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This proves the full statement of L190. Write $V=\mathbb R^P\times\mathbb R^P$ and

$$\ell_u(m,k)=\sum_{v\in P}m_vL_v-k_uL_u, \qquad C=\{(m,k):m_v\ge0,\ \ell_u(m,k)\le0\ (u,v\in P)\}.$$

Unique factorization implies that $L_p,L_q,L_r$ are linearly independent over $\mathbb Q$. The coefficient vector of each $\ell_u$ therefore has rational rank three. Multiplying a vector by a nonzero real scalar preserves this rank: scalar multiplication is a $\mathbb Q$-linear isomorphism between the two coordinate spans. Consequently no nonzero normal of rational rank at most two has the same kernel as $\ell_u$.

Fix $u\in P$. There is a relatively open, nonempty part $F^\circ$ of the hyperplane $H=\ker\ell_u$ on which every $m_v$ is positive and all the inequalities $\ell_v<0$ for $v\ne u$ are strict. For example, start at the point with all $m_v=1$, with $k_u=\sum_vL_v/L_u$, and with $k_v=2\sum_wL_w/L_v$ for $v\ne u$. A small neighbourhood within $H$ has the required properties. All its $k_v$ are positive as well.

Suppose a proposed finite indicator identity uses the normals $a_1,\ldots,a_M$, after listing every defining inequality of every region. Each $H\cap\ker a_i$ is a proper linear subspace of $H$, by the rank observation. Finitely many proper linear subspaces cannot cover a nonempty open subset of $H$. Choose $x\in F^\circ$ outside their union.

There is a ball $U$ around $x$ in $V$ on which all coordinates $m_v,k_v$ are positive, all $\ell_v$ with $v\ne u$ are negative, and each $a_i$ has one fixed, nonzero sign. Both sides of $H$ meet $U$. Choose $x_{\rm in},x_{\rm out}\in U$ with

$$\ell_u(x_{\rm in})<0<\ell_u(x_{\rm out}).$$

For each positive integer $N$, round the coordinates of $Nx_{\rm in}$ and $Nx_{\rm out}$ to nearest integers, obtaining $z_{{\rm in},N}$ and $z_{{\rm out},N}$. The rounding errors are bounded independently of $N$. Thus, for any fixed linear form $a$,

$$a\cdot z_{{\rm in},N}=N(a\cdot x_{\rm in})+O(1), \qquad a\cdot z_{{\rm out},N}=N(a\cdot x_{\rm out})+O(1).$$

For all sufficiently large $N$, both rounded vectors have positive coordinates. Every $\ell_v(z_{{\rm in},N})$ is negative, hence less than $L_v$, so $z_{{\rm in},N}\in\Omega$. In contrast, $\ell_u(z_{{\rm out},N})>L_u$, so $z_{{\rm out},N}\notin\Omega$.

For every defining inequality with normal $a_i$ and constant $b_i$, the values at these two vectors tend to the same signed infinity. They therefore either both satisfy that inequality or both fail it, whether the comparison is strict or non-strict. Because there are only finitely many inequalities, for all sufficiently large $N$ the two vectors have identical membership in every region $E_j$. The proposed right-hand side then takes equal values at them, whereas $\mathbf1_\Omega$ takes values one and zero. This contradiction proves the indicator assertion.

Finally, the coordinate span of a normal is unchanged by an invertible rational linear transformation: each transformed coefficient is a rational linear combination of the old ones, and the inverse gives the reverse inclusion. Translation changes only the constant term. A product of planar polyhedra, each with only one irrational slope, and rational polyhedra has defining normals of rational rank at most two, even when the slopes differ between the planar factors. The indicator assertion therefore applies to every finite signed partition into their invertible affine rational images.