2. A proof of Theorem 1.2

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

if and only if , for and for some .

(2.5) |

Clearly if and only if , and if and only if and . In particular, for any there exist unique that satisfy (2.1) and (2.2).

(2.8) |

for each

By using Lemma 2.2 and Remark 2.1 for we have for each

for each and , and hence we have .

for any with .

for any with .

for any with .

for any with and .

and .

Let , then and .

We assume that .

Suppose that the following is valid for .

Suppose that there exists such that

Suppose that there exists such that for and .

We define for , , for , and

(2.14) |

and | (2.15) |

Let and . Clearly . By Lemma 2.1 we have , and hence we have of this lemma.

Suppose that . Then by the same method used in we get of this lemma.

Suppose that . Then let , , and for and .

Let and . Then we have .

Then at least one of the following (1), (2), (3) and (4) is true.

for some with .

for some with .

for some with .

for some with and .

If , we define by for and for . Then we have and . Therefore we have (1).

If , we define by for and for . Then we have and . Therefore we have (2).

Next we suppose that and .

*By Lemma 2.4
For
there exists
that satisfies one of the following two conditions
and
.
satisfies
and
, then by
the fact that
for
we have
for
, and hence by the fact that
we have
. Therefore we have
Let
, then
and
. By the fact that
for
we have
for
, and hence by the fact that
we have
. Let
, then we have
.*

Next we are going to define the function
for a state
of chocolates.

is the set of all states that can be reached from the state
in one step (directly).

Therefore we have . It is easy to see that

. It is clear that , since we cannot move to from . Note that , since it will take 2 steps to reach from .

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

(2.16) |

(2.17) |

Since ,by Theorem 2.1 .

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

By Theorem 2.3 and 2.4 we finish the proof of Theorem 1.2. If we start the game with a state
, then by Theorem 2.3 any option by us leads to a state
in
. From this state
by Theorem 2.4 our opponent can choose a proper option that leads to a state in
. Note that any option reduces some of the numbers in the coordinates. In this way our opponent can always reach a state in
, and finally he wins by reaching
. Therefore
is the set of L states.

If we start the game with a state
, then by Theorem 2.4 we can choose a proper option leads to a state
in
. From
any option by our opponent leads to a state in
. In this way we win the game by reaching
. Therefore
is the set of W states.