Read-backs

Throughout, a board is a finite multiset of natural numbers (repetitions allowed and counted, order irrelevant). For a board $s$ and a value $v$, $\#_v(s)$ denotes the multiplicity of $v$ in $s$, $\prod s$ the product of all entries of $s$ (with repetition), and $|s|$ the number of entries of $s$ (with repetition). For $x, p \in \mathbb{N}$, $v_p(x)$ denotes the exponent of $p$ in the prime factorization of $x$; this is $0$ whenever $p$ is not prime, and $0$ for every $p$ when $x = 0$ or $x = 1$.

Three auxiliary notions are used and are expanded inline below wherever they occur; they are recorded here once.

One move. $s \to t$ ("$t$ is obtained from $s$ by one move") means: there exist natural numbers $m$ and $n$ with

$$m > 1, \qquad n > 1, \qquad m \in s, \qquad n \in (s \setminus \{m\}),$$

where $s \setminus \{m\}$ is $s$ with exactly one copy of $m$ deleted, and such that

$$t \;=\; \{\gcd(m,n)\} \;\uplus\; \{\operatorname{lcm}(m,n) \,/\, \gcd(m,n)\} \;\uplus\; \big((s \setminus \{m\}) \setminus \{n\}\big).$$

That is: delete one copy of $m$ and one copy of $n$ from $s$ and insert the two values $\gcd(m,n)$ and $\operatorname{lcm}(m,n)/\gcd(m,n)$. The division is truncated natural-number division. Note that $m$ and $n$ are values: the condition $n \in s \setminus \{m\}$ permits $m = n$ when that value occurs at least twice in $s$, and it forces $s$ to contain at least two entries greater than $1$ (counted with multiplicity). The two inserted values are not required to exceed $1$. Every move preserves the number of entries: $|t| = |s|$.

Reachability. $s \rightsquigarrow t$ means $t$ is obtained from $s$ by a finite (possibly empty) chain of moves — the reflexive–transitive closure of $\to$. In particular $s \rightsquigarrow s$ always holds.

Terminal. $t$ is terminal means there is no board $u$ with $t \to u$. Since the target board in the definition of a move is produced by an explicit equation, this is the same as saying no admissible pair $m, n$ can be selected at all: $t$ does not contain two entries (with multiplicity) both greater than $1$.

Big part. $\operatorname{big}(t)$ is the sub-multiset of $t$ consisting of exactly those entries that are greater than $1$, with their multiplicities.

Exponent gcd. For $p \in \mathbb{N}$ and a board $s$, $G_p(s)$ is obtained by replacing each entry $x$ of $s$ by $v_p(x)$ and folding the resulting multiset of natural numbers with $\gcd$, starting from the value $0$; since $\gcd(0,a) = a$, this is the greatest common divisor of the numbers $v_p(x)$ over all entries $x$ of $s$, with $G_p(\emptyset) = 0$. Entries equal to $0$ or $1$ contribute the value $0$ and therefore do not change the result. If $p$ is not prime then $v_p(x) = 0$ for every $x$ and $G_p(s) = 0$ for every board $s$.

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IMO2026P1.move_ones_or_prod

For all boards $s$ and $t$ (both implicit arguments, i.e. universally quantified over all finite multisets of natural numbers), if $t$ is obtained from $s$ by one move — that is, if there exist $m, n \in \mathbb{N}$ with $m > 1$, $n > 1$, $m \in s$, $n \in s \setminus \{m\}$ and

$$t = \{\gcd(m,n)\} \uplus \{\operatorname{lcm}(m,n)/\gcd(m,n)\} \uplus \big((s \setminus \{m\}) \setminus \{n\}\big)$$

— then both of the following hold:

1. $\#_1(s) \le \#_1(t)$: the multiplicity of the value $1$ in $t$ is at least its multiplicity in $s$;
2. $\#_1(s) < \#_1(t) \ \ \text{or}\ \ \prod t < \prod s$: either that multiplicity strictly increases, or the product of all entries of $t$ is strictly smaller than the product of all entries of $s$.

Conditionality. The statement is conditional on a single move existing and on $t$ being one of its outcomes: the hypothesis is the move relation between the two given boards. For any pair $(s,t)$ that is not related by a move — in particular for every terminal $s$, for the empty board, and for every board none of whose entries exceeds $1$ — the statement asserts nothing whatsoever. Reachability (multi-step) does not appear; nothing is claimed about chains of two or more moves.

Counts and products. Conjunct 1 states that $\#_1$ cannot decrease across a move; no upper bound on its increase is given. The disjunction in conjunct 2 is the ordinary inclusive "or": both disjuncts are permitted to hold simultaneously, and the statement does not say which one holds in any particular case. The comparisons $<$ are strict; the comparison in conjunct 1 is non-strict. Products are taken in $\mathbb{N}$, so $\prod t < \prod s$ can never hold when $\prod s = 0$; consequently, for a board $s$ containing an entry $0$ (which does not prevent a move, since $0$ is never chosen as $m$ or $n$), the conclusion forces the first disjunct, $\#_1(s) < \#_1(t)$. The product of the empty multiset is $1$ (the empty product), but the hypothesis forces $|s| \ge 2$ and $|t| = |s| \ge 2$, so neither board is empty under the hypothesis and this convention is never invoked.

Degenerate cases. Empty board, a board all of whose entries are $1$, a board with at most one entry greater than $1$, and any already-terminal board: in each of these no move exists, so the hypothesis fails and the statement is vacuous for them. The hypotheses are jointly satisfiable (e.g. $s = \{2,3\}$, $t = \{1,6\}$), so the statement is not vacuous overall.

Not asserted. Nothing about $|t|$ versus $|s|$; nothing about the multiplicity of $0$ or of any value other than $1$; no claim that $\prod t \le \prod s$ in general (only the disjunct); no claim that $\prod t$ divides $\prod s$; no claim that the process terminates, that moves can be iterated, or that either quantity is monotone along chains of moves; no claim about the entries of $t$ (e.g. that they are positive, or greater than $1$); no claim that $m, n$ are uniquely determined or that $t$ is unique for given $s$.

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