# Objections and counterobjections: The Nash approach as a model of haggling and dispute resolution

One possible criticism of the axiomatic bargaining approach is that the axioms are often too abstract or difficult to justify in terms of economic behaviour. Why is one particular set of axioms more acceptable or desirable than another, and why should parties in a bargaining situation obey them instead of another set of axioms? Fortunately, however, in the case of the Nash solution, there is an alternative, more direct and more intuitive approach to axiomatic bargaining, which yields very similar results.^{1} Moreover, in the context of law and economics, the approach is a very natural one.

Let us sketch this approach. Suppose that the factory and the residents are in a legal dispute in which their disagreement points are assumed to be zero, and that there is a proposal on the table from the residents to share the gains from trade as follows:

An objection by the factory to any proposed agreement n* is a pair *(p _{F},p)* for which

*pn*where 0 <

_{F}>n*,*p <*1. The interpretation of an objection is that by threatening to walk away from the bargaining table, the factory can cause negotiations to break down with probability 1 - p, and if this were to occur the factory would receive nothing. However, if negotiations do not break down (which occurs with probability

*p*) the factory will receive

*n*

_{F}. The expected value to the factory of this objection is therefore

*pn*which to be something that the factory wishes to raise, must be higher than the payoff that the factory would receive under the proposal n* that is currently on the table. If the factory can find such a pair

_{F},*(n*

_{F}, p), then it is said to have an objection to n*.

The residents are said to have a counterobjection to the factory's objection *(n _{F}*,p) if pn* >

*n*The interpretation is that even in this new, risky situation that the factory has created by lodging an objection, the residents can profitably insist on the original proposal. If negotiations were to break down, the residents would also receive nothing, but if negotiations do not break down (which occurs with probability

_{R}.*p*) and the residents insist on the original proposal, then they will receive n*. The expected value of this countergambit is therefore pnR, which is higher than

*n*the payoff that the residents would receive under the objection conveyed by the factory. If the residents can find such a pair

_{R},*(n*

**,*

*p)*

*,*then they are said to have a counterobjection to the factory's objection

*(*

*n*

*,p).*

_{F}It turns out that the Nash solution described above can also be characterised as the set of all agreements such that the residents can counterobject to *every* objection of the factory, and the factory can counterobject to *every *objection by the residents. In other words, Nash's axioms, if we wish, can be dispensed with altogether and we can instead think about the proposed solution in terms of objections and counterobjections.

The requirement that every possible objection can be met with a counterobjection is a very restrictive requirement which narrows down the set of possible outcomes to a single point. Suppose, for example, that the disagreement point is (0,0), and that the players are negotiating over a surplus of *p =* 1. Then the point (/,/) is the only point at which the residents can counterobject to every objection of the factory, and the factory can counterobject to every objection by the residents.

To see the link between the Nash solution and the objection/coun- terobjection result, suppose that there is some other outcome which does not maximise the product of the parties' payoffs, but which, for every objection that the factory had, the residents could make a counterobjection. Let this solution be (p,1 - p), and suppose it does not maximise the product of the payoffs. We will show that this leads to a contradiction. Suppose that the factory objects to this outcome, nominating an alternative split, *p' > p.* Choose *p* arbitrarily, with the only restric-

*n*

tion being that that *p **>* . Then, by assumption, the residents have a

*П*

counterobjection, so *p**>*-*—* as well. But this implies that — >-.

1 *— **n n**'* 1*— **n*

*n* 1 *— **%**'*

If not (that is, if we instead had ? < ) it would be possible to find a

п 1 — *n*

*p* which could be 'squeezed' in between *—* and ^{1 — n} , which would be an

*n**'* 1 — *n*

objection for the factory, but which would not allow the possibility of a

counterobjection by the residents. But since * ^{—}* >

^{1}—

^{П}*—*, and

*n*

*'*was cho

*—*1 —

*n*

sen arbitrarily, this means that n(1 — *n**)* must maximise the product of the parties' payoffs. We are therefore left with a contradiction, and it must therefore be the case that the Nash solution is the only outcome which has the property that for every objection that the factory can make, the residents can make a counterobjection.