A_2$.

By Theorem 2.3 and 2.4 we finish the proof of Theorem 1.2. If we start the game with a state $\{x,y,z\}\in A_{2}$ , then by Theorem 2.3 any option by us leads to a state $\{p,q,r\}$ in $B_2$ . From this state $\{p,q,r\}$ by Theorem 2.4 our opponent can choose a proper option that leads to a state in $A_2$ . Note that any option reduces some of the numbers in the coordinates. In this way our opponent can always reach a state in $A_2$ , and finally he wins by reaching $\{0,0,0\}\in A_{2}$ . Therefore $A_2$ is the set of L states.
If we start the game with a state $\{x,y,z\}\in B_{2}$ , then by Theorem 2.4 we can choose a proper option leads to a state $\{p,q,r\}$ in $A_2$ . From $\{p,q,r\}$ any option by our opponent leads to a state in $B_2$ . In this way we win the game by reaching $\{0,0,0\}$ . Therefore $B_2$ is the set of W states.