#### Proof

Let *S* = {*x* : *A*_{m} < *x* ≤ *A*_{n}, and *x* ∈ ℤ^{even}}. As the above, ♯ (*S*) =
${A}_{j}^{\prime}$
/2, ♯_{A} = ♯(*A*∩ *S*) = *n*−*m*, ♯_{B} = ♯(*B* ∩ *S*) = ♯(*S*)− ♯A. Then 2(*n*−*m*) =
${A}_{j}^{\prime}$
−2 ♯_{B}. The minimum value of 2(*n*−*m*) is reached at the maximum value of ♯_{B}, that is, when there is a smallest distance of 2*s* + *t* + *δ*_{t}+2 between consecutive elements of *B* ∩ *S*.

Since
${A}_{j}^{\prime}$
is even, and also 2*s*+ *t*+*δ*_{t}+2 is even, dividing
${A}_{j}^{\prime}$
by 2*s*+*t*+*δ*_{t}+2 gives
${A}_{j}^{\prime}$
= (2*s*+*t*+*δ*_{t}+2)*q* + 2*r*, *r* ∈ {1, 2, 3,…, *s*+(*t* + *δ*_{t})/2}. Note that for *r* ≥ 2, there may be an additional element in *B* ∩ *S*. Thus
$$2(n-m)\ge (2s+t+{\delta}_{t})q+\left\{\begin{array}{}2r\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\text{if}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}r\in \{0,1\},\\ 2(r-1)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{if}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}r\in \{3,4,\dots ,s+(t+{\delta}_{t})/2\}.\end{array}\right.$$

Now
$${B}_{j}^{\prime}={B}_{n}-{B}_{m}=\sum _{i=m}^{n-1}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}({B}_{i+1}-{B}_{i})\ge (2s+t+{\delta}_{t}+2)(n-m).$$

If *r* ∈ {0, 1},
$$\begin{array}{}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(n-m)\\ \phantom{\rule{2em}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}(s+(t+{\delta}_{t})/2+1)((2s+t+{\delta}_{t})q+2r)\\ \phantom{\rule{2em}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(s+(t+{\delta}_{t})/2){A}_{j}^{\prime}+2r\\ \phantom{\rule{2em}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}(s+(t+{\delta}_{t})/2){A}_{j}^{\prime}.\end{array}$$

If *r* ∈ {2, 3,⋯, *s* +(*t*+*δ*_{t})/2},
$$\begin{array}{}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime}\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(n-m)\\ \phantom{\rule{2em}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}(s+(t+{\delta}_{t})/2)(2s+t+{\delta}_{t})q+(2s+t+{\delta}_{t}+2)(r-1)\\ \phantom{\rule{2em}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(s+(t+{\delta}_{t})/2){A}_{j}^{\prime}+2r-(2s+t+{\delta}_{t}+2)\\ \phantom{\rule{2em}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}(s+(t+{\delta}_{t})/2){A}_{j}^{\prime}+2-(2s+t+{\delta}_{t}).\end{array}$$

In either case, the following inequality holds:
$$s{A}_{j}^{\prime}+(t+{\delta}_{t})j={B}_{j}^{\prime}\ge (s+(t+{\delta}_{t})/2){A}_{j}^{\prime}+2-(2s+t+{\delta}_{t})$$
Therefore,
$${A}_{j}^{\prime}\le 2(j+1+\frac{2s-2}{t+{\delta}_{t}}).$$ □

#### Proof

Again let *S* = {*x* : *A*_{m} < *x* ≤ *A*_{n}, *x* ∈ ℤ^{even}}. From the proof of Lemma 4.3 we have 2(*n*−*m*) =
${A}_{j}^{\prime}$
−2♯*B*. By Lemma 2.2,
${A}_{j}^{\prime}$
≥ 2*j* for all *j* ∈ ℤ^{+}. Thus Eq. (9) implies that there are three possibilities: (a)
${A}_{j}^{\prime}$
= 2*j*, (b)
${A}_{j}^{\prime}$
= 2*j*+2 and (c)
${A}_{j}^{\prime}$
= 2*j*+4.

*Case (a)*
${A}_{j}^{\prime}$
= 2*j*. By definition,
${B}_{j}^{\prime}$
= (2*s* + *t* +*δ*_{t})*j*. By Lemma 2.3,
${A}_{j}^{\prime}$
= 2*j* means that 2 ≤ *A*_{n}−*A*_{m} = 2*j* ≤ 2*s* + *t* +*δ*_{t}−2.

There must be up to one *B*_{i} ∈ *S*, i.e., ♯_{B}∈{0, 1}. For if there are more than two elements in *B* ∩ *S*, then there is a gap less than 2*s* + *t* +*δ*_{t}−2 between the adjacent elements of *B* ∩ *S*, but it contradicts with the fact that *B*_{i+1}−*B*_{i} ∈ {2*s* + *t* +*δ*_{t}+2, 4*s*+ *t* +*δ*_{t}}.

So suppose ♯_{B} = 0. Then *n*−*m* =
${A}_{j}^{\prime}$
/2 = *j*, this implies that *A*_{i}−*A*_{i−1} = 2 for all *m* < *i* ≤ *n*. Accordingly, *B*_{i}−*B*_{i−1} = 2*s* + *t* +*δ*_{t}+2 for *m* < *i* ≤ *n*. Thus *B*_{n}−*B*_{m} = (2*s* + *t* +*δ*_{t}+2)(*n*−*m*)>(2*s* + *t* +*δ*_{t}) *j* =
${B}_{j}^{\prime}$
, This contradicts our assumption.

Now suppose ♯_{B} = 1. Then *n*−*m* = (
${A}_{j}^{\prime}$
−2)/2 = *j*−1. Let *B*_{i0} denote the only element of *B* ∩ *S* with *A*_{λ} < *B*_{i0} < *A*_{λ+1}, where *m*<*λ*<*n*.
$$\dots ,\stackrel{2(\lambda -m)}{\stackrel{\u23de}{{A}_{m},\dots ,A\lambda}},\phantom{\rule{thinmathspace}{0ex}}\underset{4}{\underset{\u23df}{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{{i}_{0}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}}},\stackrel{2(n-\lambda -1)}{\stackrel{\u23de}{{A}_{\lambda +1},\dots ,{A}_{n},}}\dots $$

Therefore,
$$\begin{array}{}{B}_{n}-{B}_{m}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\sum _{m\le i<\lambda}({B}_{i+1}-{B}_{i})\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}({B}_{\lambda +1}\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}{B}_{\lambda})\phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}\sum _{\lambda <i\le n-1}\phantom{\rule{thinmathspace}{0ex}}({B}_{i+1}\phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}{B}_{i})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(\lambda -m)+(4s+t+{\delta}_{t})+(2s+t+{\delta}_{t}+2)(n-\lambda -1)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(n-m-1)+4s+t+{\delta}_{t}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(j-2)+4s+t+{\delta}_{t}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t})j+2j-(t+{\delta}_{t}+4)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime}+(2j-(t+{\delta}_{t}+4)).\end{array}$$

If *s* ∈ {1, 2}, then 2*j*−(*t* + *δ*_{t}+4) ≤ 2*s* + *t* +*δ*_{t}−2−(*t* + *δ*_{t}+4) = 2*s*−6 < 0. Thus for any *t* ∈ ℤ^{+},
${B}_{j}^{\prime}$
= *B*_{n}−*B*_{m} is impossible.

If *s* ≥ 3, then
${B}_{j}^{\prime}$
= *B*_{n}−*B*_{m} if and only if 2 ≤ 2*j* = *t* + *δ*_{t}+4 ≤ 2*s* + *t* +*δ*_{t}−2, which is true for any *t* ∈ ℤ^{+}. Thus the only possibility is *j* = (*t* + *δ*_{t})/2+2.

*Case (b)*
${A}_{j}^{\prime}$
= 2*j*+2. In this case,
${B}_{j}^{\prime}$
= *s*
${A}_{j}^{\prime}$
+(*t* + *δ*_{t})*j* = (2*s* + *t* +*δ*_{t})*j*+2*s*. By Lemma 2.3,
${A}_{j}^{\prime}$
= 2*j* + 2 means that
$$2s+t+{\delta}_{t}\le {A}_{n}-{A}_{m}-2=2j\le 4s+2t+2{\delta}_{t}-4.$$(10)

We claim that ♯_{B}∈{1, 2}. Indeed, if ♯_{B} = 0, then *n*−*m* =
${A}_{j}^{\prime}$
/2 = *j*+1. By the same argument as for the case (a), we get
${B}_{j}^{\prime}$
= *B*_{n}−*B*_{m} = (2*s* + *t* +*δ*_{t}+2)(*n*−*m*) = (2*s* + *t* +*δ*_{t}+2)(*j*+1)>
${B}_{j}^{\prime}$
, a contradiction. If ♯_{B} ≥ 3, then there is a gap less than (*A*_{n}−*A*_{m})_{max}/2 between the neighbouring elements of *B* ∩ *S*. But (*A*_{n}−*A*_{m})/2 ≤ 2*s* + *t* +*δ*_{t}−1, so this also cannot happen by reason of *B*_{i+1}−*B*_{i} ∈{2*s* + *t* +*δ*_{t}+2, 4*s*+ *t* +*δ*_{t}}. We first suppose ♯_{B} = 1. Then *n*−*m* = (
${A}_{j}^{\prime}$
- 2)/2 = *j*. In the same manner of case (a), we can see that
$$\begin{array}{}{B}_{n}-{B}_{m}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(n-m-1)+4s+t+{\delta}_{t}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(j-1)+4s+t+{\delta}_{t}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)j+2s+2j-2\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime}+2j-2\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\ge \phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime}+(2s+t+{\delta}_{t}-2)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}>\phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime},\end{array}$$
a contradiction.

Next suppose ♯_{B} = 2. Then *n*−*m* = (
${A}_{j}^{\prime}$
−4)/2 = *j*−1. Let *B*_{j0}, *B*_{i0+1} ∈ *B* ∩ *S* with *A*_{λ0} < *B*_{j0} < *A*_{λ0+1}, and *A*_{λ1} < *B*_{i0+1} < *A*_{λ1+1}, where *m* < *λ*_{0} < *λ*_{1} < *n*.
$$\dots ,\stackrel{2({\lambda}_{0}-m)}{\stackrel{\u23de}{{A}_{m},\cdots ,{A}_{{\lambda}_{0}}}},\underset{4}{\underset{\u23df}{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{{i}_{0}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}}},\stackrel{2({\lambda}_{1}-{\lambda}_{0}-1)}{\stackrel{\u23de}{{A}_{{\lambda}_{0}+1},\dots ,{A}_{{\lambda}_{1}}}},\underset{4}{\underset{\u23df}{{B}_{{i}_{0}+1}}},\stackrel{2(n-{\lambda}_{1}-1)}{\stackrel{\u23de}{{A}_{{\lambda}_{1}+1},\dots ,{A}_{n}}},\dots $$

Similarly, we have
$$\begin{array}{}{B}_{n}-{B}_{m}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(n-m-2)+2(4s+t+{\delta}_{t})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t}+2)(j-3)+2(4s+t+{\delta}_{t})\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}(2s+t+{\delta}_{t})j+2s+2j-(t+{\delta}_{t}+6)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}{B}_{j}^{\prime}+2j-(t+{\delta}_{t}+6).\end{array}$$(11)

It is worth to mention that *B*_{j0} ≥ *A*_{m} + 2 and *B*_{i0} + 1 ≤ *A*_{n} − 2. Consequently, 2*j* + 2 =
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
= *A*_{n} − *A*_{m} ≥ *B*_{i0} + 1 − *B*_{i0}+4 ≥ 2*s*+ *t* + *δ*_{t} + 6. It follows Eq. (10) that
$$\begin{array}{}2s+t+{\delta}_{t}+4\le 2j\le 4s+2t+2{\delta}_{t}-4.\end{array}$$(12)

If *s* = 1, *t* ∈{3, 4}, there is no such *j* satisfying Eq. (12).

If *s* = 1, *t* > 4, by Eq. (11), *B*_{n} − *B*_{m} =
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
if and only if *j* = (*t* + *δ*_{t})/2+3, which satisfies Eq. (12).

If *s* ≥ 2, then 2*j* ≥ 2*s* + *t* + *δ*_{t} + 4 ≥ *t* + *δ*_{t} + 8 > *t* + *δ*_{t} + 6, thus
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
= *B*_{n} − *B*_{m} is impossible.

*Case (c)*
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
= 2*j* + 4. On account of Eq. (9) this case can happen only when *t* + *δ*_{t} = 2*s* − 2. Thus we have
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
= *sA*
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
+(*t* + *δ*_{t})*j* = (4*s* − 2)*j* + 4*s*. And the condition 2*s* + *t* + *δ*_{t} = 4*s*− 2 > 4 implies *s* ≥ 2. We below consider *s* = 2 and *s* ≥ 3, respectively.

*(c-i) s* = 2 (and also *t* + *δ*_{t} = 2). Now
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
= 6*j* + 8. By Lemma 2.3,
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
= 2*j* + 4 implies that
$\begin{array}{}{B}_{2}^{\prime}-2\le {A}_{n}-{A}_{m}-4=2j\le {B}_{3}^{\prime}-6.\end{array}$
Recall that
$\begin{array}{}{A}_{0}^{\prime}={B}_{0}^{\prime}=0,{A}_{1}^{\prime}=\text{mex}\{0,1\}=2,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{1}^{\prime}=2s+t+{\delta}_{t}=6.\end{array}$
Thereby,
$\begin{array}{}{A}_{2}^{\prime}=\text{mex}\{0,1,2,3,6,7\}=4\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{2}^{\prime}=4s+2(t+{\delta}_{t})=12.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{Also}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{A}_{3}^{\prime}=\text{mex}\{0,1,2,3,4,5,6,7,12,13\}=8,\end{array}$
and so
$\begin{array}{}{B}_{3}^{\prime}=22.\end{array}$
Thus 14 ≤ *A*_{n} − *A*_{m} ≤ 24, and 5 ≤ *j* ≤ 8. In this case, we also have the fact that *B*_{i+1} − *B*_{i} ∈ {2*s* + *t* + *δ*_{t}+2, 4*s* + *t* + *δ*_{t}}={8, 10}.

Then we claim that ♯*B* ∈ {1, 2, 3}. Indeed, if ♯*B* = 0, then *n* − *m* =
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
/2 = *j* + 2. By the same argument as for the cases (a) and (b), we have
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
= *B*_{n} − *B*_{m} = 8(*n* − *m*) = 8(*j* + 2)>
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
, a contradiction. If ♯*B* ≥ 4, then there is a gap less than (*A*_{n} − *A*_{m})/3 ≤ 8 between adjacent elements of *B* ∩ *S*, which is impossible.

Now if ♯*B* = 1, then *n* − *m* = (
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
− 2)/2=*j* + 1. As in the proof of case (a), *B*_{n} − *B*_{m} = 8(*n* − *m* − 1)+10 = 8*j* + 10 >
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
, contradicting the assumption.

If ♯*B* = 2, we know *n* − *m* = (
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
− 4)/2 = *j*. Similarly, *B*_{n} − *B*_{m} = 8(*n* − *m* − 2) + 20 = 8(*j* − 2)+20 >
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
,a contradiction.

If ♯*B* = 3, then *n* − *m* = (
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
− 6)/2 = *j* − 1. Assume the three elements in *B* ∩ *S* are *B*_{j0}, *B*_{i0 + 1}, *B*_{i0 + 2} for some *i*_{0} ∈ ℤ^{+}, then we have *B*_{i0} ≥ *A*_{m} + 2 and *B*_{i0 + 2} ≤ *A*_{n}− 2, thus
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
= *A*_{n} − *A*_{m} ≥ *B*_{i0 + 2} − *B*_{j0} + 4 ≥ 20, i.e., *j* ≥ 8. In the same manner we can see that
$$\begin{array}{}{B}_{n}-{B}_{m}=8(n-m-3)+30=8(j-4)+30>6j+8={B}_{j}^{\prime}\end{array}$$
since *j* ≥ 8 (by above 5 ≤ *j* ≤ 8, the only possibility j fact is *j* =8), giving a contradiction.

*(c-ii) s* ≥ 3. Then we have 2*s* + *t* +*δ*_{t} = 4*s* − 2 ≥ 10. Recall that
$\begin{array}{}{A}_{0}^{\prime}={B}_{0}^{\prime}=0,\phantom{\rule{thinmathspace}{0ex}}{A}_{1}^{\prime}=2,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{1}^{\prime}=2s+t+{\delta}_{t}\ge 10.\end{array}$
So
$\begin{array}{}{A}_{2}^{\prime}=\text{mex}\{0,1,2,3,{B}_{1}^{\prime},\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{B}_{1}^{\prime}+1\}=4,{B}_{2}^{\prime}=4s+2(t+{\delta}_{t})=8s-4.\end{array}$
Also
$\begin{array}{}{A}_{3}^{\prime}=\text{mex}\{0,1,2,3,4,5,{B}_{1}^{\prime},{B}_{1}^{\prime}+1,{B}_{2}^{\prime},{B}_{2}^{\prime}+1\}=6,\end{array}$
and so
$\begin{array}{}{B}_{3}^{\prime}=12s-6.\end{array}$
Lemma 9 now shows that 8*s*− 2 =$\begin{array}{}{B}_{2}^{\prime}+2\le {A}_{n}-{A}_{m}={A}_{j}^{\prime}\le {B}_{3}^{\prime}-2=12s-8,\end{array}$
i.e., 4*s* − 3 ≤ *j* ≤ 6*s*−6.

Now we claim that ♯*B* ∈ {1, 2, 3}. Otherwise, if ♯*B* = 0, then *n* − *m* =
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
/2 = *j* + 2, we have
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
= (2*s*+ *t* + *δ*_{t} + 2)(*n* − *m*) = (4*s* − 2)(*j* + 2) > (4*s* − 2)*j* + 4*s* =
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
, a contradiction. If ♯*B* ≥ 4, there is a gap less than (*A*_{n} − *A*_{m})/3 < 4*s* between neighbouring elements of *B* ∩ *S*, but *B*_{i+1} − *B*_{i} ∈ {2*s* + *t* + *δ*_{t}+2, 4*s* + *t* +*δ*_{t}}= {4*s*, 6*s* − 2}, giving a contradiction.

First suppose ♯*B* = 1. As the above, we get *B*_{n} − *B*_{m} = 4*s*(*n* − *m* − 1) + (6*s* − 2) = 4*sj* + 6*s* − 2=
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
+ 2*j* + 2*s* − 2 >
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
by reason of *j* ≥ 4*s* − 3 and *s* ≥ 3, a contradiction.

Next suppose ♯*B* = 2. Similarly, *B*_{n} − *B*_{m} = 4*s*(*n* − *m* − 2) + 2(6*s* − 2) = 4*s*(*j* − 2) + 12*s* − 4 =
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
+ 2*j* − 4 >
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
, a contradiction.

Finally suppose ♯*B* = 3. Analogously, we have *B*_{n} − *B*_{m} = 4*s*(*n* − *m* − 3) + 3(6*s* − 2) = 4*s*(*j* − 4) + 18*s* − 6 =
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
+ 2*j* − 2*s* − 6 >
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
since *j* ≥ 4*s* − 3 and *s* ≥ 3.

Therefore, if
$\begin{array}{}{A}_{j}^{\prime}\end{array}$
= 2*j* + 4, for any *s* ≥ 2 and *t* ≥ 1 with *t* + *δ*_{t} = 2*s* − 2, the hypothesis
$\begin{array}{}{B}_{j}^{\prime}\end{array}$
= *B*_{n} − *B*_{m} is impossible. □

Given two games, they have the same game positions but with possibly different move rules. We call them *equivalent* if their *P*-positions are the same. Two equivalent games certainly have also the same *N*-positions, as well as winning strategy.

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