Proposition 1: singleton occupancy reveals the adjacent-domain identifier

For every choice of natural numbers $n,k,a,b$ with $1\le k\le n$, $1\le a$, and $1\le b$, let the cells be indexed by $\{0,\ldots,n-1\}$ and let a word be a pair in $(\mathbb F_2)^a\times(\mathbb F_2)^b$, where $\mathbb F_2=\mathbb Z/2\mathbb Z$ and addition is coordinatewise. A row consists of a set $S$ of exactly $k$ cell indices, an arbitrary vector $u\in(\mathbb F_2)^a$, and an identifier vector $v\in(\mathbb F_2)^b$; its target word is $(u,v)$. A snapshot assigns to each cell either no word or a present word, with a present zero word distinct from no word. An attempt consists of a row, either no pivot or one cell index as pivot, and an arbitrary function $f$ assigning a word to every cell. For every finite ordered list $E$ of such attempts and every row $r=(S_r,u_r,v_r)$, start with no word in any cell and process $E$ in order as follows: an attempt with no pivot leaves the snapshot unchanged; an attempt with pivot $p$ leaves cells outside its row's support $S$ unchanged, sets each cell $i\in S\setminus\{p\}$ to a present word $w_i$, where $w_i$ is the word already present at $i$ immediately before this attempt if there is one and otherwise is $f(i)$, and sets $p$ to the present word $(u,v)+\sum_{j\in S\setminus\{p\}}w_j$. Assume that at each attempt, relative to the snapshot immediately before that attempt, having no pivot entails that every cell in its row's support already has a word, and having pivot $p$ entails both $p\in S$ and that $p$ has no word. Assume also that the list obtained by retaining precisely the rows of the attempts that have a pivot, in their original order and with repetitions retained, is exactly the one-element list $[r]$; thus there is exactly one attempt with a pivot and its entire row equals $r$. Then, in the snapshot $A$ obtained after processing all attempts, the second component of the sum of the words in all occupied cells equals $v_r$: $\left(\sum_{i\in O} A_i\right)_2=v_r$, where $O$ is the set of cell indices with a present word and $A_i$ denotes that word for $i\in O$ (the underlying total payload function assigns the zero word to a cell with no word). All sums use coordinatewise addition modulo $2$, and an empty sum is the zero vector of the appropriate type. The hypotheses exclude zero cells, zero shares, zero prefix or suffix length, and an empty attempt list; $k=1$ is allowed, in which case the sum used to set the pivot is empty. Since every row has a nonempty support, the initial all-empty snapshot cannot admit an attempt with no pivot, so the unique attempt with a pivot must be first; further attempts with no pivot are permitted exactly when all cells in their respective supports are already occupied. The vectors $u$, $v$, and all values of $f$ may be zero; no relation between a row's prefix vector and its identifier, and no randomness or distinctness condition on these vectors, is assumed.