Court Opinion

ID: 5436540
Source: CourtListenerOpinion
Date Created: 2022-01-08 17:55:13.978361+00
Date Added: 2024-06-11T08:31:51.540518
License: Public Domain

By the Court, Sanderson, J.:
It is unnecessary to consider the ruling of the Court below in respect to the votes of Diller and Dikeman. The facts in relation to each are substantially the same, and the ruling of the Court was the same. If we were to hold that the Court erred in not counting Biller’s vote for the contestant, we should also be compelled to hold that it erred in not counting Bikeman’s vote for the respondent. So whether their votes be counted or not will make no difference in the result.
Whitford’s vote should have been rejected. His name was not on the roll at Poverty Hill, where he voted, and his vote was therefore illegal, (Statutes 1865-6, p. 297, Sec. 29,) and must be deducted from the respondent’s tally.
*276Smith’s vote, however, was properly allowed to respondent. He was on the Great Begister and also on the poll list. He was also an elector according to the testimony. Whether the Board of Begistration required full or competent evidence of his naturalization cannot be inquired into in this case. The question here is: Is he a qualified elector of the precinct at which he voted, and was his name at the time upon the Great Begister and poll list ?
Deducting Whitford’s vote from the respondent’s tally leaves him three hundred and ninety-eight instead of three hundred and ninety-nine, as found by the Court below.
The Court below erred in counting for contestant the supposed votes of Gonsalves, Larkin and Haas, under the pretense that they would have voted for him had they been allowed to vote. In all contests of this character, the question is, which candidate received the highest number of legal votes ? The idea that the supposed votes of persons who did not vote, but who could have voted had they taken the necessary legal steps to entitle them to do so, should be counted for the candidate for whom they would have voted, is simply preposterous.
Deducting these three votes from the contestant’s tally, leaves biun three hundred and ninety-four instead of three hundred and ninety-seven, as found by the Court below. If we add to this number the votes of Jarnegan, Pittorff and Meyer, the respondent would still have a majority of one, so it is unnecessary to consider the ruling of the Court below in regard to their votes.
Judgment affirmed.