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In case (c), the x Electors can, as in case (a), supply s Candidates with vx/x votes apiece. But each of the (e − x) Electors can only use (m + 1 − s) of his votes: hence the (e − x) Electors can only give (m + 1 − s).(e − x) votes, thus supplying (m + 1 − s) Candidates with (e − x) votes apiece. Hence we must have

vx/s >	e − x;
∴ vx >	se − sx;
∴ x.(s + v) >	se;
∴ x >	se/s + v.

In case (d), the x Electors can, as in case (b), supply s Candidates with x votes apiece. And the (e − x) Electors can, as in case (c), supply (m + 1 − s) Candidates with (e − x) votes apiece. Hence we must have

x >	e − x;
∴ 2x >	e;
∴ x >	e/2.