%
\label{prop:bayesian-factorization}
% Depends: I.D21, I.T12, I.P13
Every $\tau$-Bayesian state factors
through the spectral decomposition
(Theorem~\ref{thm:spectral-decomposition}, I.T12).
That is, for every $\tau$-Bayesian state $S$,
\[
    \boxed{%
    S = S_+ \circ e_+ \;+\; S_- \circ e_-,}
\]
where $S_+$ and $S_-$
are classical probability distributions
on $\{\mathsf{T}, \mathsf{F}\}$,
corresponding to the $B$-channel
and $C$-channel respectively.

[Proof sketch]
By the bipolar decomposition
(Definition~\ref{def:bipolar-spectral-algebra}, I.D21),
every element of $[0,1]_\tau$
decomposes via the idempotents
$e_+$ and $e_-$
from the spectral decomposition.
The polarity coherence condition
(Definition~\ref{def:tau-bayesian-state}, condition~2)
forces $S$ to respect this decomposition:
the probability assigned to each truth value
is determined by the witness pair,
which splits into $B$-channel and $C$-channel components.

The component $S_+ := S \circ e_+$
assigns probability only to the outcomes
of the $B$-sector witness $w_B$.
Since $w_B$ delivers either ``confirms'' or ``refutes''
(a binary outcome),
$S_+$ is a classical probability distribution
on $\{\mathsf{T}, \mathsf{F}\}$.
Similarly, $S_- := S \circ e_-$
is a classical probability distribution
on $\{\mathsf{T}, \mathsf{F}\}$
for the $C$-sector.

The normalization condition ensures
that $S_+$ and $S_-$ combine consistently:
$S(\mathsf{T}) = S_+(\mathsf{T}) \cdot S_-(\mathsf{T})$,
$S(\mathsf{F}) = S_+(\mathsf{F}) \cdot S_-(\mathsf{F})$,
$S(\mathsf{B}) = S_+(\mathsf{T}) \cdot S_-(\mathsf{F})
+ S_+(\mathsf{F}) \cdot S_-(\mathsf{T})$,
and $S(\mathsf{N})$ absorbs the remainder.
The Boolean recovery theorem
(Proposition~\ref{prop:boolean-recovery})
then shows that each sector reduces
to classical probability.