Finite-budget bounded mixed-integer branch and bound terminates

The main tree can make at most $N$ calls because each call decreases the remaining integer budget. Every integer interval split has at least two values and produces finite children; every continuous split produces two children. Thus no one iteration creates an infinite operation sequence.

Write $M=\mathrm{\mid }Z\mathrm{\mid }\ge 1$ and let $R\le N$ be the remaining integer budget after main-tree calls. The fallback is entered only if $R\ge M$. Its allocation rule [F1] writes $R=Mq+h$ with integers $q=\lfloor R\mathrm{/}M\rfloor \ge 1$ and $0\le h<M$, and assigns $q+1$ calls to the first $h$ assignments and $q$ to the rest. Every subcall therefore has a positive integer node budget, and the allocations sum to $R$. The algorithm represents a zero-dimensional continuous box by a dummy singleton coordinate before calling suppliers, so every spatial input has dimension at least one. Each subcall terminates by [F2]; the finite enumeration and main-tree calls together use at most $N$ calls and return in finite time.