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the (e − x) Electors can give v.(e − x) votes, which, divided among (m + 1 − s) Candidates, supply them with v.(e − x)m + 1 − s votes apiece. Hence we must have
| vxs > | v.(e − x)m + 1 − s, |
| where v divides out; | |
| ∴ x.(m + 1 − s) > | se − sx; |
| ∴ x.(m + 1) > | se; |
| ∴ x > | sem + 1. |
In case (b), each of the x Electors can only use s of his v votes, since he can only give one to each Candidate: hence the x Electors can only give sx votes, thus supplying s Candidates with x votes apiece. But the (e − x) Electors can, as in case (a), supply (m + 1 − s) Candidates with v.(e − x)m + 1 − s votes apiece. Hence we must have
| x > | v.(e − x)m + 1 − s |
| ∴ x.(m + 1 − s) > | ve − vx; |
| ∴ x.(m + 1 − s + v) > | ve; |
| ∴ x > | ve(m + 1 − s + v). |
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