In the Kochen-Specker theorem, where does the assumption $v(-n)=v(n)$ come from?

I'm trying to get a better understanding of the Kochen-Specker theorem, reading the presentation in arXiv:1708.01380.

In section (2) of the above paper, the authors introduce the two KS assumptions as:

1. Let $$M\equiv \{A_i\}_i$$ be a set of observables. All these observables can be simultaneously assigned a real number. In other words, there is a mapping $$v$$ such that $$v(A_i)\in\mathbb R$$ for all $$i$$.
2. The mapping $$v$$ above satisfies, for all commuting observables $$A,B$$ we have $$v(aA+bB)=av(A)+bv(B)$$ for all $$a,b\in\mathbb C$$, and $$v(AB)=v(A)v(B)$$. Moreover, there is some observable $$X$$ such that $$v(X)\neq0$$ (this is the improved form of the axioms, as referred to in the paper).

I'll refer to these assumptions with $$\textbf{KS}1$$ and $$\textbf{KS}2$$, respectively.

The authors then proceed to prove a few consequences of $$\textbf{KS}2$$. In particular, that $$v(I)=1$$, $$v(P)\in\{0,1\}$$ for all projectors $$P$$. They also observe that $$\sum_i P_i=I$$ implies $$\sum_i v(P_i)=1$$ (Eq. (2.4) in the paper).

Now this is where I don't quite follow the presentation. The authors seem to claim (around Eq. (2.5) in the paper) that the $$\textbf{KS}2$$ condition can be reduced to the following constraints for $$v$$: $$\sum_i v(n_i)=1, \qquad v(n_i)\in\{0,1\},\qquad v(-n)=v(n),\tag{2.5}$$ for any orthonormal basis of vectors $$n_i$$.

Where does the third condition, $$v(-n)=v(n)$$ come from? I can see how the first two conditions follow from $$\textbf{KS}2$$, but not the third one.

The conditions in Eq. (2.5) don't actually follow from $$\textbf{KS}2$$ at all. In Eq. (2.5), we are instead rewriting the statement $$v(P)\in\{0,1\}$$ when $$P$$ is a projection, using a slight abuse of notation. This is pointed out in the sentence between (2.4) and (2.5) in the paper, where the authors write
(...) It is customary to identify the projectors $$P_i=|n_i\rangle\!\langle n_i|$$ with the corresponding unit vectors $$n_i$$ (defined up to a sign), with the $$n_i$$ forming a basis for the Hilbert space, and in $$d$$ dimensions write (...)
In other words, in (2.5) we are actually referring to some other function, call it $$\tilde v$$, that is defined on unit vectors $$n\in S^{d-1}$$, rather than observables, like $$v$$. We define such function so that $$v(P_n)=\tilde v(n)$$ for any $$P_n\equiv|n\rangle\!\langle n|$$ and $$n\in S^{d-1}$$.
Then, the condition $$v(P_n)\in\{0,1\}$$ becomes $$\tilde v(n)\in\{0,1\}$$, the condition $$\sum_i v(P_{n_i})=1$$ becomes $$\sum_i \tilde v(n_i)=1$$, and, finally, the fact that $$P_{-n}=P_n$$ implies that $$\tilde v(-n)=\tilde v(n)$$. In other words, the latter condition is consequence of defining a function on vectors, rather than on the operators projecting onto such vectors.