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A game whose players' utilities are specified separately, without assuming a fixed sum. Improving one player's utility need not worsen another's. A finite perfect-information game can be solved by backward induction using entire utility tuples and each moving player's own coordinate, as in max-n search.

Suppose every terminal pair satisfies $|U_A+U_B|\le ϵ$. 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 $u\le ϵ-y$. Therefore $ϵ-y<\alpha$ 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 $x+y\ge -ϵ$, so $u\le x+2ϵ$ and $x+2ϵ\le \alpha$ is a weaker sufficient test. At $ϵ=0$ 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.

Propagate an entire utility pair from a child chosen to maximize the component of the player whose turn it is. This is the two-player case of max-n search. At a terminal node let $V(s)=(U_A(s),U_B(s))$. At an A-node chooseand at a B-node replace $V_A$ by $V_B$. Do not maximize the two coordinates independently: their separate maxima may belong to incompatible outcomes. Fix a tie rule, such as the first child in the specified order. The rule recursively gives a pure strategy subgame perfect equilibrium in a finite perfect-information game: backward induction makes the moving player's selected continuation a best response at every node.^[ref]

In the original Figure 2, each of the three children of the root belongs to B. The left B-node prefers its first leaf because $1>0$, the middle B-node prefers its first leaf because $2>-2$, and the right B-node prefers its second leaf because $3>1$. Their propagated utility pairs, from left to right, areAt the root A compares $1,-1,-1$ and chooses the left child. Thus the root pair is $(1,1)$, with play following the root's left branch and then that B-node's first branch. No tie affects this outcome.

Consider a B-node whose parent is an A-node with an already completed incumbent outcome having A-coordinate $\alpha$. At the B-node, let $(x,y)$ be the best completed child pair found so far according to B's coordinate. Every remaining child can be skipped, and this B-option discarded by its parent, ifUse a strict inequality if ties must remain open; equality is safe under the fixed rule that the parent retains its already selected child on an A-coordinate tie. This is a nearly-zero-sum pruning rule.

To prove it, note first that a completed max-n search value is some terminal utility pair: each backup copies one entire child pair. Thus the nearly-zero-sum inequality holds at every completed subtree, not just at leaves. Since B maximizes its own coordinate, the ultimately selected pair $(u,v)$ at this B-node has $v\ge y$. ThereforeThe already available A-option is at least as good, so remaining children of this B-node cannot alter the parent's selected continuation under the stated tie rule. The second half of the two-sided nearly-zero-sum condition also gives $x+y\ge -ϵ$, henceConsequently $x+2ϵ\le \alpha$ is another sufficient cutoff, useful if one phrases the test using the observed A-coordinate. The pair-based test is sharper: the value of $y$ can certify a cutoff even when the conservative $x+2ϵ$ test cannot.

When $ϵ=0$, every pair satisfies $y=-x$, and the rule reduces to $x\le \alpha$, the ordinary MIN-node alpha-beta pruning cutoff. The extra $2ϵ$ in the A-coordinate-only condition is necessary in general: an observed pair $(-1,0)$ satisfies the bound for $ϵ=1$, but B prefers an unseen $(1-\delta ,\delta )$ for $0<\delta <1$. A increases by almost $2ϵ$; using just $x+ϵ$ as the upper bound would be invalid.

Implementation must discard the ruled-out B-option at its parent rather than return $(x,y)$ as an exact completed utility pair. The proof uses an actual completed alternative at that parent; it does not authorize passing arbitrary distant ancestor bounds through general-sum game backups as if every scalar alpha-beta pruning invariant still held. With these local certified cutoffs, an induction over completed subtrees preserves the selected pair and hence the root outcome.