Case 1

In this subsection we study the case that $z = 2 (mod \ 4)$ and $y$ is even.
An example of this case is Example 5.3.
The argument here looks complicated, but it is a generalization of argument used in Example 5.3. If you read Example 5.3 and the proof here side by side, the logic we use looks clear enough.

Figure 5.19  

In Fig. 5.19 we suppose that Predictions 5.1, 5.2, 5.3,5.4 and 5.5 are valid for $z=0,1,2,...,4m+1$ and for $y = 0,1,2,...,z$ . We also suppose that these predictions are valid for $z=4m+2$ and $y=0,1,...,2k-5$ . Under these assumptions we prove that $X= G1R(\{2k-4,4m+2\}) = 4m+k$ .

For $z=n=4m+1 = 1 (mod \ 4)$ by Prediction 5.1 the first number from the right end of the row is $6m+2$ . If we start with $6m+2$ and go leftward picking up every other number, then we have $4m+2 \leftarrow ... 6m\leftarrow 6m+1 \leftarrow 6m+2$ which is an arithmetic sequence with common difference of $-1$ .
On the other hand if we start with $1$ that is the second number from the right end of the row and go leftward picking up every other number, then we have an arithmetic sequence with common difference of $3$ whose length is $\lceil (4m+1)/4 \rceil$ $=m+1$ , and hence the last number of this sequence is $3\lceil n/4 \rceil-2$ $=3(m+1)-2 = 3m+1$ . This number $3m+1$ is the first number of an arithmetic sequence with common difference of $1$ and the last number of this sequence is $4m+1$ .
By Prediction 5.5 for $y = n=2k-4$ the number at the top of the column is $\frac{3n}{2}$ $=3k-6$ . If we move down picking up every other number, then we have an arithmetic sequence with common difference of $2$ .
On the other hand if we start with 1 that is the second number of the column and move down picking up every other number, then we have $1,4,7,10,13,...,5k-12,5k-10,...,4m+k-2,...$ that starts as an arithmetic sequence with common difference of $3$ such that it has $\lfloor \frac{n}{2} \rfloor +1$ $=k-1$ terms and the last number is $3(k-2) +1=3k-5$ , then after this number it becomes an arithmetic sequence with common difference of $2$ whose first number is $3k-5$ , and after that we have $3k-3,3k-1,...,4m-k+3$ .
In Fig. 5.19 we assume that

$$G1R(\{2m,4m+1\}) = 3m+1,$$ (5.30)

$$G1R(\{2k-4,4k-7\}) = 3k-5,$$ (5.31)

$$4m+1 > 4k-7 2k-4 < 2m.$$ (5.32)

Note that $G1R(\{2m,4m+1\})$ and $G1R(\{2k-4,4k-7\})$ are printed in red on blue rectangles in Fig. 5.19. These two Grundy numbers are important, because $G1R(\{2m,4m+1\})$ is the last term in the arithmetic sequence of common difference of 3 in the row when $z = 4m+1$ , and $G1R(\{2k-4,4k-7\})$ is the last term in the arithmetic sequence of common difference of 3 in the column when $y = 2k-4$ . By (5.30) and (5.32) $X= G1R(\{2k-4,4m+2\})$ is under Grundy number $G1R(\{2k-4,4k-7\})$ , and it is on the right side of Grundy number $G1R(\{2m,4m+1\})$ . By Lemma 5.1 we have only to prove that $4m+k$ is the smallest number that does not belong to $L1(\{2k-4,4m+2\})$ , if we are to prove that $X= G1R(\{2k-4,4m+2\}) = 4m+k$ .
By Definition 5.2 $L1(\{2k-4,4m+2\})$ can be separated into three parts $L11(\{2k-4,4m+2\})$ , $L12(\{2k-4,4m+2\})$ and $L13(\{2k-4,4m+2\})$ .
$L11(\{2k-4,4m+2\}) = \{G1R(\{0,0\}),G1R(\{1,1\}),...,G1R(\{2k-4,2k-4\})\}$

$$= \{0,2,3,5,...,3k-7,3k-6\}.$$ (5.33)

$L12(\{2k-4,4m+2\}) = \{G1R(\{2k-4,2k-3\}),G1R(\{2k-4,2k-2\}),...,G1R(\{2k-4,4m+1\})\}$ $= \{1,3k-4,4,...,4m+k-2,4m-k+3\}$

$$=\{1,4,...,3k-5\}$$ (5.34)

$$\cup \{3k-3,3k-1,...,4m-k+3\}$$ (5.35)

$$\cup \{3k-4,3k-2,...,4m+k-2\}.$$ (5.36)

By using predictions for $z=4m+2$ and $y=0,1,...,2k-5$
$L13(\{2k-4,4m+2\}) = \{G1R(\{0,4m+2\}),G1R(\{1,4m+2\}),...,G1R(\{2k-5,4m+2\})\}$
$=\{4m+2,4m+1,4m+3,...,4m-k+4\}$

$$=\{4m+2,4m+3,...,4m+k-1\}$$ (5.37)

$$\cup \{4m+1,4m,...,4m-k+4\}.$$ (5.38)

By (5.33) and (5.34) we have

$$L11(\{2k-4,4m+2\}) \cup L12(\{2k-4,4m+2\}) \supset \{0,1,2,3,...,3k-5\}.$$ (5.39)

By (5.35) and (5.36) we have

$$L12(\{2k-4,4m+2\}) \supset \{3k-4,3k-3,3k-2,...,4m-k+3\}.$$ (5.40)

By (5.37) and (5.38)

$$L13(\{2k-3,4m+2\})=\{4m-k+4,4m-k+5,...,4m+k-1\}.$$ (5.41)

By (5.39), (5.40) and (5.41) we have $L1(\{2k-4,4m+2\}) = \{0,1,2,3,...,4m+k-1\}$ and $4m+k$ is the smallest number that does not belong to $L1(\{2k-4,4m+2\})$ . Therefore $X= G1R(\{2k-4,4m+2\}) = 4m+k$ .