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If an admissible smaller feasible set with the same disagreement point contains the original chosen vector, deleting the other outcomes must leave the choice unchanged. The Nash bargaining solution satisfies this axiom because its unique product maximizer remains feasible.

The essential two-person bargaining domain consists of a compact convex set of feasible utility vectors $F$, a disagreement point $d\in F$, and at least one feasible vector strictly exceeding $d$ coordinatewise. The Nash bargaining solution selects a jointly feasible improvement using a rule characterized by Pareto efficiency, bargaining symmetry, positive affine invariance in bargaining and bargaining independence of irrelevant alternatives.

The product of the two players' gains over the disagreement point is maximized over the individually rational feasible set in the Nash bargaining solution. Positive affine changes of utility multiply this product by a positive constant and leave the selected payoff correspondence unchanged after transforming coordinates.

The Pareto frontier consists of feasible payoff vectors that no other feasible vector weakly improves in every coordinate and strictly improves in at least one. A Nash bargaining solution with positive gains lies on this frontier, because increasing either gain without decreasing the other raises its Nash product.

Use the usual essential two-person Nash bargaining problem: $F\subset {\mathbb{R}}^2$ is a compact convex set, $d\in F$ is the disagreement point, and some $u\in F$ satisfies $u_i>d_i$ for both players. The Nash bargaining solution isThe positive maximum exists by compactness and essentiality. On positive gains, maximizing this Nash product is equivalent to maximizing $\log (u_1-d_1)+\log (u_2-d_2)$, a strictly concave function. Convexity then gives a unique maximizer. The essentiality and compactness hypotheses matter: for example, with $F=[0,1]\times \{0\}$ and $d=0$, the product is zero everywhere and its argmax alone is not a single-valued definition.

The rule satisfies all four axioms. Pareto efficiency: a feasible vector dominating the chosen vector with at least one strict improvement would increase its positive Nash product. Bargaining symmetry: if the problem is unchanged by swapping players, uniqueness makes the answer unchanged, so the two payoffs agree. Positive affine invariance in bargaining: for $v_i=a_i{u}_i+b_i$ with $a_i>0$, gains transform to $a_i(u_i-d_i)$ and the product is multiplied by the positive constant $a_1{a}_2$, preserving its maximizer. Bargaining independence of irrelevant alternatives: if $G\subseteq F$ is another admissible feasible set containing the chosen vector and the same disagreement point, that vector remains the unique product maximizer over $G$.

To prove characterization, let $f$ be any feasible single-valued rule satisfying these axioms, and let $u^{\ast }=N(F,d)$. Normalize payoffs by the positive affine transformationThe transformed set $H$ has disagreement point zero and product maximizer $e=(1,1)$. For any $z\in H$, convexity puts $e+t(z-e)$ in $H$; for sufficiently small $t>0$ both gains remain positive. The one-sided derivative of the product at its maximum is therefore nonpositive:Thus $H\subseteq \{z:z_1+z_2\le 2\}$. By compactness choose $M\ge 0$ so every coordinate of every $z\in H$ is at least $-M$. The supporting triangle for Nash bargaining isIt contains $H$, is compact, convex, symmetric and essential, and contains disagreement zero. Symmetry forces $f(T_M,0)$ onto the diagonal; Pareto efficiency then forces it to be $e=(1,1)$. Since $e\in H\subseteq T_M$, bargaining independence of irrelevant alternatives gives $f(H,0)=e$. Undoing the normalization by positive affine invariance in bargaining gives $f(F,d)=u^{\ast }$. Hence the four axioms uniquely characterize the Nash bargaining solution on this domain.

Transforming each utility by an independent positive affine map must transform the chosen payoff by the same map. The Nash product changes only by a positive factor, so the Nash bargaining solution has this invariance.