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Case 5
Next we study the case that
and
is odd.
An example of this case is Example 5.6.
In Fig. 5.23 we suppose that Predictions 5.1, 5.2, 5.3,5.4 and 5.5 are valid for
and for
. We also suppose that these predictions are valid for
and
. Under these assumptions we prove that
.
Remark 5.4
In this case we assume similar conditions that are written in Remark 5.3.
Note that some Grundy numbers
with
are presented in the table. We do not assume the values of these Grundy numbers to prove this prediction, but the authors present these numbers to show the relative positions of
in the table of Grundy numbers.
Here we assume that
, which is the last number of the sequence
that is an arithmetic sequence with common difference of
and its length is
.
Therefore the position of
is in the midst of the the sequence
, and this condition make Case 5 very different from Case 1, Case 2 and Case 3.

(5.61) 

(5.62) 
Clearly

(5.63) 

(5.64) 
By using Prediction 5.3 for
and

(5.65) 

(5.66) 
By (5.61) and (5.63) we have

(5.67) 
By (5.62) and the fact that
we have

(5.68) 
By (5.65) and (5.66) we have

(5.69) 
By using predictions for
and
we know that the sequence of (5.69) starts as an arithmetic sequence with common difference of
whose length is
, and hence the last term of this arithmetic sequence is
. The next term in (5.69) is
, and after that we have an arithmetic sequence with common difference of
up until it reaches
.
Therefore by (5.68), (5.67) and (5.69) we have
.
It is clear that there is no
in
, and hence we have
.
Proof
The outline of the proof is almost clear from Case 1, Case 2, Case 3, Case 4 and Case 5.
Theorem 5.1
The list of Lstates of the game with inequality
is the union of the following lists
,
,
,
,
and
.
.
,
and
.
and
.
,
.
,
and
.
,
and
.
,
and
.
Proof
This is direct from Prediction 5.5.
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