disagreement by a relative disagreement situation because, when an agent i informs her opponent j that they are in relative disagreement, then j knows that i proposed one of i's CAF that is a restriction of her CAF and j accepts it. Instead, in 1-n MN, the previous situations may raise: the auctioneer may not accept the proposal of one of the negotiating agents who said that they are in relative disagreement, because the proposal is not shared by the other agents. The set of deductive rules for the auctioneer is an extension of those in Table 3 with those in Table 5. In particular, all the added rules are violations and they represent the changing of the negotiation situation from relative disagreement to the relations of absolute disagreement, essence disagreement and compatibility, and from agreement to all the possible relations between agents: absolute disagreement, essence disagreement, compatibility, relative disagreement and agreement. The rule (Ag-Ag) is not only for the auctioneer but also for negotiating agents and it is used by them whenever the auctioneer proposes an acceptable angle.

Moreover, the system transition rules are different from the 1-1 MN ones (Table 4)
because the agreement and disagreement conditions are different. In particular, the test of the agreement condition needs to count the number of agreeing agents. The 1-n MN ends in:
- *disagreement* when all the agents involved are in stubbornness and no agreement is found yet;
- *agreement* when all the agents or an acceptable part of them, i.e. α agents where α is the degree of sharing, agree about a common angle. In all the other cases, the MN continues. The system transition rules for 1-n MN
are in Table 6. The following example shows a simple 1-n MN ending negatively or positively depending on the sharing degree decided in front of the beginning of the MN.

Example 5 Let Alice, Bob and Charles be three agents negotiating the definition of the term "vehicle". Suppose that the initial viewpoint of Alice is flex 0A = has3wheels ∧ hasSteeringWheel ∧ *hasMotor* and her stubbornness knowledge is stubA = (has3wheels ∨ has4wheels) ∧ *hasSteeringWheel*
∧(hasMotor ∨ has2bicyclePedals ∨ *has4bicyclePedals*)
Suppose that Bob's initial viewpoint is flex 0B = has2wheels ∧ hasHandlebar ∧ *has2bicyclePedals* and that his stubbornness knowledge is stubB = (has2wheels ∨ has4wheels) ∧ (hasSteeringWheel ∨ *hasHandlebar* ) ∧
(hasMotor ∨ has2bicyclePedals ∨ *has4bicyclePedals*)
and that the initial viewpoint of Charles is flex 0C = has4wheels ∧ hasHandlebar ∧ hasTowBar