Yes, the former. For example, for âat most n are differentâ, if we do this for each pair:

```
(1 - c) Ă (c) = 0 // c is boolean
(ÎŒ) Ă (c) = (a - b) // a â b â c â 0
```

then for pairs that are different we must have c = 1, and for pairs that are the same we can choose either c = 0 or c = 1. So we can arbitrarily pick m - n of the pairs that are the same and set c = 0 for those; then set c = 1 for all the other (n) pairs. Then the sum of the câs will be n.

Similarly for âat least n are differentâ, we do this for each pair:

```
(1 - c) Ă (c) = 0 // c is boolean
(a - b) Ă (Î») = (c) // a = b â c = 0
```

then for pairs that are the same we must have c = 0, and for pairs that are different we can choose either c = 0 or c = 1. So we can arbitrarily pick n of the pairs that are different and set c = 1 for those; then set c = 0 for all the other (m - n) pairs. Then the sum of the câs will be n.

Incidentally, thereâs no need for a separate constraint to prove that the sum is n (in any of the three cases). Just substitute n - âc_{1âŠm-1} for all occurrences of c_{0}.