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\label{prop:non-boolean}
The earned topos $\mathcal{E}_\tau$ has the following logical character:
\begin{enumerate}
    \item $\Omega_\tau \cong \mathbf{2} \times \mathbf{2}$
          is a Boolean algebra:
          the complement law
          $v \lor \lnot v = \mathsf{T}$ holds for all $v$.
    \item Material implication is \textbf{paraconsistent}:
          $\mathsf{B} \Rightarrow \mathsf{F} = \mathsf{N} \neq \mathsf{T}$,
          so contradictions do not explode.
    \item The subobject complement law fails:
          there exist subobjects $S \hookrightarrow X$
          with $S \cup S^c \neq X$.
\end{enumerate}

\textbf{(1).}
In $\Omega_\tau$:
$\mathsf{B} \lor \lnot \mathsf{B}
= \mathsf{B} \lor \mathsf{N}
= \mathsf{T}$
(since $\lnot \mathsf{B} = \mathsf{N}$
by Proposition~\ref{prop:negation-B},
Chapter~\ref{ch:explosion-barrier}).
Similarly,
$\mathsf{N} \lor \lnot \mathsf{N}
= \mathsf{N} \lor \mathsf{B} = \mathsf{T}$.
The complement law holds for all four values;
$\Omega_\tau$ is a Boolean algebra ($\mathbf{2} \times \mathbf{2}$).

\textbf{(2).}
The Heyting implication in $\Omega_\tau$
does not satisfy ex falso quodlibet:
$\mathsf{B} \Rightarrow \mathsf{F} = \mathsf{N} \neq \mathsf{T}$.
This is the explosion barrier (Theorem~I.T13,
Chapter~\ref{ch:explosion-barrier}):
from $\mathrm{val}(P) = \mathsf{B}$
one cannot derive $\mathrm{val}(Q) = \mathsf{T}$
for arbitrary~$Q$.
The lattice is Boolean, but the implication is paraconsistent ---
the two properties coexist because paraconsistency
is a property of the implication connective,
not of the lattice complement.

\textbf{(3).}
Let $S \hookrightarrow X$ have
$\chi_S^{-1}(\mathsf{B}) \neq \varnothing$.
The complement $S^c$ is classified by $\lnot \circ \chi_S$:
$S^c(c) = \{x : \chi_S(x) = \mathsf{F}\}$.
Then $S(c) \cup S^c(c)
= \chi_S^{-1}(\mathsf{T}) \cup \chi_S^{-1}(\mathsf{F})
\subsetneq X(c)$,
because elements with
$\chi_S(x) \in \{\mathsf{B}, \mathsf{N}\}$
belong to neither $S$ nor $S^c$.