The reverse-mex condition for subtraction Nim with s_3=s_1+s_2

Complete proof for all additive subtraction sets and all positions.

Larsson–Manabe–Miyadera, arXiv:2605.14321, Conjecture 2, considers the ordinary subtraction game with

$$S=\{a,b,a+b\},\qquad 1\le a<b.$$

Let $p$ be any eventual period of its Sprague–Grundy sequence. For a losing position $w$, define

$$d=\operatorname{dist}(w) =\min\{j\ge1:(w-ja)\bmod p\text{ is losing}\}.$$

The conjecture asserts that whenever $d\ge4$ is even,

$$(w+b)\bmod p\text{ is winning}, \qquad (w+b-2a)\bmod p\text{ is losing}.$$

A prefix issue must first be handled: the conjecture includes losing positions before the eventual Grundy preperiod. We use the prior result of Bhagat–Larsson–Manabe–Yamashita, arXiv:2601.18715, equation (1), Theorem 3, and Corollary 4. Their equation (1) gives

$$o_{\mathrm{wall}}(x) =o_{\mathrm{sink}}(x+\max S+1) \qquad(x\ge0),$$

and their Theorem 3 proves that the additive sink sequence is purely periodic. Consequently the ordinary additive-game outcome has a global period $P$:

$$o_{\mathrm{wall}}(x+P)=o_{\mathrm{wall}}(x) \qquad(x\ge0).$$

Although $p$ is initially assumed to be only an eventual Grundy period, it is in fact a global outcome period. For any $x\ge0$, choose $t$ so large that $x+tP$ and $x+p+tP$ lie in the eventual Grundy tail. Then

$$o(x+p)=o(x+p+tP)=o(x+tP)=o(x).$$

Thus all source-prescribed residue classes, including those represented by initial-prefix positions, have well-defined outcomes. Choose sufficiently large representatives so every subtraction below is legal. The ordinary recursion on the cycle is

$$x\text{ losing}\iff x-a,\ x-b,\ x-a-b\text{ are all winning}.$$

Now fix a losing $w$ and set

$$z_j=w+b-ja\pmod p.$$

The first queried position $z_0=w+b$ is winning: its $b$-move reaches the losing position $w$.

By minimality of $d$,

$$w-ja\text{ is winning}\quad(1\le j<d), \qquad w-da\text{ is losing}.$$

The three followers of $z_j$ are

$$z_j-a=z_{j+1}, \qquad z_j-b=w-ja, \qquad z_j-(a+b)=w-(j+1)a.$$

Therefore $z_{d-1}$ is winning, since its $(a+b)$-move reaches $w-da$. For $1\le j\le d-2$, its latter two followers are both winning, so the entire recursion reduces to

$$z_j\text{ losing}\iff z_{j+1}\text{ winning}.$$

Backward induction yields the stronger exact alternating-strip formula

$$\boxed{ z_j\text{ losing} \iff d-1-j\text{ is odd} \qquad(1\le j\le d-1).}$$

When $d\ge4$ is even, $d-3$ is odd, and therefore

$$\boxed{z_2=w+b-2a\text{ is losing},\qquad z_0=w+b\text{ is winning}.}$$

These are exactly both parts of the conjectured condition for every $a<b$, every permitted eventual Grundy period, and every losing position, including the entire initial preperiod.

The earlier sink-subtraction paper supplies global outcome periodicity; the alternating-strip argument above proves the distinct subsequent conjecture.