A move adds a 1 or shrinks the product (positive boards)

Part (a) of the source, both bullets — corrected. A move never destroys a $1$, and it either creates one (when $\gcd(m,n) = 1$) or strictly shrinks the product of the board (when $\gcd(m,n) > 1$). Together these give the monovariant behind termination: the count of $1$s can rise at most as far as the board's size, and the product is a positive integer that cannot fall forever.

This supersedes IMO2026P1.move_ones_or_prod, which is false. That card omitted the positivity hypothesis and so quantified over boards containing $0$, which the problem never produces. On $s = \{0, 4, 6\}$ with $m = 4$, $n = 6$ the move gives $t = \{2, 6, 0\}$: both boards contain $0$, so both products are $0$ and the product cannot strictly decrease, while neither board contains a $1$, so the count cannot strictly increase either. Both disjuncts fail and the statement is refuted.

hpos closes exactly that gap and costs nothing the problem needs: the starting board has every entry greater than $1$, and $\gcd(m,n)$ and $\operatorname{lcm}(m,n)/\gcd(m,n)$ of positive numbers are positive, so positivity is preserved along every move.