SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof. For the bin packing setting, consider an instance with  $n$  agents and  $n + 1$  items. The capacity of each agent's bins is 1, i.e.,  $c_{i} = 1$  for all  $i \in N$ . Each item is very tiny so that every agent can pack all items in just one bin, e.g.,  $s_{i,j} = \frac{1}{n + 1}$  for any  $i \in N$  and  $e_j \in M$ . Therefore, we have  $v_{i}(M) = 1$  and  $\mathsf{PROP}_i = \frac{1}{n}$  for each agent  $i \in N$ . By the pigeonhole principle, at least one agent  $i$  receives two or more items in any allocation of  $M$ . After removing any item  $e \in A_i$ , agent  $i$  still needs one bin to pack the remaining items. Hence, we have  $v_{i}(A_{i} \backslash \{e\}) = 1 = n \cdot \mathsf{PROP}_{i}$  for any  $e \in A_i$ .
For the job scheduling setting, consider an instance with  $2n$  agents and  $2n + 1$  items where each agent possesses  $2n$  machines with the same speed of 1, and the size of each item is 1 for every agent. It can be easily seen that for every agent  $i \in N$ , the maximum completion time of her machines is minimized when assigning two items to one machine and one item to each of the remaining  $2n - 1$  machines. Therefore, we have  $v_{i}(M) = 2$  and  $\mathrm{PROP}_i = \frac{2}{2n} = \frac{1}{n}$  for any  $i \in N$ . Similarly, by the pigeonhole principle, at least one agent  $i$  receives two or more items in any allocation of  $M$ . This implies that  $v_{i}(A_{i} \backslash \{e\}) = 1 = n \cdot \mathrm{PROP}_{i}$  for any  $e \in A_{i}$ , thus completing the proof.