Theorem 6 ([21]) Unweighted 1**-approval elections are computationally vulnerable to bribery by unpriced** voters.

Proof: **In this case, one simply bribes the voters that vote for the current winner to vote for our**
distinguished candidate, p, until either p **wins, in which case we accept, or until we run out of voters to**
bribe, in which case we reject.

Theorem 7 ([21]) Unweighted 1-veto elections are computationally vulnerable to bribery **by unpriced voters.**
Proof: In this case, we simply bribe voters whom veto p **to veto for the current winner instead.**
We now demonstrate how the seemingly unrelated problem of Set Cover relates to bribery of election systems, and prove the resistance to bribery of some approval-based election systems.

In [14] a reduction is given from X3C to control in approval elections **by adding or deleting voters. The**
reduction operates by encoding the elements as candidates and voters as subsets of elements, by approving these elements. A number of buffer candidates and voters are added to enforce the constraints of the problem of X3C.

We demonstrate how to modify this result to the cases of k-approval and k**-veto, where either the number**
of candidates approved or vetoed is fixed, for appropriately valued k**. The reduction requires the usage of**
additional buffer candidates and voters, as in the reduction given in **[14], each voter approves a different** number of candidates.

Theorem 8 Unweighted k**-approval elections are computationally resistant to bribery by unpriced voters for**
k ≥ 3.

Proof: We first show resistance for the case of 3-approval elections. Cases for k > **3 can be shown by a**
simple change. We make our reduction from Exact 3-Set Cover (X3C).

Let
$$S=\{s_{1},\ldots,s_{3m}\}$$
and
$T_{1}=\{t_{1,1},t_{1,2},t_{1,3}\},\ldots,T_{n}=\{t_{n,1},t_{n,2},t_{n,3}\}$.  
be an instance of X3C. Without loss of generality, we assume n ≥ m.

We construct our k**-approval election as follows. Our candidate set will be**
C = {p, p′, p′′} ∪ {s1, . . . , s3m} ∪ {b1**, . . . , b** 3(nm+m−n)
2
}.

For each 3-set Ti = {ti,1, ti,2, ti,3} we construct a voter who approves of {ti,1, ti,2, ti,3}**. For 1** ≤ i ≤
3(nm+m−n)
2 we add voters that approve of bi and two members of s1, . . . , s3m**. We choose these votes in a**
way such that each s candidate receives exactly n + 1 approvals in total. Finally, we add n − m **votes that**
approve of {p, p′, p′′}**. We set our bribery quota** q to m.