= Nearly-zero-sum pruning
{title2=$\epsilon-y\le\alpha$}
Suppose every terminal pair satisfies $|U_A+U_B|\le\epsilon$. At a B-node with immediate A-parent, let $\alpha$ be the A-coordinate of a completed alternative already retained by the parent. If a completed B-child supplies B-coordinate $y$, then every eventual B-selected pair $(u,v)$ has $v\ge y$ and hence $u\le\epsilon-y$. Therefore $\epsilon-y<\alpha$ permits discarding this B-option and skipping its remaining children. Equality is also safe if the parent keeps its existing option on a tie.
The inequality remains valid at completed subtrees because <max-n search> copies one terminal pair. If the observed pair is $(x,y)$, then $x+y\ge-\epsilon$, so $u\le x+2\epsilon$ and $x+2\epsilon\le\alpha$ is a weaker sufficient test. At $\epsilon=0$ this reduces to the usual MIN-node <alpha-beta pruning> cutoff. A discarded option is not assigned a fabricated exact pair; the proof uses a certified alternative at the immediate parent.
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