how to show vector is an Imputation of the core? can anyone help me on this? I would like to show that the vector reward: $(v(1,2)+v(1,3)-v(1,2,3), v(1,2,3)-v(1,3), v(1,2,3)-v(1,2))$ is an imputation of the core Knowing that: $v(1,2)+v(1,3)+v(2,3)\le2(1,2,3)$ and $v(1,2)+v(1,3)-v(1,2,3)\ge v(1)$ I know that it needs to respect the group and individual rationality but I do not know how to start. Can anyone give an hint? Thanks

An imputation $\mathbf{x}$ is in the core of a three person game, if it satisfies the following set of inequalities

$$v(12) + v(13) +v (23) \le 2 v(123) \qquad v(1) + v(23) \le v(123) \qquad v(2) + v(13) \le v(123) \qquad v(3) + v(12) \le v(123).$$

Now, the following imputation is given

$$\mathbf{x}=(x_{1},x_{2},x_{3})=(v(12)+v(13)-v(123),v(123)-v(13),v(123)-v(12)). $$

We have to check that these inequalities are satisfied by $\mathbf{x}$ to conclude that this imputation belongs to the core

Under the mentioned assumptions and that the game is super-additive, we obtain now

$$ v(2) \le x_{2} = v(123) - v(13) \qquad v(3) \le x_{3} = v(123) - v(12). $$

Moreover, it holds that

$$ \sum_{i \in N} x_{i} = v(12) + v(13) - v(123) + v(123) -v(13)+v(123)-v(12) = v(123).$$

Similar, we get

$$x(12) = x_{1}+x_{2} = v(12) \qquad x(13) = x_{1}+x_{3} = v(13) \qquad x(23) = x_{2}+x_{3} = v(23). $$

This shows that the imputation $\mathbf{x}$ is in the core.