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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

vx/s >	v.(e − x)/m + 1 − s,
	⁠where v divides out;
∴ x.(m + 1 − s) >	se − sx;
∴ x.(m + 1) >	se;
∴ x >	se/m + 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).