of $\Omega$ that, for any outcome $abc$, replaces every letter by its opposite. For example, $g(+-+)= -+-$. It is easy to see that $g$ is an automorphism of the observation space $T$. By Theorem \[thm:sym\], the average $R(e) = \frac12(Q(ge) + Q(e))$ is a signed probability distribution that extends ${\mathcal P}$ to all events.

Since $Q\{3,6\} = \frac14(1-\sqrt2)$, we have: $$\begin{aligned}
&R(3) = R(6) = \frac12(Q(3)+Q(6)) = \frac12Q\{3,6\} = \frac18(1-\sqrt2)\\
&R(1) = R(8) = \frac18(2-\sqrt2) - \frac18(1-\sqrt2) = \frac18
  &&\text{by \eqref{AC}}\\
&R(2) = R(7) = \frac14 - \frac18 = \frac18 &&\text{by \eqref{AB}}\\
&R(4) = R(5) = \frac14 - \frac18(1-\sqrt2) = \frac18(1+\sqrt2) 
  &&\text{by \eqref{AB}}\end{aligned}$$

To prove that $R$ is consistent with ${\mathcal P}$, it suffices to check that these probabilities satisfy the constraints , and where ${\mathcal P}\{k,l\}$ is replaced with $R(k)+R(l)$.

The constraints and the first two of the constraints are satisfied in a trivial way (because of the way they have been used to compute the outcome probabilities). $R(2)+R(4) = R(7)+R(5) = \frac18 + \frac18(1+\sqrt2) = \frac18(2+\sqrt2) ={\mathcal P}\{2,4\} = {\mathcal P}\{5,7\}$, and so the remaining two constraints are satisfied. We check the constraints. $$\begin{aligned}
&R(1)+R(5) = R(8)+R(4) = \frac18+\frac18(1+\sqrt2) = \frac18(2+\sqrt2) 
= {\mathcal P}\{1,5\} = {\mathcal P}\{4,8\}\\
&R(2)+R(6) 