Invariance Reduction Process conjecture for sub-games of SetNim

Counterexample: a five-heap game with an invariant vector reduces to a nontrivial, merge-irreducible four-heap game with no invariant vector.

Consider the SetNim game $G$ on heaps $(a,b,c,d,e)$ with maximal move sets

$$\{a,b\},\qquad\{a,c\},\qquad\{a,d\},\qquad\{b,c,e\}.$$

Its complete set of losing positions is

$$\mathcal P(G)= \{(0,b,c,d,e):d=b+c+e\} \cup \{(a,b,c,d,e):a>0,\ b=c,\ a+d=b+e\}. \tag{1}$$

Here is a direct proof. No legal move joins two positions in the first family, since it cannot decrease both $d$ and $b+c+e$. Within the second family, moves on $\{b,c,e\}$ preserving $b=c$ decrease $b+e$, moves on $\{a,d\}$ decrease $a+d$, and moves on $\{a,b\}$ or $\{a,c\}$ cannot preserve both equalities. A move from the first family to the second would increase $a$. A move from the second family to the first would have to set $a=0$; using $\{a,b\}$ or $\{a,c\}$ would require a negative remaining heap, while using $\{a,d\}$ would require increasing $d$.

Conversely, every position outside (1) reaches (1). If $a=0$, lower $d$ to $b+c+e$ when it is larger, or lower heaps in $\{b,c,e\}$ to total $d$ when it is smaller. Suppose $a>0$, and put

$$L=a+d,\qquad m=\min(b,c).$$

If $L\le m+e$, set $t=\max(0,L-e)$ and move

$$(b,c,e)\longmapsto(t,t,L-t),$$

reaching the second family. Otherwise $L>m+e$; by symmetry suppose $m=b\le c$. There are three exhaustive cases:

1. If $d<b+e$, move $(a,c)\mapsto(b+e-d,b)$, reaching the second family.
2. If $b+e\le d\le b+c+e$, move $(a,c)\mapsto(0,d-e-b)$, reaching the first family.
3. If $d>b+c+e$, move $(a,d)\mapsto(0,b+c+e)$, reaching the first family.

The case $c\le b$ is symmetric. This proves (1) globally.

Both families in (1) depend on $d,e$ only through $d-e$. Therefore

$$z=(0,0,0,1,1)$$

is an invariant vector of $G$ for every admissible translation.

Start from the non-losing position

$$p=(1,1,2,2,1).$$

The prescribed invariance reduction subtracts $z$, giving

$$p-z=(1,1,2,1,0).$$

Zero reduction at heap $e$ produces the four-heap subgame $H$ with maximal move sets

$$\{a,b\},\qquad\{a,c\},\qquad\{a,d\},\qquad\{b,c\}. \tag{2}$$

The four heap-incidence vectors in these move sets are

$$(1,1,1,0),\ (1,0,0,1),\ (0,1,0,1),\ (0,0,1,0),$$

which are pairwise distinct. Thus no merge reduction is available, and $H$ is genuinely nontrivial.

Setting $e=0$ in (1) gives

$$\mathcal P(H)= \{(0,b,c,d):d=b+c\} \cup \{(a,b,b,d):a>0,\ d=b-a\}. \tag{3}$$

Any nonzero binary invariant vector must itself be a losing position, by applying invariance to the zero position. Formula (3) leaves exactly three candidates:

$$u_1=(0,0,1,1),\qquad u_2=(0,1,0,1),\qquad u_3=(1,1,1,0).$$

But $u_3\in\mathcal P(H)$, whereas

$$u_3+u_1=(1,1,2,1)\notin\mathcal P(H), \qquad u_3+u_2=(1,2,1,1)\notin\mathcal P(H).$$

These rule out $u_1,u_2$, and the first identity also rules out $u_3$ by starting at $u_1$. Hence $H$ has no invariant vector.

Therefore $G$ has an invariant vector but its prescribed nontrivial invariance-reduction subgame $H$ has none, contradicting Conjecture 2 of arXiv:2604.02587.