We note that the ontic completion of $(\overline{\mathfrak{S}}_A \hat{\otimes} \overline{\mathfrak{S}}_B, \star)$ can be defined because the basic conditions are fulfilled.

Our aim is now to check all the basic requirements addressed in subsection 4.1.

**Lemma 4.33.** $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$ is a down complete Inf semi-lattice. The lowest bound is given by $\perp_{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} = \perp_{\overline{\mathfrak{S}}_A \hat{\otimes} \overline{\mathfrak{S}}_B}$.

*Proof.* Direct consequence of the construction of the ontic completion $((\overline{\mathfrak{S}}_A \hat{\otimes} \overline{\mathfrak{S}}_B, \star))_c$. $\square$

**Lemma 4.34.** $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$ admits a real structure given by $(\overline{\mathfrak{S}}_A \hat{\otimes} \overline{\mathfrak{S}}_B, \star)$.

*Proof.* Direct consequence of Theorem 3.21 for $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$ defined as the ontic completion of $(\overline{\mathfrak{S}}_A \hat{\otimes} \overline{\mathfrak{S}}_B, \star)$. $\square$

**Lemma 4.35.** There exists a map $\iota^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B}$ from $\overline{\mathfrak{S}}_A \times \overline{\mathfrak{S}}_B$ to $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$.

*Proof.* Trivial $\square$

**Lemma 4.36.**

$$ \begin{gathered} \forall \left( \bigcap_{i \in I}^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} \sigma_{i,A} \hat{\otimes} \sigma_{i,B} \right) \in \overline{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B}, \quad \forall \left( \bigcap_{j \in J}^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} \sigma'_{j,A} \hat{\otimes} \sigma'_{j,B} \right) \in \overline{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B}, \\ \left( \forall \iota_A \in \mathfrak{S}_A, \iota_B \in \mathfrak{S}_B, \bigwedge_{i \in I} \varepsilon_{\iota_A}^{\mathfrak{S}_A}(\sigma_{i,A}) \cdot \varepsilon_{\iota_B}^{\mathfrak{S}_B}(\sigma_{i,B}) = \bigwedge_{j \in J} \varepsilon_{\iota_A}^{\mathfrak{S}_A}(\sigma'_{j,A}) \cdot \varepsilon_{\iota_B}^{\mathfrak{S}_B}(\sigma'_{j,B}) \right) \\ \Rightarrow \left( \bigcap_{i \in I}^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} \sigma_{i,A} \hat{\otimes} \sigma_{i,B} = \bigcap_{j \in J}^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} \sigma'_{j,A} \hat{\otimes} \sigma'_{j,B} \right). \end{gathered} \quad (4.69) $$

*Proof.* Direct consequence of the definition of the tensor product. $\square$

**Lemma 4.37.** We have the two required bimorphic properties for the tensor product. Explicitly, for any $\{\sigma_{i,A} | i \in I\} \subseteq \overline{\mathfrak{S}}_A$, $\{\sigma_{j,B} | j \in J\} \subseteq \overline{\mathfrak{S}}_B$, $\sigma_A \in \overline{\mathfrak{S}}_A$ and $\sigma_B \in \overline{\mathfrak{S}}_B$, we have

$$ (\bigcap_{i \in I}^{\mathfrak{S}_A} \sigma_{i,A}) \hat{\otimes} \sigma_B = \bigcap_{i \in I}^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} (\sigma_{i,A} \hat{\otimes} \sigma_B), \quad (4.70) $$

$$ \sigma_A \hat{\otimes} (\bigcap_{i \in I}^{\mathfrak{S}_B} \sigma_{i,B}) = \bigcap_{i \in I}^{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} (\sigma_A \hat{\otimes} \sigma_{i,B}). \quad (4.71) $$

*Proof.* Direct consequence of Lemma 4.36. $\square$

**Lemma 4.38.** We have the two following partial traces

$$ \begin{gathered} \zeta_{(1)}^{\mathfrak{S}_A \mathfrak{S}_B} : \quad \overline{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B} \rightarrow \overline{\mathfrak{S}_A} \\ \bigcap_{i \in I}^{\overline{\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B}} \sigma_{i,A} \widetilde{\otimes} \sigma_{i,B} \mapsto \bigcap_{i \in I}^{\overline{\mathfrak{S}_A}} \sigma_{i,A} \end{gathered} \quad (4.72) $$

*Proof.* Direct consequence of Theorem 4.11. $\square$

**Theorem 4.39.** The tensor product $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$ defined in Definition 4.32 satisfies the requirements addressed in subsection 4.1 for the description of generic bipartite experiments.

*Proof.* According to subsection 4.1, this is a direct consequence of Lemma 4.33, Lemma 4.34, Lemma 4.35, Lemma 4.37, Lemma 4.36, Lemma 4.38. $\square$
We note that the minimal tensor product had already proven to be adequate for the description of bipartite deterministic experiments, especially because of the property saying that the minimal tensor product of simplex spaces of states is itself a simplex space of states (Theorem 4.22).

Now, the tensor product $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$ defined in Definition 4.32 reveals to be adequate for the description of bipartite completely indeterministic experiments on a compound system build from two completely indeterministic individual systems described respectively by the ontic completions $((\mathfrak{S}_A, \star))_c$ and $((\mathfrak{S}_B, \star))_c$. This fact is exhibited in the following theorem.

**Theorem 4.40.** If $(\mathfrak{S}_A, \star)$ and $(\mathfrak{S}_B, \star)$ are completely indeterministic spaces of states, then $(\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B, \star)$ is also a completely indeterministic space of states.

*Proof.* Let us consider $(\mathfrak{S}_A, \star)$ and $(\mathfrak{S}_B, \star)$ two completely indeterministic spaces of states. Let us consider $\sigma := \sigma_A \tilde{\otimes} \sigma_B, \lambda := \lambda_A \tilde{\otimes} \lambda_B \in (\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B)^{\text{pure}}$ such that $\sigma \perp \lambda$. We recall that $\sigma \perp \lambda$ means $\lambda_A \tilde{\otimes} \lambda_B \sqsubseteq_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} (\sigma_A^* \tilde{\otimes} \perp_{\mathfrak{S}_B} \sqcap_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} \perp_{\mathfrak{S}_A} \tilde{\otimes} \sigma_B^*)$ which means $\lambda_A \sqsupseteq_{\mathfrak{S}_A} \sigma_A^*$ or $\lambda_B \sqsupseteq_{\mathfrak{S}_B} \sigma_B^*$, i.e. $\sigma_A \perp \lambda_A$ or $\sigma_B \perp \lambda_B$. Let us distinguish the three different cases.

(1) $\sigma_A \perp \lambda_A$ and $\sigma_B \perp \lambda_B$. $(\mathfrak{S}_A, \star)$ and $(\mathfrak{S}_B, \star)$ being two completely indeterministic spaces of states, we deduce that there exists $\kappa_A \in \mathfrak{S}_A^{\text{pure}}$ and $\kappa_B \in \mathfrak{S}_B^{\text{pure}}$ such that $\kappa_A \not\perp \lambda_A$, $\kappa_A \not\perp \sigma_A$, and $\kappa_B \not\perp \lambda_B$, $\kappa_B \not\perp \sigma_B$, i.e. $\kappa_A \not\sqsupseteq_{\mathfrak{S}_A} \sigma_A^*$ and $\kappa_A \not\sqsupseteq_{\mathfrak{S}_A} \lambda_A^*$ and $\kappa_B \not\sqsupseteq_{\mathfrak{S}_B} \lambda_B^*$ and $\kappa_B \not\sqsupseteq_{\mathfrak{S}_B} \sigma_B^*$. Then, we have $(\kappa_A \tilde{\otimes} \kappa_B) \not\sqsupseteq_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} (\sigma_A^* \tilde{\otimes} \perp_{\mathfrak{S}_B} \sqcap_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} \perp_{\mathfrak{S}_A} \tilde{\otimes} \sigma_B^*)$ and $(\kappa_A \tilde{\otimes} \kappa_B) \not\sqsupseteq_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} (\lambda_A^* \tilde{\otimes} \perp_{\mathfrak{S}_B} \sqcap_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} \perp_{\mathfrak{S}_A} \tilde{\otimes} \lambda_B^*)$, i.e. $(\kappa_A \tilde{\otimes} \kappa_B) \not\perp (\sigma_A \tilde{\otimes} \sigma_B)$ and $(\kappa_A \tilde{\otimes} \kappa_B) \not\perp (\lambda_A \tilde{\otimes} \lambda_B)$.

(2) $\sigma_A \perp \lambda_A$ and $\sigma_B \perp \lambda_B$. $(\mathfrak{S}_A, \star)$ being a completely indeterministic spaces of states, we deduce that there exists $\kappa_A \in \mathfrak{S}_A^{\text{pure}}$ such that $\kappa_A \not\perp \lambda_A$, $\kappa_A \not\perp \sigma_A$, i.e. $\kappa_A \not\sqsupseteq_{\mathfrak{S}_A} \sigma_A^*$ and $\kappa_A \not\sqsupseteq_{\mathfrak{S}_A} \lambda_A^*$. Let us fix $\kappa_B := \lambda_B$. We then note that $\kappa_B \not\sqsupseteq_{\mathfrak{S}_B} \lambda_B^*$ and $\kappa_B \not\sqsupseteq_{\mathfrak{S}_B} \sigma_B^*$. We then have $(\kappa_A \tilde{\otimes} \kappa_B) \not\sqsupseteq_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} (\sigma_A^* \tilde{\otimes} \perp_{\mathfrak{S}_B} \sqcap_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} \perp_{\mathfrak{S}_A} \tilde{\otimes} \sigma_B^*)$ and $(\kappa_A \tilde{\otimes} \kappa_B) \not\sqsupseteq_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} (\lambda_A^* \tilde{\otimes} \perp_{\mathfrak{S}_B} \sqcap_{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} \perp_{\mathfrak{S}_A} \tilde{\otimes} \lambda_B^*)$, i.e. $(\kappa_A \tilde{\otimes} \kappa_B) \not\perp (\sigma_A \tilde{\otimes} \sigma_B)$ and $(\kappa_A \tilde{\otimes} \kappa_B) \not\perp (\lambda_A \tilde{\otimes} \lambda_B)$.

(3) idem.

This concludes the proof. $\square$

## 4.8. Morphisms for the bipartite indeterministic experiments

Let us consider a morphism $f$ (resp. $g$) from a states space $\mathfrak{S}_{A_1} := \mathfrak{J}_{\mathfrak{S}_{A_1}}^c$ (resp. $\mathfrak{S}_{B_1} := \mathfrak{J}_{\mathfrak{S}_{B_1}}^c$) to another states space $\mathfrak{S}_{A_2} := \mathfrak{J}_{\mathfrak{S}_{A_2}}^c$ (resp. $\mathfrak{S}_{B_2} := \mathfrak{J}_{\mathfrak{S}_{B_2}}^c$).

We can define the morphism $(f \tilde{\otimes} g)$ from the states space $\hat{S}_{A_1 B_1} := \mathfrak{J}_{\hat{S}_{A_1 B_1}}^c$ with $\overline{S}_{A_1 B_1} = \overline{\mathfrak{S}}_{A_1} \tilde{\otimes} \overline{\mathfrak{S}}_{B_1}$ to the states space $\hat{S}_{A_2 B_2} := \mathfrak{J}_{\hat{S}_{A_2 B_2}}^c$ with $\overline{S}_{A_2 B_2} = \overline{\mathfrak{S}}_{A_2} \tilde{\otimes} \overline{\mathfrak{S}}_{B_2}$ by

$$ \Theta_{\overline{\mathfrak{S}}_{A_2} \tilde{\otimes} \overline{\mathfrak{S}}_{B_2}}((f \tilde{\otimes} g)(\xi)) = \tag{4.73} \\ \qquad c l_c^{\overline{\mathfrak{S}}_{A_2} \tilde{\otimes} \overline{\mathfrak{S}}_{B_2}} (\{\sqcap_{i \in I}^{\overline{\mathfrak{S}}_{A_2 B_2}} f(\omega_{i,A_1}) \tilde{\otimes} g(\omega_{i,B_1}) | (\sqcap_{i \in I}^{\overline{\mathfrak{S}}_{A_1 B_1}} \omega_{i,A_1} \tilde{\otimes} \omega_{i,B_1}) \in \Theta_{\overline{\mathfrak{S}}_{A_1} \tilde{\otimes} \overline{\mathfrak{S}}_{B_1}}(\xi) \}) $$
**5. Ontic completions and contextuality**

In this section, we will clarify the concept of compatibility between real measure-
ment operators and construct the concept of operational description, which will lead
us to establish the concept of contextuality. Throughout our demonstration, we will
observe how generalized spaces of states with hidden states are not only obtained
naturally by a completion procedure from real indeterministic spaces of states (see
Section 3), but are also linked to *contextual* empirical models associated with the
operational descriptions of these real spaces of states. The notions relative to con-
textuality presented here are directly inspired by the notions developed in [1].

**5.1. Notions relative to contextuality**

During this subsection, the states/effects Chu Space ($\mathfrak{S}$, $\mathfrak{E}_\mathfrak{S}$, $\varepsilon^{\mathfrak{S}}$) will be equipped
with a real structure denoted by $(\mathfrak{S}, \star)$. As before, $\mathfrak{E}_\mathfrak{S}$ is defined as the sub Inf
semi-lattice of $\mathfrak{E}_\mathfrak{S}$ formed by the elements of

$$
\{ l_{(\sigma, \sigma')} | \sigma, \sigma' \in \mathfrak{S} \setminus \{\perp_{\mathfrak{S}}\}, \sigma' \stackrel{\blacksquare}{=} \mathfrak{S} | \sigma^* \} \cup \{ l_{(\sigma, \cdot)} | \sigma \in \mathfrak{S} \} \cup \{ l_{(\cdot, \sigma)} | \sigma \in \mathfrak{S} \} \cup \{ l_{(\cdot, \cdot)} \}
$$

We recall that, $\mathfrak{B}$ being a simplex, the $N$-th tensor product $\mathfrak{B}^{\otimes N}$ is also a simplex (cf Lemma 4.22). The space $\mathfrak{B}^{\otimes N}$ is then naturally equipped with a real structure (cf formula (2.107)). Hence, the reduced space of effects $\mathfrak{E}_{\mathfrak{B}^{\otimes N}}$ is naturally defined.

**Definition 5.1.** $N$ real measurement maps $\phi_1, \cdots, \phi_N \in \overline{\mathfrak{M}}_{\mathfrak{S}}$ are said to be jointly compatible iff there exists a map $\Psi_{(\phi_1, \cdots, \phi_N)}$, called joint morphism, defined from $\mathfrak{S}$ to $\mathfrak{B}^{\otimes N}$ and satisfying

$$
\forall \sigma_1, \sigma_2 \in \mathfrak{S}, \quad \Psi_{(\phi_1, \dots, \phi_N)}(\sigma_1 \sqcap_{\mathfrak{S}} \sigma_2) = \Psi_{(\phi_1, \dots, \phi_N)}(\sigma_1) \sqcap_{\mathfrak{B}^{\otimes N}} \Psi_{(\phi_1, \dots, \phi_N)}(\sigma_2), \quad (5.1)
$$

$$
\Psi^*_{(\phi_1, \dots, \phi_N)} (\bar{\mathfrak{E}}_{\bar{\mathfrak{B}}^{\otimes N}}) \subseteq \bar{\mathfrak{E}}_{\mathfrak{S}}, \qquad (5.2)
$$

$$
\forall i = 1, \dots, N, \quad \phi_i = \zeta_{(i)}^{\mathfrak{B}^{\otimes \mathfrak{B}}} \circ \Psi_{(\phi_1, \dots, \phi_N)}. \qquad (5.3)
$$

We recall that $\zeta_{(i)}^{\mathfrak{B}^{\otimes \mathfrak{B}}}$ is the partial trace projecting the tensor product $\mathfrak{B}^{\otimes N}$ on its $i$-th component. Here, we have denoted by $\Psi^*_{(\phi_1, \dots, \phi_N)}$ the right component of the Chu morphism $(\Psi_{(\phi_1, \dots, \phi_N)}, \Psi^*_{(\phi_1, \dots, \phi_N)})$.

**Definition 5.2.** A compatibility context $U$ is defined to be a subset of $\bar{\mathfrak{E}}_\mathfrak{S}$ such that the elements of $\{m_I | I \in U\}$ form a family of jointly compatible real measurements. A compatibility cover $\mathcal{C}$ is defined as a family of maximal compatibility contexts covering the whole set of real measurements. In other words,

$$
\left\{
\begin{array}{l}
(\cup_{C \in \mathcal{C}} C) = \bar{\mathfrak{E}}_{\mathfrak{S}} \\
\forall C, C' \in \mathcal{C}, \ C \subseteq C' \Rightarrow C = C' \\
\forall C \in \mathcal{C}, \text{ the elements of } \{m_I \mid I \in C\} \text{ are jointly compatible.}
\end{array}
\right.
\qquad (5.4)
$$

Let us briefly clarify the situation for $\mathfrak{S}$ being a deterministic space of states.

**Theorem 5.3.** Let $\mathfrak{S}$ be a simplex.

We note that $\mathfrak{S} = \bar{\mathfrak{S}}, \bar{\mathfrak{E}}_{\mathfrak{S}} = \mathfrak{E}_{\mathfrak{S}}$ and all measurements are real.
Let us consider any two measurement morphisms $m_{l(\sigma_1, \sigma'_1)}$ and $m_{l(\sigma_2, \sigma'_2)}$. These morphisms are necessarily compatible.

*Proof.* We define the map $\Psi$ as a map from $\mathfrak{S}$ to $\mathfrak{B}\tilde{\otimes}\mathfrak{B}$ which satisfies, for any $\sigma \in \mathfrak{S}$,

$$ \Psi(\sigma) := \bigcap_{\sigma' \in \underline{\sigma}} \mathfrak{B}\tilde{\otimes}\mathfrak{B} \Psi(\sigma') \quad (5.5) $$

and for any $\sigma \in \mathfrak{S}^{\text{pure}} = \text{Max}(\mathfrak{S})$

$$ \left\{ \begin{array}{ll} \Psi(\sigma) = \mathbf{Y} \tilde{\otimes} \mathbf{Y} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \mathbf{Y} \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \mathbf{Y}, \\ \Psi(\sigma) = \mathbf{Y} \tilde{\otimes} \mathbf{Y} \sqcap_{\mathfrak{B}\tilde{\otimes}\mathfrak{B}} \mathbf{Y} \tilde{\otimes} \mathbf{N} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \mathbf{Y} \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \perp, \\ \Psi(\sigma) = \mathbf{Y} \tilde{\otimes} \mathbf{Y} \sqcap_{\mathfrak{B}\tilde{\otimes}\mathfrak{B}} \mathbf{N} \tilde{\otimes} \mathbf{Y} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \perp \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \mathbf{Y}, \\ \Psi(\sigma) = \mathbf{N} \tilde{\otimes} \mathbf{N} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \mathbf{N} \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \mathbf{N}, \\ \Psi(\sigma) = \mathbf{N} \tilde{\otimes} \mathbf{N} \sqcap_{\mathfrak{B}\tilde{\otimes}\mathfrak{B}} \mathbf{N} \tilde{\otimes} \mathbf{Y} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \mathbf{N} \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \perp, \\ \Psi(\sigma) = \mathbf{N} \tilde{\otimes} \mathbf{N} \sqcap_{\mathfrak{B}\tilde{\otimes}\mathfrak{B}} \mathbf{Y} \tilde{\otimes} \mathbf{N} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \perp \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \mathbf{N}, \\ \Psi(\sigma) = \mathbf{Y} \tilde{\otimes} \mathbf{N} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \mathbf{Y} \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \mathbf{N}, \\ \Psi(\sigma) = \mathbf{N} \tilde{\otimes} \mathbf{Y} & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \mathbf{N} \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \mathbf{Y}, \\ \Psi(\sigma) = \perp \tilde{\otimes} \perp & \text{if } \varepsilon_{l(\sigma_1, \sigma'_1)}^{\mathfrak{S}}(\sigma) = \perp \text{ and } \varepsilon_{l(\sigma_2, \sigma'_2)}^{\mathfrak{S}}(\sigma) = \perp. \end{array} \right. \quad (5.6) $$

We observe that $\Psi(m_{l(\sigma_1, \sigma'_1)}, m_{l(\sigma_2, \sigma'_2)})$ is unambiguously defined by (5.5) because $\mathfrak{S}$ is a simplex and then satisfies (2.102).

Secondly, $\Psi(m_{l(\sigma_1, \sigma'_1)}, m_{l(\sigma_2, \sigma'_2)})$ is a morphism by construction.

Endly, we can check easily for any $\sigma \in \mathfrak{S}$ (the proof begins on each expressions of (5.6) by a simple computation and is extended to the whole set of elements of $\mathfrak{S}$ using the expression (5.5) and the homomorphic properties of $m_{l(\sigma_1, \sigma'_1)}$ and $m_{l(\sigma_2, \sigma'_2)}$)

$$ (\zeta_{(1)}^{\mathfrak{B}\tilde{\otimes}} \circ \Psi(m_{l(\sigma_1, \sigma'_1)}, m_{l(\sigma_2, \sigma'_2)}))(\sigma) = m_{l(\sigma_1, \sigma'_1)}(\sigma) \quad (5.7) $$

$$ (\zeta_{(2)}^{\mathfrak{B}\tilde{\otimes}} \circ \Psi(m_{l(\sigma_1, \sigma'_1)}, m_{l(\sigma_2, \sigma'_2)}))(\sigma) = m_{l(\sigma_2, \sigma'_2)}(\sigma). \quad (5.8) $$

$\Psi$ is then the joint morphism $\Psi(m_{l(\sigma_1, \sigma'_1)}, m_{l(\sigma_2, \sigma'_2)})$. $\square$

**Theorem 5.4.** If $\mathfrak{S}$ is a simplex then the whole set of elements of $\{m_l | l \in \mathfrak{C}_{\mathfrak{S}}\}$ are jointly compatible measurement maps. In other words, the compatibility cover $\mathcal{C}$ is then reduced to the singleton $\{\mathfrak{C}_{\mathfrak{S}}\}$.

*Proof.* Straightforward using the generalization of the construction used in previous Theorem for the joint-morphism. $\square$

Let us now come back to the generic case for $\mathfrak{S}$. We intent to clarify the structure of contexts.
**Lemma 5.5.** If the $N$-uple of real measurement maps $(\mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$ are jointly compatible, then the $(N+1)$-uple of real measurement maps $(\mathfrak{m}_{l_1}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$ are jointly-compatible, for any $l_1, \cdots, l_N \in \mathfrak{C}$.

*Proof.* Let us consider that the $N$-uple of real measurement maps $(\mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$ are jointly compatible. There exists a joint morphism $\Psi_{(\mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})}$ satisfying the required conditions mentioned in Definition 5.1. Using the map $\varphi$ defined as a morphism from $\mathfrak{B}$ to $\mathfrak{B}^{\otimes 2}$ by

$$ \varphi(\mathbf{Y}) := \mathbf{Y} \otimes \mathbf{Y} \qquad \varphi(\mathbf{N}) := \mathbf{N} \otimes \mathbf{N} \qquad (5.9) $$

we then build a map denoted $\Phi$ from $\mathfrak{S}$ to $\mathfrak{B}^{\otimes(N+1)}$ by

$$ \Phi := (\varphi \otimes id^{\otimes (N-1)}) \circ \Psi_{(\mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})} \qquad (5.10) $$

It is easy to check that $\Phi$ is a joint morphism for the $(N+1)$-uple of measurement maps $(\mathfrak{m}_{l_1}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$. In other words,

$$ \Phi = \Psi_{(\mathfrak{m}_{l_1}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})}. \qquad (5.11) $$

□

**Lemma 5.6.** If the $N$-uple of real measurement maps $(\mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$ are jointly compatible, then the $(N+1)$-uple of real measurement maps $(\mathfrak{m}_{\overline{l_1}}, \mathfrak{m}_{\overline{l_1}}, \cdots, \mathfrak{m}_{\overline{l_N}})$ are also jointly-compatible.

*Proof.* We first note that

$$ \mathfrak{m}_{\overline{l}} = \omega \circ \mathfrak{m}_{l} \qquad (5.12) $$

$$ \text{with } \forall \alpha \in \mathfrak{B}, \omega(\alpha) := \bar{\alpha}. \qquad (5.13) $$

On another part, using the Lemma 5.5, we know that there exists a joint morphism for the $(N+1)$-uple of measurement maps $(\mathfrak{m}_{l_1}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$, it is denoted $\Psi_{(\mathfrak{m}_{l_1}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})}$.

We then define the map denoted $\Phi'$ from $\mathfrak{S}$ to $\mathfrak{B}^{\otimes(N+1)}$ by

$$ \Phi := (\omega \otimes id^{\otimes N}) \circ \Psi_{(\mathfrak{m}_{l_1}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})}. \qquad (5.14) $$

We check easily that $\Phi$ is a joint morphism for the $(N+1)$-uple of real measurement maps $(\mathfrak{m}_{\overline{l_1}}, \mathfrak{m}_{\overline{l_1}}, \cdots, \mathfrak{m}_{\overline{l_N}})$. In other words,

$$ \Phi' = \Psi_{(\mathfrak{m}_{\overline{l_1}}, \mathfrak{m}_{\overline{l_1}}, \cdots, \mathfrak{m}_{\overline{l_N}})}. \qquad (5.15) $$

□

**Lemma 5.7.** For any $\eta \in \mathfrak{B}$, we define the morphism $\tau_{\eta}$ as the constant map from $\mathfrak{S}$ to $\mathfrak{B}$ sending any state to $\eta$.

If the $N$-uple of real measurement maps $(\mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$ are jointly compatible, then the $(N+1)$-uple of real measurement maps $(\tau_{\eta}, \mathfrak{m}_{l_1}, \cdots, \mathfrak{m}_{l_N})$ are jointly compatible.
*Proof.* Direct consequence of Lemma 5.5 using the property

$$ \tau_{\eta} = \tau_{\eta} \circ m_{l}. \qquad (5.16) $$

□

**Lemma 5.8.** For any $\eta \in \mathfrak{B}$, if the $N$-uple of real measurement maps $(m_{l_1}, \cdots, m_{l_N})$ are jointly compatible ($N \ge 2$), then the $(N+1)$-uple of real measurement maps $(m_{(l_1 \sqcap \epsilon_{\mathfrak{G}} l_2)}, m_{l_1}, \cdots, m_{l_N})$ are jointly compatible.

*Proof.* Let us fix $N=2$ for simplicity.

First of all, we remark that, if we denote by $\Psi_{(m_{l_1}, m_{l_2})}$ the joint morphism associated to the compatible real measurement maps $(m_{l_1}, m_{l_2})$, the property (5.2), i.e. $\Psi^*_{(m_{l_1}, m_{l_2})}(\mathfrak{E}_{\mathfrak{B}\bar{\otimes}2}) \subseteq \mathfrak{E}_{\mathfrak{G}}$ implies immediately

$$ (l_1 \sqcap \epsilon_{\mathfrak{G}} l_2) \in \bar{\mathfrak{E}}_{\mathfrak{G}}. \qquad (5.17) $$

Let us then consider the morphism denoted $\Omega$ and defined from $\mathfrak{B}\bar{\otimes}\mathfrak{B}$ to $\mathfrak{B}\bar{\otimes}\mathfrak{B}\bar{\otimes}\mathfrak{B}$ as follows. For any $\sigma \in \mathfrak{B}\bar{\otimes}\mathfrak{B}$, we adopt the writing of $\sigma$ in terms of pure states $\sigma = (\bigcap_{i \in I} \mathfrak{B}\bar{\otimes}\mathfrak{B}) \alpha_i \tilde{\otimes} \beta_i)$ with $\alpha_i, \beta_i \in \{\mathbf{Y}, \mathbf{N}\}$, then $\Omega(\sigma)$ is defined by

$$ \Omega(\bigcap_{i \in I} \mathfrak{B}\bar{\otimes}\mathfrak{B}) \alpha_i \tilde{\otimes} \beta_i) := \bigcap_{i \in I} \mathfrak{B}\bar{\otimes}\mathfrak{B}\bar{\otimes}\mathfrak{B}) (\alpha_i \tilde{\otimes} \alpha_i \tilde{\otimes} \beta_i \sqcap \mathfrak{B}\bar{\otimes}\mathfrak{B}\bar{\otimes}\mathfrak{B}) \beta_i \tilde{\otimes} \alpha_i \tilde{\otimes} \beta_i) \quad (5.18) $$

The map $\Omega$ is build as a morphism. Moreover, we have immediately

$$ \zeta_{(2)}^{\mathfrak{B}\bar{\otimes}\mathfrak{B}} \circ \Omega \circ \Psi_{(m_{l_1}, m_{l_2})} = \zeta_{(1)}^{\mathfrak{B}\bar{\otimes}} \circ \Psi_{(m_{l_1}, m_{l_2})} = m_{l_1}, \quad (5.19) $$

$$ \zeta_{(3)}^{\mathfrak{B}\bar{\otimes}\mathfrak{B}} \circ \Omega \circ \Psi_{(m_{l_1}, m_{l_2})} = \zeta_{2}^{\mathfrak{B}\bar{\otimes}} \circ \Psi_{(m_{l_1}, m_{l_2})} = m_{l_2}, \quad (5.20) $$

$$ \zeta_{(1)}^{\mathfrak{B}\bar{\otimes}\mathfrak{B}} \circ \Omega \circ \Psi_{(m_{l_1}, m_{l_2})} = m_{(l_1 \sqcap \epsilon_{\mathfrak{G}} l_2)}. \quad (5.21) $$

Then, $\Omega \circ \Psi_{(m_{l_1}, m_{l_2})}$ is a joint morphism for the triple $(m_{(l_1 \sqcap \epsilon_{\mathfrak{G}} l_2)}, m_{l_1}, m_{l_2})$. As a result, $m_{l_1}, m_{l_2}$ and $m_{(l_1 \sqcap \epsilon_{\mathfrak{G}} l_2)}$ are jointly compatible. □

We can then summarize Lemma 5.8, Lemma 5.6 and Lemma 5.7 in terms of the following theorem on contexts :

**Theorem 5.9.** Let $C \in \mathcal{C}$ designate a maximal compatibility context, we have

$$ \mathfrak{Y}_{\mathfrak{E}_{\mathfrak{G}}} \in C \text{ and } \perp_{\mathfrak{E}_{\mathfrak{G}}} \in C \text{ and } \overline{\mathfrak{Y}_{\mathfrak{E}_{\mathfrak{G}}}} \in C, \quad (5.22) $$

$$ \forall l \in C, \quad \overline{l} \in C, \qquad (5.23) $$

$$ \forall l_1, l_2 \in C, \quad (l_1 \sqcap \epsilon_{\mathfrak{G}} l_2) \in C. \qquad (5.24) $$

We note the following fact, implicit in (5.24) and clarified in Lemma 5.8 :

$$ \forall l_1, l_2 \in \bar{\mathfrak{E}}_{\mathfrak{G}}, \quad (l_1 \sqcap \epsilon_{\mathfrak{G}} l_2) \in \mathfrak{E}_{\mathfrak{G}} \setminus \bar{\mathfrak{E}}_{\mathfrak{G}} \Rightarrow (\mathfrak{N}C \in \mathcal{C} \mid l_1, l_2 \in C). \quad (5.25) $$

We now intent to introduce our central ontic notion.
**Definition 5.10.** Once is known the compatibility cover $\mathcal{C}$, we define an *operational description* $\mathcal{D}$ to be a family of maps $(\mathcal{D}^C)_{C \in \mathcal{C}}$, where each map $\mathcal{D}^C$ is an element of $\mathfrak{B}^C$ (i.e. a counterfactual statement in $\mathfrak{B}$ is attributed to any of the jointly compatible measurements $m_I$ for any $I \in C$), these maps being assigned to satisfy the following conditions

$$ \forall C \in \mathcal{C}, \mathcal{D}^C(\mathfrak{D}) \bar{\mathcal{E}}_{\mathcal{E}} = \mathbf{Y} \qquad (5.26) $$

$$ \forall C \in \mathcal{C}, \forall I \in C, \mathcal{D}^C(\overline{I}) = \overline{\mathcal{D}^C(I)}, \qquad (5.27) $$

$$ \forall C \in \mathcal{C}, \forall \{l_i | i \in I\} \subseteq C, \mathcal{D}^C(\bigcap_{i \in I} \mathcal{E}_{\mathcal{E}} l_i) = \bigwedge_{i \in I} \mathcal{D}^C(l_i). \qquad (5.28) $$

**Definition 5.11.** If the operational description $\mathcal{D} := (\mathcal{D}^C)_{C \in \mathcal{C}}$ satisfies the following supplementary condition

$$ (\textit{sheaf condition}) \quad \forall C, C' \in \mathcal{C}, \forall I \in C \cap C', \mathcal{D}^C(I) = \mathcal{D}^C(I'), \qquad (5.29) $$

it will be designated as a *coherent operational description*.

The set of coherent operational descriptions associated to the compatibility cover $\mathcal{C}$ will be denoted $\mathcal{D}$ and called *the empirical model* for the space of states $\mathfrak{S}$.

**Definition 5.12.** The empirical model $\mathcal{D}$ associated to the space of states $\mathfrak{S}$ is an Inf semi-lattice by defining its infima as follows:

$$ \forall (\mathcal{D}_1^C)_{C \in \mathcal{C}}, (\mathcal{D}_2^C)_{C \in \mathcal{C}} \in \mathcal{D}; \forall C \in \mathcal{C}, \forall I \in C, (\mathcal{D}_1^C \sqcap_{\mathfrak{D}} \mathcal{D}_2^C)(I) := \mathcal{D}_1^C(I) \wedge \mathcal{D}_2^C(I). \qquad (5.30) $$

**Definition 5.13.** If an operational description $\mathcal{D} := (\mathcal{D}^C)_{C \in \mathcal{C}}$ is such that there exists a real state $\sigma \in \bar{\mathfrak{S}}$ satisfying

$$ \forall C \in \mathcal{C}, \forall I \in C \quad \mathcal{D}^C(I) = \varepsilon_I^{\mathfrak{S}}(\sigma), \qquad (5.31) $$

we will say that $\mathcal{D}$ is *defined globally*. It will then be denoted $(\mathcal{D}^C_{(\sigma)})_{C \in \mathcal{C}}$. Note that if $\mathcal{D}$ is defined globally, then it is trivially a coherent operational description.

**Definition 5.14.** The empirical model $\mathcal{D}$ is said to be *non-contextual* iff every $\mathcal{D} \in \mathcal{D}$ is defined globally, otherwise it is said to be *contextual*.

**Theorem 5.15.** If $\mathfrak{S}$ is a simplex then every coherent operational description $\mathcal{D}$ are defined globally. The empirical model associated to a simplex space of states is then non-contextual.

*Proof.* Direct consequence of Theorem 5.4 and Theorem 2.15. $\square$

<figure><img src="image_5.png" /></figure>

It appears clearly that an operational description $\mathcal{D}$ defines a collection of preparation procedures. More explicitly, each complete context $U$ leads to the preparation procedure selecting the samples which return the correct result (i.e. the correct counterfactual statement) to the family of jointly-compatible measurements $(m_I)_{I \in U}$.

In an empirical model, the aim of a preparation procedure is to select an ontic state for the system (or at least a mixture of individual ontic states) compatible with all the observed data. The knowledge of this ontic state is necessary and exactly sufficient to fix the result returned by any of the measurements that can be realized simultaneously (i.e. in a jointly-compatible way) on the system.
**Definition 5.16.** The *ontic state* associated to the operational description $\mathfrak{D} := (\mathfrak{D}^C)_{C \in \mathcal{C}} \in \mathfrak{D}$ is a family $(\Sigma_{\mathfrak{D}}^C)_{C \in \mathcal{C}} \in \bar{\mathfrak{S}}^{\mathcal{G}}$ satisfying the following conditions

$$ \forall C \in \mathcal{C}, \forall l \in C, \mathfrak{m}_l(\Sigma_{\mathfrak{D}}^C) = \mathfrak{d}^C(l), \qquad (5.32) $$

$$ \forall C \in \mathcal{C}, \Sigma_{\mathfrak{D}}^{C} = \bigcap^{\mathcal{G}} \{ \sigma^{C} \in \bar{\mathfrak{S}} | \forall l \in C, \mathfrak{m}_{l}(\sigma^{C}) \geq \mathfrak{d}^{C}(l) \}. \qquad (5.33) $$

**Theorem 5.17.** Let us fix a coherent operational description $\mathfrak{D} := (\mathfrak{D}^C)_{C \in \mathcal{C}} \in \mathfrak{D}$ associated to the compatibility cover $\mathcal{C}$. Let us also fix a compatibility context $C$. Then,

$$ \exists \sigma \in \bar{\mathfrak{S}} \mid (\varepsilon_l^{\mathcal{G}}(\sigma) = \mathfrak{d}^C(l), \forall l \in C). \qquad (5.34) $$

More explicitly, we will fix explicitly the solution $\sigma_{\mathfrak{D}}^C$ of the equation (5.34) defined by

$$ \mathfrak{I}_{\mathfrak{D}}^{C} := \bigcap^{\bar{\mathfrak{S}}^{\mathcal{G}}} \{ l \in C \mid \mathfrak{d}^{C}(l) = \mathbf{Y} \} \qquad (5.35) $$

$$ \sigma_{\mathfrak{D}}^{C} := \bigcap^{\mathcal{G}} (\varepsilon_{l_{\mathfrak{D}}}^{\mathcal{G}})^{-1}(\mathbf{Y}) \in \bar{\mathfrak{S}} \qquad (5.36) $$

The family $(\sigma_{\mathfrak{D}}^C)_{C \in \mathcal{C}} \in \bar{\mathfrak{S}}^{\mathcal{G}}$ is the *ontic state associated to* $\mathfrak{D}$.

*Proof.* Let us consider the effect $l_{\mathfrak{D}}^C := \bigcap^{\bar{\mathfrak{S}}^{\mathcal{G}}} \{ l \in C \mid \mathfrak{d}^C(l) = \mathbf{Y} \}$. Note that the infimum in the definition of $l_{\mathfrak{D}}^C$ exists because $\bar{\mathfrak{S}}^{\mathcal{G}}$ is a down-complete Inf semi-lattice and because $\mathfrak{D}^{\mathcal{G}} \in \{ l \in C \mid \mathfrak{d}^C(l) = \mathbf{Y} \} \neq \emptyset$. Moreover, due to the property (5.24), we deduce that $l_{\mathfrak{D}}^C$ is in $C$.

Secondly, $\mathfrak{d}^C(l_{\mathfrak{D}}^C) = \mathbf{Y}$ because of the property (5.28).

Thirdly, we note that, for any $l \in C$, we have $l \supseteq_{\mathcal{G}} l_{\mathfrak{D}}^C$ implies $\mathfrak{d}^C(l) = \mathbf{Y}$ because $\mathfrak{d}^C(l) \wedge \mathbf{Y} = \mathfrak{d}^C(l) \wedge \mathfrak{d}^C(l_{\mathfrak{D}}^C) = \mathfrak{d}^C(l \sqcap_{\mathcal{G}} l_{\mathfrak{D}}^C) = \mathfrak{d}^C(l_{\mathfrak{D}}^C) = \mathbf{Y}$. Conversely, the property $\mathfrak{d}^C(l) = \mathbf{Y}$ for $l \in C$ implies $l \supseteq_{\bar{\mathfrak{S}}^{\mathcal{G}}} l_{\mathfrak{D}}^C$ from the explicit definition of $l_{\mathfrak{D}}^C$.

As a result, we have $\{ l \in C \mid \mathfrak{d}^C(l) = \mathbf{Y} \} = (\uparrow^C l_{\mathfrak{D}}^C)$.

We also note that, due to properties (5.26) and the definition of $l_{\mathfrak{D}}^C$, we have $l_{\mathfrak{D}}^C = l_{\mathfrak{D}} \sqcap_{\mathcal{G}} \mathfrak{D}^{\mathcal{G}}$. Hence, there must exist a state $\sigma_{\mathfrak{D}}^C \in \mathfrak{S}$ such that $l_{\mathfrak{D}}^C = l_{(\sigma_{\mathfrak{D}}^C, \cdot)}$. In other words, $\sigma_{\mathfrak{D}}^C := \bigcap^{\mathcal{G}} (\varepsilon_{l_{\mathfrak{D}}}^{\mathcal{G}})^{-1}(\mathbf{Y})$.

We can summarize these results by the following equivalence :

$$ \forall C \in \mathcal{C}, \forall l \in C, \quad \mathfrak{d}^C(l) = \mathbf{Y} \Leftrightarrow \varepsilon_l^{\mathcal{G}}(\sigma_{\mathfrak{D}}^C) = \mathbf{Y}. \qquad (5.37) $$

Using (2.29) and (5.27), we deduce also : $\varepsilon_l^{\mathcal{G}}(\sigma_{\mathfrak{D}}^C) = \mathbf{N} \Leftrightarrow \varepsilon_l^{\mathcal{G}}(\sigma_{\mathfrak{D}}^C) = \mathbf{Y} \Leftrightarrow \mathfrak{d}^C(l) = \mathbf{Y} \Leftrightarrow \mathfrak{d}^C(l) = \mathbf{N}$. As a final conclusion, we have for any $l \in C$ the equality $\varepsilon_l^{\mathcal{G}}(\sigma_{\mathfrak{D}}^C) = \mathfrak{d}^C(l)$.

Note then that the defining property (5.32) is identical to the property (5.34) satisfied by $\sigma_{\mathfrak{D}}^C$ for any $C \in \mathcal{C}$.

Let us now consider a family $(\rho^C)_{C \in \mathcal{C}} \in \mathfrak{S}^{\mathcal{G}}$ satisfying $\forall C \in \mathcal{C}, \forall l \in C, \varepsilon_l^{\mathcal{G}}(\rho^C) \geq \mathfrak{d}^C(l)$.

We have then in particular $\varepsilon_{l_{\mathfrak{D}}}^{\mathcal{G}}(\rho^C) \geq \mathfrak{d}^C(l_{\mathfrak{D}}^C) = \mathbf{Y}$. Then $\rho^C \supseteq_{\mathcal{G}} \bigcap^{\mathcal{G}} (\varepsilon_{l_{\mathfrak{D}}}^{\mathcal{G}})^{-1}(\mathbf{Y}) =$
$\sigma_{\mathfrak{D}}^C$. This is the defining property (5.33).

As a conclusion, $(\sigma_{\mathfrak{D}}^C)_{C \in \mathcal{C}}$ is the ontic state associated to $\mathfrak{D}$.

<figure><img src="image_6.png" /></figure>

## 5.2. Hidden states and contextuality

During this subsection, $((\mathfrak{S}', \star))_c := (\mathfrak{S}', \mathcal{Q}_c(\mathfrak{S}'), cl_c^{\mathfrak{S}', \star})$ will define the ontic completion of $(\mathfrak{S}', \star)$. We will denote equivalently by $\mathfrak{S}$ the space of states $\mathcal{J}_{\mathfrak{S}'}^c$, and we will then denote by $\overline{\mathfrak{S}}$ the space of states $\mathfrak{S}'$. As shown before, the states/effects Chu Space $(\mathfrak{S}, \mathfrak{E}_{\mathfrak{S}}, \mathfrak{E}^{\mathfrak{S}})$ is equipped with a real structure given simply by $(\overline{\mathfrak{S}}, \star)$. As before, $\overline{\mathfrak{E}}_{\mathfrak{S}}$ is defined as the sub Inf semi-lattice of $\mathfrak{E}_{\mathfrak{S}}$ generated by the elements of

$$ \{ l_{(\sigma, \sigma')} | \sigma, \sigma' \in \overline{\mathfrak{S}} \setminus \{\perp_{\mathfrak{S}}\}, \sigma' \overline{=}_{\overline{\mathfrak{S}}} \sigma^* \} \cup \{ l_{(\sigma, \sigma)} | \sigma \in \overline{\mathfrak{S}} \} \cup \{ l_{(\sigma, \sigma')} | \sigma \in \overline{\mathfrak{S}} \} \cup \{ l_{(\sigma, \sigma')} | \sigma \in \overline{\mathfrak{S}} \} $$

**Definition 5.18.** In the following, we will adopt the following notation :

$$ \mathfrak{U}_{\overline{\mathfrak{S}}} := (\{l_{(\sigma, \sigma')} | \sigma \in \overline{\mathfrak{S}}\} \cup \{l_{(\sigma, \sigma)} | \sigma \in \overline{\mathfrak{S}}\} \cup \{l_{(\sigma, \sigma')}\}). \quad (5.38) $$

Let us consider a maximal compatibility context $C$ in $\mathcal{C}$. We are necessarily in one of the two following cases :

* Maximal compatibility context of Type 1 :

$$ C \subseteq \mathfrak{U}_{\overline{\mathfrak{S}}} \quad (5.39) $$

* Maximal compatibility context of Type 2 :

$$ \exists \sigma, \sigma' \in \overline{\mathfrak{S}} \setminus \{\perp_{\mathfrak{S}}\}, \sigma' \overline{=}_{\overline{\mathfrak{S}}} \sigma^* \mid l_{(\sigma, \sigma')} \in C. \quad (5.40) $$

**Lemma 5.19.** Let us consider a compatibility context denoted $C$ such that $C \subseteq \mathfrak{U}_{\overline{\mathfrak{S}}}$. We have necessarily

$$ \exists \omega \in \overline{\mathfrak{S}} \mid C \subseteq (\{l_{(\sigma, \sigma')} | \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega\} \cup \{l_{(\sigma, \sigma)} | \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega\} \cup \{l_{(\sigma, \sigma')}\}), \quad (5.41) $$

or

$$ \exists \gamma, \gamma', \omega, \omega' \in \overline{\mathfrak{S}} \mid \gamma \overline{=}_{\overline{\mathfrak{S}}} \gamma^*, \omega \overline{=}_{\overline{\mathfrak{S}}} \gamma, \omega' \overline{=}_{\overline{\mathfrak{S}}} \gamma' $$

$$ C \subseteq (\{l_{(\sigma, \sigma')} | \gamma \sqsubseteq_{\overline{\mathfrak{S}}} \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega\} \cup \{l_{(\sigma, \sigma)} | \gamma \sqsubseteq_{\overline{\mathfrak{S}}} \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega\} \cup \{l_{(\sigma, \sigma')}\} \cup \\ \{l_{(\sigma, \sigma')} | \gamma' \sqsubseteq_{\overline{\mathfrak{S}}} \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega'\} \cup \{l_{(\sigma, \sigma)} | \gamma' \sqsubseteq_{\overline{\mathfrak{S}}} \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega'\}) \quad (5.42) $$

*Proof.* Let us consider a compatibility context denoted $C$ such that $C \subseteq \mathfrak{U}_{\overline{\mathfrak{S}}}$ and let us consider the following subset of $\mathcal{P}(\overline{\mathfrak{S}})$

$$ \mathfrak{Q}(\overline{\mathfrak{S}}) := \{ U \subseteq C \mid \forall \sigma, \sigma' \in \overline{\mathfrak{S}} \text{ s.t. } l_{(\sigma, \sigma')}, l_{(\sigma', \sigma)} \in U \text{ we have } \sigma' \overline{=}_{\overline{\mathfrak{S}'}} \sigma^* \} \quad (5.43) $$

and let us consider an element $V$ which is maximal for inclusion in $\mathfrak{Q}(\overline{\mathfrak{S}})$. As a trivial consequence of the requirement (5.2) (more explicitly, of the property (5.17)), we know that $\omega := \sqcup_{\overline{\mathfrak{S}}} \{\sigma | l_{(\sigma, \sigma)} \in V\}$ exists in $\overline{\mathfrak{S}}$.

If we denote $C_{\omega} := (\{l_{(\sigma, \sigma')} | \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega\} \cup \{l_{(\sigma, \sigma)} | \sigma \sqsubseteq_{\overline{\mathfrak{S}}} \omega\} \cup \{l_{(\sigma, \sigma')}\})$, we note that $C_{\omega}$ is a compatibility context (the joint morphism is built straightforwardly) and $V \subseteq C_{\omega}$ by construction.

We will now distinguish two cases : (1) $C \setminus V = \emptyset$, and (2) $C \setminus V \neq \emptyset$.

The first case is treated immediately, we obtain (5.41).

Let us then focus on the second case. For any $\alpha \in \overline{\mathfrak{S}} \setminus \{\perp_{\mathfrak{S}}\}$ such that $l_{(\alpha, \alpha)} \in C \setminus V$, we have necessarily, due to the requirement (5.2) (more explicitly, of the
property (5.17)), for any $\sigma \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\}$ such that $l_{(\sigma,.)} \in V$ the property $\alpha^* \sqsubseteq_{\bar{\mathfrak{S}}} \sigma$, i.e. $\alpha^* \sqsubseteq_{\bar{\mathfrak{S}}} \bigcap^{\mathfrak{G}} \{\sigma \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} | l_{(\sigma,.)} \in V\}$. If we denote by $\gamma := \bigcap^{\mathfrak{G}} \{\sigma \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} | l_{(\sigma,.)} \in C \setminus V\}$ and $\gamma' := \bigcap^{\mathfrak{G}} \{\sigma \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} | l_{(\sigma,.)} \in C \setminus V\}$, we have $\gamma' \sqsupseteq_{\bar{\mathfrak{S}}} \gamma^*$.
On another part, and for the same reasons as before, if we denote $\omega' := \sqcup^{\mathfrak{G}} \{\sigma \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} | l_{(\sigma,.)} \in C \setminus V\}$, then $C_{\omega'}$ is a compatibility context (the joint morphism is built straightforwardly) and $(C \setminus V) \subseteq C_{\omega'}$ by construction.
This concludes the proof. $\square$

**Lemma 5.20.** In the following, we will adopt the following notation

$$ \forall \omega \in \bar{\mathfrak{S}}^{\text{pure}} \mid C_{\omega}^{(1)} := (\{l_{(\sigma,.)} | \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,\sigma)} | \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,..)}\}). \quad (5.44) $$

Let C be a maximal compatibility context of type 1. Then,

$$ \exists! \omega \in \bar{\mathfrak{S}}^{\text{pure}} \mid C = C_{\omega}^{(1)}. \quad (5.45) $$

*Proof.* Trivial consequence of Lemma 5.19. $\square$

**Lemma 5.21.** Let C be a maximal compatibility context of type 2 and let $\gamma, \gamma' \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\}$ with $\gamma' \sqsupseteq_{\bar{\mathfrak{S}}} \gamma^*$ and such that $l_{(\gamma,\gamma')} \in C$. Then,

$$ \exists \lambda, \lambda', \omega, \omega' \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} \mid \lambda' \sqsupseteq_{\bar{\mathfrak{S}}} \lambda^*, \omega \sqsupseteq_{\bar{\mathfrak{S}}} \gamma \sqsupseteq_{\bar{\mathfrak{S}}} \lambda, \omega' \sqsupseteq_{\bar{\mathfrak{S}}} \gamma' \sqsupseteq_{\bar{\mathfrak{S}}} \lambda' \\ (C \cap \mu_{\bar{\mathfrak{S}}}) \subseteq (\{l_{(\sigma,.)} | \lambda \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,\sigma)} | \lambda \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,..)}\} \cup \\ \{l_{(\sigma,.)} | \lambda' \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega'\} \cup \{l_{(.,\sigma)} | \lambda' \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega'\}) \quad (5.46) $$

*Proof.* Let C be a maximal compatibility context of type 2 and let $\gamma, \gamma' \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\}$ with $\gamma' \sqsupseteq_{\bar{\mathfrak{S}}} \gamma^*$ and such that $l_{(\gamma,\gamma')} \in C$.

Let us first note that $(C \cap \mu_{\bar{\mathfrak{S}}})$ is a compatibility context included in $\mu_{\bar{\mathfrak{S}}}$. Secondly, we remark that $l_{(\gamma,.)}, l_{(\gamma',.)} \in (C \cap \mu_{\bar{\mathfrak{S}}})$. Indeed, we recall that $l_{(\gamma,.)} = l_{(\gamma,\gamma')} \sqcap_{\bar{\mathfrak{S}}} \mathfrak{Y}_{\mathfrak{C}_{\bar{\mathfrak{S}}}}$ and $l_{(\gamma',.)} = l_{(\gamma,\gamma')} \sqcap_{\bar{\mathfrak{S}}} \overline{\mathfrak{Y}_{\mathfrak{C}_{\bar{\mathfrak{S}}}}} \text{ and we have Theorem 5.9 for the maximal compatibility context C. Moreover, we have } \neg\gamma\gamma' \bar{\mathfrak{S}}\text{. We are then in the case (5.42) and not in the case (5.41). Because of Lemma 5.19, we know that}$

$$ \exists \lambda, \lambda', \omega, \omega' \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} \mid \lambda' \sqsupseteq_{\bar{\mathfrak{S}}} \lambda^*, \omega \sqsupseteq_{\bar{\mathfrak{S}}} \lambda, \omega' \sqsupseteq_{\bar{\mathfrak{S}}} \lambda' \\ (C \cap \mu_{\bar{\mathfrak{S}}}) \subseteq (\{l_{(\sigma,.)} | \lambda \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,\sigma)} | \lambda \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,..)}\} \cup \\ \{l_{(\sigma,.)} | \lambda' \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega'\} \cup \{l_{(.,\sigma)} | \lambda' \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega'\}) \quad (5.47) $$

and we must obviously have $\lambda \sqsubseteq_{\bar{\mathfrak{S}}} \gamma$ and $\lambda' \sqsubseteq_{\bar{\mathfrak{S}}} \gamma'$, because $l_{(\gamma,.)}, l_{(\gamma',.)} \in C$. $\square$

**Lemma 5.22.** Let C be a maximal compatibility context of type 2 such that there exists $\gamma \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\}$ with $l_{(\gamma,\gamma^*)} \in C$. Then,

$$ \exists \omega, \omega' \in \bar{\mathfrak{S}} \setminus \{\perp_{\bar{\mathfrak{S}}}\} \mid \omega \sqsupseteq_{\bar{\mathfrak{S}}} \gamma, \omega' \sqsupseteq_{\bar{\mathfrak{S}}} \gamma^* \\ (C \cap \mu_{\bar{\mathfrak{S}}}) \subseteq (\{l_{(\sigma,.)} | \gamma \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,\sigma)} | \gamma \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega\} \cup \{l_{(.,..)}\} \cup \\ \{l_{(\sigma,.)} | \gamma^* \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega'\} \cup \{l_{(.,\sigma)} | \gamma^* \sqsubseteq_{\bar{\mathfrak{S}}} \sigma \sqsubseteq_{\bar{\mathfrak{S}}} \omega'\}) \quad (5.48) $$

*Proof.* We obviously take Lemma 5.21 as a departure point.

If $\gamma' = \gamma^*$ we have then $\gamma^* = \gamma' \sqsupseteq_{\bar{\mathfrak{S}}} \lambda' \sqsupseteq_{\bar{\mathfrak{S}}} \lambda^* \sqsupseteq_{\bar{\mathfrak{S}}} \gamma^*$. This concludes the proof. $\square$
In the following, $\mathscr{C}$ will designate the compatibility cover and we will consider a coherent operational description denoted $\mathscr{D} := (\partial^C)_{C \in \mathscr{C}}$. We will denote by $(\Sigma_{\mathscr{D}}^C)_{C \in \mathscr{C}}$ the corresponding ontic state.

**Lemma 5.23.** For any pair $C_{\omega}^{(1)}, C_{\omega'}^{(1)}$ of maximal compatibility contexts of type 1 associated to $\omega, \omega' \in \bar{\mathscr{S}}^{\text{pure}}$ (see Lemma 5.20), we have

$$ \left\{ \begin{array}{l} \omega \sqcap_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}} \sqsubseteq_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} \\ \omega' \sqcap_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} \sqsubseteq_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}} \end{array} \right. \qquad (5.49) $$

*Proof.* By definition of the ontic state, we have

$$ \forall I \in C_{\omega}^{(1)}, \quad \partial_{\omega}^{C_{\omega}^{(1)}}(I) = \varepsilon_{I}^{\mathscr{S}}(\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}}) \qquad (5.50) $$

The consistency of $\mathscr{D}$ implies

$$ \forall I \in C_{\omega}^{(1)} \cap C_{\omega'}^{(1)}, \quad \varepsilon_{I}^{\mathscr{S}}(\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}}) = \varepsilon_{I}^{\mathscr{S}}(\Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}}) \qquad (5.51) $$

But,

$$ C_{\omega}^{(1)} \cap C_{\omega'}^{(1)} = (\{l_{(\sigma,.)} | \sigma \sqsubseteq_{\bar{\mathscr{S}}} \omega \sqcap_{\bar{\mathscr{S}}} \omega'\} \cup \{l_{(.,\sigma)} | \sigma \sqsubseteq_{\bar{\mathscr{S}}} \omega \sqcap_{\bar{\mathscr{S}}} \omega'\} \cup \{l_{(.,.)}\}). \qquad (5.52) $$

Hence,

$$ (\sigma \sqsubseteq_{\bar{\mathscr{S}}} \omega \sqcap_{\bar{\mathscr{S}}} \omega' \text{ and } \sigma \sqsubseteq_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}}) \Rightarrow (\varepsilon_{l_{(\sigma,.)}}^{\mathscr{S}}(\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}}) = \mathbf{Y} = \varepsilon_{l_{(\sigma,.)}}^{\mathscr{S}}(\Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}}) \Rightarrow \sigma \sqsubseteq_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}}. \qquad (5.53) $$

In other words, noting that $\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} \sqsubseteq_{\bar{\mathscr{S}}} \omega$, we obtain $\omega' \sqcap_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} \sqsubseteq_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}}$. The other inequality is obtained in the same way. $\square$

**Lemma 5.24.** If there exists $\omega, \omega' \in \bar{\mathscr{S}}^{\text{pure}}$ such that $\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}}$ and $\Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}}$ admit a common upper-bound in $\bar{\mathscr{S}}$, then for any $\omega'' \in \Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} \sqcup_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}}$ we have $\Sigma_{\mathscr{D}}^{C_{\omega}''} \sqsupseteq_{\bar{\mathscr{S}}} (\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} \sqcup_{\bar{\mathscr{S}}} \Sigma_{\mathscr{D}}^{C_{\omega'}^{(1)}})$

*Proof.* Straightforward. $\square$

**Lemma 5.25.** For any coherent operational description $\mathscr{D}$ the subset $U_{\mathscr{D}}$ of $\bar{\mathscr{S}}$ defined by $U_{\mathscr{D}} := \{\Sigma_{\mathscr{D}}^{C_{\omega}^{(1)}} | \omega \in \bar{\mathscr{S}}^{\text{pure}}\}$ satisfies

$$ c l_{c}^{\bar{\mathscr{S}}} (U_{\mathscr{D}}) = U_{\mathscr{D}} \qquad (5.54) $$

*Proof.* As a strict rewriting of property (5.49), we note that we have

$$ \forall \omega \in \bar{\mathscr{S}}^{\text{pure}}, \exists \alpha \in U_{\mathscr{D}} \mid \alpha \sqsubseteq_{\bar{\mathscr{S}}} \omega \text{ and } \forall \beta \in U_{\mathscr{D}}, (\omega \sqcap_{\bar{\mathscr{S}}} \beta) \sqsubseteq_{\bar{\mathscr{S}}} \alpha. \qquad (5.55) $$

and then

$$ \mathrm{Max}\{\sqcup_{\kappa \in U_{\mathscr{D}}} (\omega \sqcap_{\bar{\mathscr{S}}} \kappa) | \omega \in \bar{\mathscr{S}}^{\text{pure}}\} \sqsubseteq U_{\mathscr{D}}. \qquad (5.56) $$
Now, we note that

$$ (V \sqsubseteq U_{\mathcal{D}} \text{ and } \hat{V}^{\overline{\mathcal{G}}}) \Rightarrow \exists \omega \in \overline{\mathcal{G}}^{\text{pure}} | V \sqsubseteq \{ (\omega \sqcap_{\overline{\mathcal{G}}} \kappa) | \kappa \in U_{\mathcal{D}} \} \quad (5.57) $$

and then

$$ \mathrm{Max}\{\sqcup_{\overline{\mathcal{G}}} V \mid V \sqsubseteq U_{\mathcal{D}} \text{ and } \hat{V}^{\overline{\mathcal{G}}}\} \sqsubseteq \mathrm{Max}\{\sqcup_{\kappa \in U_{\mathcal{D}}} (\omega \sqcap_{\overline{\mathcal{G}}} \kappa) \mid \omega \in \overline{\mathcal{G}}^{\text{pure}}\}. \quad (5.58) $$

In other words

$$ c^{\overline{\mathcal{G}}}(U_{\mathcal{D}}) \sqsubseteq U_{\mathcal{D}} \quad (5.59) $$

But we have also (see property (3.59))

$$ c^{\overline{\mathcal{G}}}(U_{\mathcal{D}}) \succeq U_{\mathcal{D}} \quad (5.60) $$

Then, by iteration, we deduce the announced result. $\square$

**Lemma 5.26.** The subset of $\overline{\mathcal{G}}$ defined by $\{\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} | \omega \in \overline{\mathcal{G}}^{\text{pure}}\}$ satisfies

$$ \mathfrak{D} \gamma \in \overline{\mathcal{G}} \mid \{ \gamma, \gamma^* \} \sqsubseteq \{ \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} | \omega \in \overline{\mathcal{G}}^{\text{pure}} \}. \quad (5.61) $$

*Proof.* Let us suppose that there exists $\gamma \in \overline{\mathcal{G}}$ such that $\{\gamma, \gamma^*\} \sqsubseteq \{\Sigma_{\mathcal{D}}^{C_{\sigma}^{(1)}} | \sigma \in \overline{\mathcal{G}}^{\text{pure}}\}$, and let us exhibit a contradiction. We denote by $\omega$ and $\omega'$ the elements of $\overline{\mathcal{G}}^{\text{pure}}$ such that $\gamma \sqsubseteq_{\overline{\mathcal{G}}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}$ and $\gamma^* \sqsubseteq_{\overline{\mathcal{G}}} \Sigma_{\mathcal{D}}^{C_{\omega'}^{(1)}}$. There exists a maximal compatibility context of type 2 denoted $C$ such that $l_{(\gamma, \gamma^*)} \in C$. Applying Theorem 5.9 to the maximal context $C$ we deduce easily that $l_{(\gamma, \cdot)} \in C$ and $l_{(\cdot, \gamma^*)} \in C$. Due to the sheaf condition satisfied by $\mathcal{D}$, we know that

$$ \mathcal{D}^C(l_{(\gamma, \cdot)}) = \mathcal{D}^C_{\omega}(l_{(\gamma, \cdot)}) = \varepsilon_{l_{(\gamma, \cdot)}}^{\overline{\mathcal{G}}} (\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}) = \mathbf{Y} \quad (5.62) $$

$$ \mathcal{D}^C(l_{(\cdot, \gamma^*)}) = \mathcal{D}^C_{\omega'}(l_{(\cdot, \gamma^*)}) = \varepsilon_{l_{(\cdot, \gamma^*)}}^{\overline{\mathcal{G}}} (\Sigma_{\mathcal{D}}^{C_{\omega'}^{(1)}}) = \mathbf{N} \quad (5.63) $$

but we have also

$$ \mathcal{D}^C(l_{(\gamma, \cdot)}) = \mathcal{D}^C(\mathfrak{Z}) \varepsilon_{\overline{\mathcal{G}}} \sqcap_{\varepsilon_{\overline{\mathcal{G}}}} l_{(\gamma, \gamma^*)} = \mathbf{Y} \wedge \mathcal{D}^C(l_{(\gamma, \gamma^*)}) \quad (5.64) $$

$$ \mathcal{D}^C(l_{(\cdot, \gamma^*)}) = \mathcal{D}^C(\overline{\mathfrak{Z}}) \varepsilon_{\overline{\mathcal{G}}} \sqcap_{\varepsilon_{\overline{\mathcal{G}}}} l_{(\gamma, \gamma^*)} = \mathbf{N} \wedge \mathcal{D}^C(l_{(\gamma, \gamma^*)}). \quad (5.65) $$

From (5.62) and (5.64) we deduce that necessarily $\mathcal{D}^C(l_{(\gamma, \gamma^*)}) = \mathbf{Y}$, but using (5.65) we deduce $\mathcal{D}^C(l_{(\cdot, \gamma^*)}) = \mathbf{N} \wedge \mathbf{Y} = \perp$ which contradicts (5.63). We have then obtained the announced contradiction. This concludes the proof. $\square$

**Definition 5.27.** We will say that the operational description $\mathcal{D}$ is admissible iff $\{\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} | \omega \in \overline{\mathcal{G}}^{\text{pure}}\}$ is admissible, i.e. iff $\{\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} | \omega \in \overline{\mathcal{G}}^{\text{pure}}\} \in \mathcal{Q}_c(\overline{\mathcal{G}})$.

**Theorem 5.28.** Every operational description $\mathcal{D}$ are admissible.

*Proof.* Direct consequence of Lemma 5.26 and Lemma 5.25. $\square$
**Lemma 5.29.** As long as the operational description $\mathcal{D}$ is admissible (see Theorem 5.28), the supremum

$$ \Sigma_{\mathcal{D}} := \bigsqcup_{\omega \in \mathcal{G}^{\text{pure}}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} \quad (5.66) $$

is well defined as an element of $\mathcal{G}$. Then, we have

$$ \forall \omega \in \mathcal{G}^{\text{pure}}, \quad \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} = \omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}} \quad (5.67) $$

*Proof.* From Theorem 5.28, we know that $\{\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} | \omega \in \mathcal{G}^{\text{pure}}\}$ is in $\mathcal{Q}_c(\mathcal{G})$ and, then,

$$ \Sigma_{\mathcal{D}} := cl_c^{\mathcal{G}'}(\{\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} | \omega \in \mathcal{G}^{\text{pure}}\}) \quad (5.68) $$

$$ = \bigsqcup_{\omega \in \mathcal{G}^{\text{pure}}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} \quad (5.69) $$

is well defined as an element of $\mathcal{G}$.

Then, $\forall \omega' \in \mathcal{G}^{\text{pure}}$, $\omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}}^{C_{\omega'}^{(1)}} \subseteq_{\mathcal{G}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}$ implies $\omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}} \subseteq_{\mathcal{G}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}$.

On another part, $\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} \subseteq_{\mathcal{G}} \Sigma_{\mathcal{D}}$ implies $\omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} \subseteq_{\mathcal{G}} \omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}}$.

Endly, we recall that $\omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} = \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}$.

As a conclusion, we obtain $\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}} = \omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}}$.

Conversely, we easily check that the expression (5.67) satisfies the constraints (5.49). $\square$

<figure><img src="image_7.png" /></figure>

**Lemma 5.30.** Let us consider a maximal compatibility context of type 1 denoted $C_{\omega}^{(1)}$. We have

$$ \forall l \in C_{\omega}^{(1)}, \quad \vartheta^{C_{\omega}^{(1)}}(l) = \varepsilon_{l}^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \quad (5.70) $$

*Proof.* Let us compute $\vartheta^{C_{\omega}^{(1)}}(l)$ for any $l \in C_{\omega}^{(1)}$. First of all, we note that, for any $\alpha \sqsubseteq_{\mathcal{G}} \omega$, we have

$$ \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) = \varepsilon_{\mathcal{D}}^{\mathcal{G}} \varepsilon_{\mathcal{G} \sqcap \varepsilon_{\mathcal{G}}} l(\alpha, \cdot)(\Sigma_{\mathcal{D}}) = \mathbf{Y} \wedge \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \quad (5.71) $$

and then

$$ \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \in \{\mathbf{Y}, \perp\}. \quad (5.72) $$

As a consequence, we have

$$ \vartheta^{C_{\omega}^{(1)}}(l(\alpha, \cdot)) = \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}) = \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\omega \sqcap_{\mathcal{G}} \Sigma_{\mathcal{D}}) = \mathbf{Y} \wedge \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) = \varepsilon_{l(\alpha, \cdot)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \quad (5.73) $$

We can check analogously

$$ \vartheta^{C_{\omega}^{(1)}}(l(\cdot, \alpha)) = \varepsilon_{l(\cdot, \alpha)}^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \quad (5.74) $$

<figure><img src="image_8.png" /></figure>
**Lemma 5.31.** Let us consider a maximal compatibility context of type 2 denoted by C. We have

$$ \forall I \in C, \quad \mathcal{D}^C(I) = \varepsilon_I^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \qquad (5.75) $$

*Proof.* Let us consider a maximal compatibility context of type 2 denoted by C and let us choose $\lambda, \lambda' \in \mathcal{G} \setminus \{\perp_{\mathcal{G}}\}$ such that $\lambda' \sqsubseteq_{\mathcal{G}} \lambda^*$ and $\omega, \omega' \in \mathcal{G}^{\text{pure}}$ such that $\omega \sqsubseteq_{\mathcal{G}} \lambda, \omega' \sqsubseteq_{\mathcal{G}} \lambda'$ with

$$ (C \cap \mathfrak{U}_{\mathcal{G}}) \subseteq (\{l_{(\sigma,.)} | \lambda \sqsubseteq_{\mathcal{G}} \sigma \sqsubseteq_{\mathcal{G}} \omega\} \cup \{l_{(.,\sigma)} | \lambda \sqsubseteq_{\mathcal{G}} \sigma \sqsubseteq_{\mathcal{G}} \omega\} \cup \{l_{(.,..)}\} \cup \\ \{l_{(\sigma,.)} | \lambda' \sqsubseteq_{\mathcal{G}} \sigma \sqsubseteq_{\mathcal{G}} \omega'\} \cup \{l_{(.,\sigma)} | \lambda' \sqsubseteq_{\mathcal{G}} \sigma \sqsubseteq_{\mathcal{G}} \omega'\}) \quad (5.76) $$

We note that

$$ \forall I_{(\alpha, \alpha')} \in C, \quad (\omega \sqsubseteq_{\mathcal{G}} \alpha \sqsubseteq_{\mathcal{G}} \lambda \text{ and } \omega' \sqsubseteq_{\mathcal{G}} \alpha' \sqsubseteq_{\mathcal{G}} \lambda'). \quad (5.77) $$

We also note that, for any $l_{(\alpha, \alpha')} \in C$, we have

$$ \begin{aligned} \mathbf{Y} \wedge \mathcal{D}^C(l_{(\alpha, \alpha')}) &= \mathcal{D}^C(\mathfrak{Z} \mathfrak{e}_{\mathcal{G}} \sqcap_{\mathfrak{e}_{\mathcal{G}}} l_{(\alpha, \alpha')}) \\ &= \mathcal{D}^C(l_{(\alpha,.)}) \\ &= \mathcal{D}^{C(1)}_{\omega}(l_{(\alpha,.)}) \\ &= \varepsilon_{l_{(\alpha,.)}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \\ &= \varepsilon_{\mathfrak{Z} \mathfrak{e}_{\mathcal{G}} \sqcap_{\mathfrak{e}_{\mathcal{G}}} l_{(\alpha, \alpha')}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \\ &= \mathbf{Y} \wedge \varepsilon_{l_{(\alpha, \alpha')}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \end{aligned} \qquad (5.78) $$

and

$$ \begin{aligned} \mathbf{N} \wedge \mathcal{D}^C(l_{(\alpha, \alpha')}) &= \mathcal{D}^C(\overline{\mathfrak{Z} \mathfrak{e}_{\mathcal{G}}} \sqcap_{\mathfrak{e}_{\mathcal{G}}} l_{(\alpha, \alpha')}) \\ &= \mathcal{D}^C(l_{(., \alpha')}) \\ &= \mathcal{D}^{C(1)}_{\omega'}(l_{(., \alpha')}) \\ &= \varepsilon_{l_{(., \alpha')}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \\ &= \varepsilon_{\mathfrak{Z} \mathfrak{e}_{\mathcal{G}} \sqcap_{\mathfrak{e}_{\mathcal{G}}} l_{(\alpha, \alpha')}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}) \\ &= \mathbf{N} \wedge \varepsilon_{l_{(\alpha, \alpha')}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \end{aligned} \qquad (5.79) $$

From these two relations, we deduce immediately

$$ \mathcal{D}^C(l_{(\alpha, \alpha')}) = \varepsilon_{l_{(\alpha, \alpha')}}^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \qquad (5.80) $$

□

**Theorem 5.32.** Let us consider an operational description $\mathcal{D}$. We have then necessarily the following property :

$$ \exists \Sigma_{\mathcal{D}} := (\bigsqcup_{\omega \in \mathcal{G}^{\text{pure}}} \Sigma_{\mathcal{D}}^{C(1)} ) \mid \forall C \in \mathcal{C}, \forall I \in C, \mathcal{D}^C(I) = \varepsilon_I^{\mathcal{G}}(\Sigma_{\mathcal{D}}). \quad (5.81) $$

*Proof.* Direct consequence of Lemma 5.29, Lemma 5.30, Lemma 5.31. □
**Theorem 5.33.** The map from $\mathcal{D}$ to $\mathcal{S}$, sending $\mathcal{D}$ to $\Sigma_{\mathcal{D}} := (\bigsqcup_{\omega \in \mathcal{S}^{\text{pure}}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}})$ is an isomorphism.

*Proof.* First of all, we note that we can effectively associate to any $\mathcal{D} \in \mathcal{D}$ an element $\Sigma_{\mathcal{D}}$ satisfying

$$ \forall C \in \mathcal{C}, \forall I \in C, \quad \mathcal{D}^C(I) = \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}}). \qquad (5.82) $$

This point is guaranteed by Lemma 5.29, Lemma 5.30 and Lemma 5.31. More explicitly, we have $\Sigma_{\mathcal{D}} := (\bigsqcup_{\omega \in \mathcal{S}^{\text{pure}}} \Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}) \in \mathcal{S}$ where the state $\Sigma_{\mathcal{D}}^{C_{\omega}^{(1)}}$ is defined in reference to (5.35) (5.36) and Lemma 5.20.

Note that the supremum defining $\Sigma_{\mathcal{D}}$ exists in $\mathcal{S}$ because $\mathcal{D}$ is admissible.

Secondly, this map is clearly injective. Indeed, let us consider a pair $\mathcal{D}, \mathcal{D}'$ of coherent operational descriptions such that $\Sigma_{\mathcal{D}} = \Sigma_{\mathcal{D}'}$. We have obviously $\forall C \in \mathcal{C}, \forall I \in C, \mathcal{D}^C(I) = \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}}) = \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}'}) = \mathcal{D}'^C(I)$.

Endly, this map is also clearly a surjection. Indeed, we can associate, to any $\alpha \in \mathcal{S}$, a coherent operational description $\mathcal{D}_{\alpha} := (\mathcal{D}_{(\alpha)}^{C})_{C \in \mathcal{C}}$ by $\forall C \in \mathcal{C}, \forall I \in C, \mathcal{D}_{(\alpha)}^{C}(I) := \varepsilon_I^{\mathcal{S}}(\alpha)$. We have then, for any $\omega \in \mathcal{S}^{\text{pure}}$,

$$ \Sigma_{\mathcal{D}_{\alpha}}^{C_{\omega}^{(1)}} = \alpha \sqcap_{\mathcal{S}} \omega. \qquad (5.83) $$

We then build as usual $\Sigma_{\mathcal{D}_{\alpha}} := (\bigsqcup_{\omega \in \mathcal{S}^{\text{pure}}} \Sigma_{\mathcal{D}_{\alpha}}^{C_{\omega}^{(1)}})$. We have by construction $\forall C \in \mathcal{C}, \forall I \in C, \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}_{(\alpha)}}) = \varepsilon_I^{\mathcal{S}}(\alpha)$. Using (2.69), we then conclude that $\Sigma_{\mathcal{D}_{(\alpha)}} = \alpha$.

Endly, we derive easily the homomorphic property from the properties of the evaluation map $\varepsilon^{\mathcal{S}}$:

$$ \forall C \in \mathcal{C}, \forall I \in C, \quad (\mathcal{D}^C \sqcap_{\mathcal{D}} \mathcal{D}'^C)(I) = \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}}) \wedge \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}'}) = \varepsilon_I^{\mathcal{S}}(\Sigma_{\mathcal{D}} \sqcap_{\mathcal{S}} \Sigma_{\mathcal{D}'}). \qquad (5.84) $$

□

**6. The multidimensional linear indeterministic space of states (the multi-quantum-bit system)**

The previous section established a general proposal for the tensor product of indeterministic spaces of states. In the next section, we will see that this construction is sufficient to achieve our main objective (i.e. to exhibit the main typically quantum properties in our quantum logic framework). The present section is devoted to a detailed analysis of the space of states obtained as the iterated tensor product of the elementary indeterministic space of states associated with the quantum bit. Not only can this space of states be simply described using the ontic completion tools developed above, but we will also demonstrate how close this space is to an irreducible Hilbert geometry.
**6.1. Preliminary remarks**

To begin this subsection, we consider the two copies of one-dimensional indeter-
ministic spaces of states $\bar{\mathcal{S}}_A = \mathfrak{Z}'_{NA}$ and $\bar{\mathcal{S}}_B = \mathfrak{Z}'_{NB}$ and we consider the space of
states $\bar{S}_{AB} := \bar{\mathcal{S}}_A \otimes \bar{\mathcal{S}}_B$. We denote $\mathcal{S}_A = \mathfrak{Z}^c_{\bar{\mathcal{S}}_A}$ and $\bar{\mathcal{S}}_B = \mathfrak{Z}^c_{\bar{\mathcal{S}}_B}$. We intent to describe
the ontic completion $\hat{S}_{AB} := \mathcal{S}_A \otimes \mathcal{S}_B$ defined according to the subsection 4.7. If
necessary, in order to shorten our formulas, we will eventually replace the notation
$\bar{S}_{AB}$ by $\bar{\mathcal{S}}$ and the notation $\hat{S}_{AB}$ by $\hat{\mathcal{S}}$.

**Theorem 6.1.** Let us consider $\mu, \nu, \phi \in \bar{\mathcal{S}}^{\text{pure}}$ and $\gamma \in \bar{\mathcal{S}}$ such that $\gamma = (\phi \sqcap_{\bar{\mathcal{S}}} \nu) \bar{\Sigma}_{\bar{\mathcal{S}}} \phi, \nu$ and $\mu^* \sqsupseteq_{\bar{\mathcal{S}}} \gamma$, we have

$$
\chi := (\mu^* \sqcup_{\mathcal{S}} \gamma) \text{ exists in } \mathcal{S}. \qquad (6.1)
$$

We have then either $\chi \in \bar{\mathcal{S}}^{\text{pure}}$ or $\chi \in \mathcal{S} \setminus \bar{\mathcal{S}}$ and then, in the last case, denoting as usual $\Theta^{\bar{\mathcal{S}}}(\chi) := \text{Max}((\downarrow_{\bar{\mathcal{S}}}, \chi) \cap \bar{\mathcal{S}})$, we have

$$
\forall \alpha \in \Theta^{\bar{\mathcal{S}}}(\chi), \quad (\exists \kappa \in \bar{\mathcal{S}}^{\text{pure}} \mid \alpha \sqsubseteq_{\bar{\mathcal{S}}} \kappa) \text{ and } \alpha \sqsubseteq_{\bar{\mathcal{S}}} \chi \qquad (6.2)
$$

$$
\forall \alpha, \beta \in \Theta^{\bar{\mathcal{S}}}(\chi), \alpha \neq \beta, \quad (\alpha \sqcap_{\bar{\mathcal{S}}} \beta) \sqsubseteq_{\bar{\mathcal{S}}} \alpha, \beta, \qquad (6.3)
$$

$$
\forall \alpha, \beta, \delta \in \Theta^{\bar{\mathcal{S}}}(\chi) \text{ distinct, } (\alpha \sqcap_{\bar{\mathcal{S}}} \beta) \neq (\alpha \sqcap_{\bar{\mathcal{S}}} \delta), \quad (6.4)
$$

Let us once again consider $\mu \in \bar{\mathcal{S}}^{\text{pure}}$ and $\gamma \in \bar{\mathcal{S}}$ such that $\mu^* \sqsubseteq_{\bar{\mathcal{S}}} \gamma$. Let us assume that there exists $\nu \in \bar{\mathcal{S}}^{\text{pure}}$ such that $\gamma = (\mu \sqcap_{\bar{\mathcal{S}}} \nu) \bar{\Sigma}_{\bar{\mathcal{S}}} \mu, \nu$ then let us here again denote $\chi = (\mu^* \sqcup_{\bar{\mathcal{S}}} \gamma)$.

If we have $\chi \in \bar{\mathcal{S}}$, then necessarily $\chi \in \bar{\mathcal{S}}^{\text{pure}}$, and if we have $\chi \in \bar{\mathcal{S}} \setminus \bar{\mathcal{S}}$ then

$$
\exists (\varphi_{\varepsilon})_{\varepsilon}, (\psi_{\varepsilon})_{\varepsilon} \in (\bar{\mathcal{S}}^{\text{pure}}) \Theta^{\bar{\mathcal{S}}}(\chi) \mid \\
\qquad \varphi_{\gamma} := \mu, \quad \psi_{\gamma} := \nu, \quad \forall \alpha \in \Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}, \quad \varphi_{\gamma}^{*} \sqsubseteq_{\bar{\mathcal{S}}} \varphi_{\alpha}, \quad \varphi_{\gamma}^{*} \sqsubseteq_{\bar{\mathcal{S}}} \psi_{\alpha}, \qquad (6.5)
$$

$$
\forall \varepsilon \in \Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}, \quad \psi_{\varepsilon}, \varphi_{\varepsilon} \sqsupseteq_{\bar{\mathcal{S}}} \varepsilon, \quad \psi_{\varepsilon} \neq \varphi_{\varepsilon} \qquad (6.6)
$$

$$
\forall \varepsilon \in \Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}, \quad \varphi_{\varepsilon}^{*} \sqsubseteq_{\bar{\mathcal{S}}} \psi_{\varepsilon}, \qquad (6.7)
$$

$$
\forall \alpha, \beta \in \Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}, \alpha \neq \beta, \quad \varphi_{\alpha}^{*} \sqsubseteq_{\bar{\mathcal{S}}} \varphi_{\beta}, \quad \varphi_{\alpha}^{*} \sqsupseteq_{\bar{\mathcal{S}}} \psi_{\beta}, \qquad (6.8)
$$

$$
\forall \alpha, \beta \in \Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}, \alpha \neq \beta, \exists \lambda_{\alpha\beta} \in \bar{\mathcal{S}}^{\text{pure}} \mid (\varphi_{\beta} \sqcap_{\bar{\mathcal{S}}} \psi_{\alpha}), (\psi_{\beta} \sqcap_{\bar{\mathcal{S}}} \varphi_{\alpha}) \bar{\Sigma}_{\bar{\mathcal{S}}} \lambda_{\alpha\beta}, \\
\text{and } \lambda_{\alpha\beta}^{*} \bar{\Sigma}_{\bar{\mathcal{S}}} \varphi_{\alpha}, \lambda_{\alpha\beta}^{*} \bar{\Sigma}_{\bar{\mathcal{S}}} \varphi_{\beta}, \qquad (6.9)
$$

$$
\forall \alpha \in \Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}, \exists \mu_{\alpha\gamma} \in \bar{\mathcal{S}}^{\text{pure}} \mid (\varphi_{\gamma} \sqcap_{\bar{\mathcal{S}}} \varphi_{\alpha}), (\psi_{\gamma} \sqcap_{\bar{\mathcal{S}}} \psi_{\alpha}) \bar{\Sigma}_{\bar{\mathcal{S}}} \mu_{\alpha\gamma} \\
\text{and } \mu_{\alpha\gamma}^{*} \sqsubseteq_{\bar{\mathcal{S}}} \psi_{\alpha}, \qquad (6.10)
$$

The collections $(\varphi_{\varepsilon})_{\varepsilon}, (\psi_{\varepsilon})_{\varepsilon} \in (\bar{\mathcal{S}}^{\text{pure}})^{\Theta^{\bar{\mathcal{S}}}(\chi)}$ satisfying the properties (6.5)(6.7)(6.8)
are unique up to a permutation on the elements of $\Theta^{\bar{\mathcal{S}}}(\chi) \setminus \{\gamma\}$.

Let us once again consider $\mu \in \bar{\mathcal{S}}^{\text{pure}}$ and $\gamma \in \bar{\mathcal{S}}$ such that $\mu^* \sqsubseteq_{\bar{\mathcal{S}}} \gamma$. If there exists $\nu \in \bar{\mathcal{S}}^{\text{pure}}$ such that $\gamma = (\mu \sqcap_{\bar{\mathcal{S}}} \nu) \bar{\Sigma}_{\bar{\mathcal{S}}} \mu, \nu$ then let us here again denote $\chi =$
($\mu^* \sqcup_{\mathfrak{S}} \gamma$). If we have $\chi \in \mathfrak{S} \setminus \bar{\mathfrak{S}}$ then
$$ \forall \beta \in \Theta_{\bar{\mathfrak{S}}} (\chi) \setminus \{\gamma\}, \exists \omega, \kappa \in \bar{\mathfrak{S}}^{\text{pure}} \mid (\omega \sqcap_{\bar{\mathfrak{S}}} \kappa) \supseteq_{\bar{\mathfrak{S}}} \omega, \kappa, \text{ and } \omega^* \sqcup_{\bar{\mathfrak{S}}} (\omega \sqcap_{\bar{\mathfrak{S}}} \kappa) = \beta \sqcup_{\bar{\mathfrak{S}}} \gamma^* \quad (6.11) $$

*Proof.* We will consider $\alpha, \alpha', \alpha'' \in \bar{\mathfrak{S}}_A^{\text{pure}}$ and $\beta, \beta', \beta'' \in \bar{\mathfrak{S}}_B^{\text{pure}}$ and we intent to compute

$$ c l_c^{\bar{\mathfrak{S}}_{AB}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}_AB} \alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) $$
with $\alpha' \neq \alpha''$ or $\beta' \neq \beta''$, and with $(\alpha \tilde{\otimes} \beta)^* \supseteq_{\bar{\mathfrak{S}}} (\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta')$.

We will distinguish different cases.

(1) $\alpha' \neq \alpha''$ and $\beta'' = \beta'$. We have $(\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta') = \perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta'$ and then $\alpha^* \tilde{\otimes} \beta'$ is the supremum of $(\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$. Hence,

$$ c l_c^{\bar{\mathfrak{S}}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) = \{ \alpha^* \tilde{\otimes} \beta' \} \in \mathcal{H}(\bar{\mathfrak{S}}) $$

(2) $\alpha' = \alpha''$ and $\beta'' \neq \beta'$. We have $(\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta') = \alpha' \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_A}$ and then $\alpha' \tilde{\otimes} \beta^*$ is the supremum of $(\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$. Hence,

$$ c l_c^{\bar{\mathfrak{S}}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) = \{ \alpha' \tilde{\otimes} \beta^* \} \in \mathcal{H}(\bar{\mathfrak{S}}) $$

(3) $\alpha'' \neq \alpha'$ or $\beta'' \neq \beta'$. We have to distinguish different sub-cases : (3A) $\alpha' \supseteq_{\bar{\mathfrak{S}}_A} \alpha^*$ or $\beta' \supseteq_{\bar{\mathfrak{S}}_A} \beta^*$. Then, $\alpha' \tilde{\otimes} \beta'$ is the supremum in $\bar{\mathfrak{S}}$ of $(\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$. And then,

$$ c l_c^{\bar{\mathfrak{S}}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) = \{ \alpha' \tilde{\otimes} \beta' \} \in \mathcal{H}(\bar{\mathfrak{S}}) $$

(3B) $\alpha'' \supseteq_{\bar{\mathfrak{S}}_A} \alpha^*$ or $\beta'' \supseteq_{\bar{\mathfrak{S}}_A} \beta^*$. Then, $\alpha'' \tilde{\otimes} \beta''$ is the supremum in $\bar{\mathfrak{S}}$ of $(\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$. And then,

$$ c l_c^{\bar{\mathfrak{S}}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'' \sqcap_{\bar{\mathfrak{S}}} \alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) = \{ \alpha'' \tilde{\otimes} \beta'' \} \in \mathcal{H}(\bar{\mathfrak{S}}) $$

(3C) $\alpha' \supseteq_{\bar{\mathfrak{S}}_A} \alpha^*$ and $\beta' \supseteq_{\bar{\mathfrak{S}}_A} \beta^*$ and $\alpha'' \supseteq_{\bar{\mathfrak{S}}_A} \alpha^*$ or $\beta'' \supseteq_{\bar{\mathfrak{S}}_A} \beta^*$. Then, we note that $(\alpha'' \tilde{\otimes} \beta'') \sqcap_{\bar{\mathfrak{S}}} (\alpha' \tilde{\otimes} \beta')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$ have no common upper-bound in $\bar{\mathfrak{S}}$. We have then to note that $(\alpha'' \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$ admit a supremum in $\bar{\mathfrak{S}}$ which is $(\alpha'' \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{S}}} (\alpha^* \tilde{\otimes} \beta')$. We note also that $(\alpha' \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta'')$ and $(\alpha^* \tilde{\otimes} \perp_{\bar{\mathfrak{S}}_B}) \sqcap_{\bar{\mathfrak{S}}} (\perp_{\bar{\mathfrak{S}}_A} \tilde{\otimes} \beta^*)$ admit a supremum in $\bar{\mathfrak{S}}$ which is $(\alpha' \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{S}}} (\alpha^* \tilde{\otimes} \beta'')$. Hence, we check directly that

$$ \begin{aligned} c_{\bar{\mathfrak{S}}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'') \sqcap_{\bar{\mathfrak{S}}} (\alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) &=: U \\ U &= \left\{ (\alpha'' \tilde{\otimes} \beta'') \sqcap_{\bar{\mathfrak{S}}} (\alpha' \tilde{\otimes} \beta'), (\alpha'' \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{S}}} (\alpha^* \tilde{\otimes} \beta'), (\alpha' \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{S}}} (\alpha^* \tilde{\otimes} \beta'') \right\} \end{aligned} \quad (6.12) $$
It is then rather easy to check that $cl_{\bar{\mathfrak{G}}_c}(\bar{\mathfrak{G}}(U)) = U$. Hence, $cl_{\bar{\mathfrak{G}}_c}(U) = U$. Endly, we check directly that $U \in \mathcal{H}(\bar{\mathfrak{G}})$ (note that this check uses explicitly the properties $\alpha' \not\supseteq_{\bar{\mathfrak{G}}_A} \alpha^*$ and $\beta' \not\supseteq_{\bar{\mathfrak{G}}_A} \beta^*$ and $\alpha'' \supseteq_{\bar{\mathfrak{G}}_A} \alpha^*$ or $\beta'' \supseteq_{\bar{\mathfrak{G}}_A} \beta^*$ assumed in case 3C). As a result

$$ 
\begin{aligned}
&cl_c^{\bar{\mathfrak{G}}} \left( \left\{ (\alpha'' \tilde{\otimes} \beta'') \sqcap_{\bar{\mathfrak{G}}_{AB}} (\alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^* \right\} \right) =: U \\
&U = \left\{ (\alpha'' \tilde{\otimes} \beta'') \sqcap_{\bar{\mathfrak{G}}} (\alpha' \tilde{\otimes} \beta'), (\alpha'' \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta'), (\alpha' \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta'') \right\}
\end{aligned} 
\quad (6.13) $$

As a conclusion of this case analysis we have proved the property (6.1). By the way we can check directly on the expressions of $cl_c^{\bar{\mathfrak{G}}}(\{(\alpha'' \tilde{\otimes} \beta'') \sqcap_{\bar{\mathfrak{G}}} (\alpha' \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta)^*\})$ established along this case analysis the properties (6.2)(6.3)(6.4) and also the properties (6.7)(6.8) if we restrict ourselves to the case $\alpha = \alpha''$ and $\beta'' = \beta$.

Let us summarize. If $\chi := (\alpha \tilde{\otimes} \beta)^* \sqcup_{\bar{\mathfrak{G}}} (\alpha \tilde{\otimes} \beta \sqcap_{\bar{\mathfrak{G}}} \alpha' \tilde{\otimes} \beta')$ is in $\bar{\mathfrak{G}} \setminus \bar{\mathfrak{G}}$, we have

$$ \Theta_{\bar{\mathfrak{G}}} (\chi) = \{\gamma, \delta, \varepsilon\} \quad (6.14) $$

$$ 
\begin{aligned}
\gamma &= (\alpha \tilde{\otimes} \beta \sqcap_{\bar{\mathfrak{G}}} \alpha' \tilde{\otimes} \beta'), & \varphi_{\gamma} &= \alpha \tilde{\otimes} \beta, & \psi_{\gamma} &= \alpha' \tilde{\otimes} \beta', \\
\delta &= (\alpha \tilde{\otimes} \beta^* \sqcap_{\bar{\mathfrak{G}}} \alpha^* \tilde{\otimes} \beta'), & \varphi_{\delta} &= \alpha \tilde{\otimes} \beta^*, & \psi_{\delta} &= \alpha^* \tilde{\otimes} \beta', \\
\varepsilon &= (\alpha' \tilde{\otimes} \beta^* \sqcap_{\bar{\mathfrak{G}}} \alpha^* \tilde{\otimes} \beta), & \varphi_{\varepsilon} &= \alpha^* \tilde{\otimes} \beta, & \psi_{\varepsilon} &= \alpha' \tilde{\otimes} \beta^*, \\
\lambda_{\delta\varepsilon} &= \alpha^* \tilde{\otimes} \beta^*, & \mu_{\gamma\delta} &= \alpha \tilde{\otimes} \beta', & \mu_{\gamma\varepsilon} &= \alpha' \tilde{\otimes} \beta.
\end{aligned}
 $$

The collections $(\varphi_{\varepsilon})_{\varepsilon} \in (\bar{\mathfrak{G}}^{\text{pure}})^{\Theta_{\bar{\mathfrak{G}}}(\chi)}$ given above satisfy the properties (6.5)(6.7)(6.8). They are unique up to a permutation on the set $\{\delta, \varepsilon\}$ because of the following basic property

$$ \forall \alpha \in \{\gamma, \delta, \varepsilon\}, \quad \underline{\alpha \bar{\mathfrak{G}}} = \{\varphi_{\alpha}, \psi_{\alpha}\} \quad (6.15) $$

inherited from (4.18).

The properties (6.9)(6.10) can be checked directly.

Concerning the property (6.11), it suffice to compute

$$ 
\begin{aligned}
&cl_c^{\bar{\mathfrak{G}}} \left( \left\{ ((\alpha \tilde{\otimes} \beta) \sqcap_{\bar{\mathfrak{G}}} (\alpha' \tilde{\otimes} \beta'))^*, (\alpha \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta') \right\} \right) = \\
&= \left\{ (\alpha^* \tilde{\otimes} \beta'^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha'^* \tilde{\otimes} \beta^*), (\alpha \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta'), (\alpha \tilde{\otimes} \beta'^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha'^* \tilde{\otimes} \beta') \right\}
\end{aligned} 
\quad (6.16) $$

and to observe then that

$$ 
\begin{aligned}
&cl_c^{\bar{\mathfrak{G}}} \left( \left\{ ((\alpha \tilde{\otimes} \beta) \sqcap_{\bar{\mathfrak{G}}} (\alpha' \tilde{\otimes} \beta'))^*, (\alpha \tilde{\otimes} \beta^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta') \right\} \right) = \\
&cl_c^{\bar{\mathfrak{G}}} \left( \left\{ (\alpha^* \tilde{\otimes} \beta'^*)^*, (\alpha \tilde{\otimes} \beta'^*) \sqcap_{\bar{\mathfrak{G}}} (\alpha'^* \tilde{\otimes} \beta^*) \right\} \right)
\end{aligned} 
\quad (6.17) $$

□

Now, we consider $N$ copies of one-dimensional indeterministic spaces of states $\bar{\mathfrak{G}}_{A_i} = \mathfrak{Z}'_{M_{A_i}}$ for $i = 1, \dots, N$ ($M_{A_i} \in \mathbb{N}$) and we now consider the real structure defined by $\bar{\mathfrak{S}}_{A_1 \cdots A_N} := \bar{\mathfrak{S}}_{A_1} \tilde{\otimes} \cdots \tilde{\otimes} \bar{\mathfrak{S}}_{A_N}$. We denote $\mathfrak{S}_{A_i} = \mathfrak{Z}_{\bar{\mathfrak{G}}_{A_i}}$ for $i = 1, \dots, N$. And we introduce, as before, the tensor product $\hat{\mathfrak{S}}_{A_1 \cdots A_N} := \mathfrak{Z}^c_{\bar{\mathfrak{G}}_{A_1} \hat{\otimes} \cdots \hat{\otimes} \bar{\mathfrak{G}}_{A_N}}$. This space of
states will be said to be the $N$-dimensional indeterministic space of states. If necessary, in order to shorten our formulas, we will denote $\bar{S}_{A_1 \dots A_N}$ simply by $\bar{\mathfrak{S}}$ and $\hat{S}_{A_1 \dots A_N}$ simply by $\hat{\mathfrak{S}}$.

For simplicity, we will restrict ourselves to the case $N = 3$ in order to clarify a central aspect of the iteration of the tensor product.

We will then consider $\sigma_{A_1}, \sigma'_{A_1}, \sigma''_{A_1} \in \bar{\mathfrak{S}}_{A_1}, \sigma_{A_2}, \sigma'_{A_2}, \sigma''_{A_2} \in \bar{\mathfrak{S}}_{A_2}, \sigma_{A_3}, \sigma'_{A_3}, \sigma''_{A_3} \in \bar{\mathfrak{S}}_{A_3}$. We will suppose that $\sigma'_{A_1} \neq \sigma''_{A_1}$ and $\sigma^*_{A_1} \neq \sigma'_{A_1}, \sigma''_{A_1}$, and $\sigma'_{A_2} \neq \sigma''_{A_2}$ and $\sigma^*_{A_2} \neq \sigma'_{A_2}, \sigma''_{A_2}$, and $\sigma'_{A_3} \neq \sigma''_{A_3}$ and $\sigma^*_{A_3} \neq \sigma'_{A_3}, \sigma''_{A_3}$.

From the following properties

$$ (\sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3})^* \sqcup_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} (\bar{\sigma}_{A_1} \otimes \sigma''_{A_2} \otimes \sigma''_{A_3} \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \sigma'_{A_3}) = \sigma^*_{A_1} \otimes \sigma''_{A_2} \otimes \sigma''_{A_3} \quad (6.18) $$

$$ (\sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3})^* \sqcup_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} (\sigma''_{A_1} \otimes \sigma''_{A_2} \otimes \sigma''_{A_3} \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \bar{\sigma}_{A_1} \otimes \sigma'_{A_2} \otimes \sigma'_{A_3}) = \sigma^*_{A_1} \otimes \sigma'_{A_2} \otimes \sigma'_{A_3} \quad (6.19) $$

we deduce

$$ \{ \sigma_{A_1}^* \otimes \sigma_{A_2}'' \otimes \sigma_{A_3}'', \sigma_{A_1}^* \otimes \sigma_{A_2}' \otimes \sigma_{A_3}' \} \subseteq \mathfrak{C}^{\bar{S}_{A_1 A_2 A_3}} \left( \left\{ (\sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3})^*, (\sigma_{A_1}'' \otimes \sigma_{A_2}'' \otimes \sigma_{A_3}'' \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \sigma_{A_1}' \otimes \sigma_{A_2}' \otimes \sigma_{A_3}' \right) \right) \quad (6.20) $$

Now we note that $\alpha := \sigma_{A_1}^* \otimes \sigma_{A_2}'' \otimes \sigma_{A_3}'$, and $\beta := \sigma_{A_1}^* \otimes \sigma_{A_2}' \otimes \sigma_{A_3}'$ are two distinct elements of $\bar{S}^{\text{pure}}_{A_1 A_2 A_3}$ and $(\alpha \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \beta) \neq \bar{\sigma}_{A_1 A_2 A_3}$ and that $\forall \gamma, \gamma' \in \bar{S}^{\text{pure}}_{A_1 A_2 A_3}$ we have $(\gamma \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \gamma') \bar{\mathfrak{C}}_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \gamma, \gamma'$. In other words, the conditions of Theorem 3.42 are satisfied. Then, $cl^{\bar{S}_{A_1 A_2 A_3}} (\{\alpha, \beta\}) \notin \mathcal{H}(\bar{S}_{A_1 A_2 A_3})$. As a result, we have obtained

$$ cl_c^{\bar{S}_{A_1 A_2 A_3}} \left( \left\{ (\sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3})^*, (\sigma_{A_1}'' \otimes \sigma_{A_2}'' \otimes \sigma_{A_3}'' \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \sigma_{A_1}' \otimes \sigma_{A_2}' \otimes \sigma_{A_3}' \right) \right) \notin \mathcal{H}(\bar{S}_{A_1 A_2 A_3}) \quad (6.21) $$

In other words, as long as $\sigma'_{A_1} \neq \sigma''_{A_1}$ and $\sigma^*_{A_1} \neq \sigma'_{A_1}, \sigma''_{A_1}$, and $\sigma'_{A_2} \neq \sigma''_{A_2}$ and $\sigma^*_{A_2} \neq \sigma'_{A_2}, \sigma''_{A_2}$, and $\sigma'_{A_3} \neq \sigma''_{A_3}$ and $\sigma^*_{A_3} \neq \sigma'_{A_3}, \sigma''_{A_3}$, we have

$$ \#\alpha \in \bar{S}_{A_1 A_2 A_3} \mid ((\sigma_{A_1}'' \otimes \sigma_{A_2}'' \otimes \sigma_{A_3}'' \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \sigma_{A_1}' \otimes \sigma_{A_2}' \otimes \sigma_{A_3}' \bar{\mathfrak{C}}_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \alpha \text{ and } \sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3} \perp \alpha). \quad (6.22) $$

If, on the contrary, the condition $(\sigma'_{A_1} \neq \sigma''_{A_1}, \sigma'_{A_2} \neq \sigma''_{A_2}, \sigma'_{A_3} \neq \sigma''_{A_3})$ is not satisfied anymore (we maintain nevertheless the conditions $\sigma^*_{A_1} \neq \sigma'_{A_1}, \sigma''_{A_1}, \sigma^*_{A_2} \neq \sigma'_{A_2}, \sigma''_{A_2}, \sigma^*_{A_3} \neq \sigma'_{A_3}, \sigma''_{A_3}$ because, if this condition is not satisfied, the result is easy to compute), we have then first of all

$$ \exists \alpha \in \bar{S}_{A_1 A_2 A_3} \mid ((\sigma_{A_1}' \otimes \sigma_{A_2}' \otimes \sigma_{A_3}' \sqcap_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \sigma_{A_1}' \otimes \sigma_{A_2}' \otimes \sigma_{A_3}' \bar{\mathfrak{C}}_{\bar{\mathfrak{S}}_{A_1 A_2 A_3}} \alpha \text{ and } \sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3} \perp \alpha). \quad (6.23) $$
i.e.

$$
\Theta^{\bar{S}_{A_1 A_2 A_3}}(\alpha) = \left\{ \left( \sigma'_{A_1} \otimes \sigma''_{A_2} \otimes \sigma''_{A_3} \sqcap_{\bar{s}_{A_1 A_2 A_3}} \sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \sigma'_{A_3} \right), \left( \sigma'_{A_1} \otimes \sigma^*_{A_2} \otimes \sigma''_{A_3} \sqcap_{\bar{s}_{A_1 A_2 A_3}} \sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \sigma^*_{A_3} \right), \right. \\
\left. \left( \sigma'_{A_1} \otimes \sigma''_{A_2} \otimes \sigma^*_{A_3} \sqcap_{\bar{s}_{A_1 A_2 A_3}} \sigma'_{A_1} \otimes \sigma^*_{A_2} \otimes \sigma'_{A_3} \right) \right\} \tag{6.24}
$$

and we have also

$$
\exists \alpha \in \tilde{S}_{A_1 A_2 A_3} \mid ((\sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \sigma''_{A_3} \sqcap_{\bar{s}_{A_1 A_2 A_3}} \sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \sigma'_{A_3}) \sqsubseteq_{\tilde{s}_{A_1 A_2 A_3}} \alpha \text{ and } \sigma_{A_1} \otimes \sigma_{A_2} \otimes \sigma_{A_3} \perp \alpha). \\
\alpha = \sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \left( (\sigma''_{A_3} \sqcap_{\bar{s}_{A_3}} \sigma'_{A_3}) \sqcup_{s_{A_3}} \sigma^*_{A_3} \right) \tag{6.25}
$$

i.e.

$$
\Theta^{\bar{S}_{A_1 A_2 A_3}} (\alpha) = \{ \sigma'_{A_1} \otimes \sigma'_{A_2} \otimes \sigma^*_{A_3} \}. \quad (6.26)
$$

The results obtained by flipping the tensors and exchanging the corresponding in-
dices are obvious.

Let us summarize the previous result, but before that, let us introduce some
notations.

**Definition 6.2.** Let us consider $\nu$ and $\phi$ two elements of $\overline{\mathfrak{S}}^{\text{pure}}$. We will define the following reflexive and symmetric binary relation

$$
\nu \nsim \phi \quad \Leftrightarrow \quad (\exists I \subseteq \{1, \dots, N\}, \mathrm{Card}(I) \ge N-2 \mid \forall i \in I, \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\nu) = \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\phi)). \quad (6.27)
$$

**Definition 6.3.** Let us fix $\nu \in \overline{\mathfrak{S}}^{\text{pure}}$ and $j, k \in \{1, \dots, N\}$ with $j < k$. Let us denote, for any $\kappa \in \overline{\mathfrak{S}}^{\text{pure}}$,

$$
\Upsilon_{\kappa}^{(j,k)} := \zeta_{(j)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\kappa) \otimes \zeta_{(k)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\kappa) \in \overline{\mathfrak{S}}_{A_j}^{\text{pure}} \otimes \overline{\mathfrak{S}}_{A_k}^{\text{pure}} \quad (6.28)
$$

and, for any $(\alpha \otimes \beta) \in \bar{\mathfrak{S}}_{A_j}^{\text{pure}} \otimes \bar{\mathfrak{S}}_{A_k}^{\text{pure}}$,

$$
\Delta_{\alpha \otimes \beta}^{(j,k); \nu} := \delta_1 \otimes \cdots \otimes \delta_{j-1} \otimes \alpha \otimes \delta_{j+1} \otimes \cdots \otimes \delta_{k-1} \otimes \beta \otimes \delta_{k+1} \otimes \cdots \otimes \delta_N \quad (6.29)
$$

with $\delta_m := \zeta_{(m)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\nu)$ for any $m \in \{1, \dots, N\}$.

**Theorem 6.4.** Let us consider $\nu$ and $\phi$ two distinct elements of $\overline{\mathfrak{S}}^{\text{pure}}$ and $\mu$ another element of $\overline{\mathfrak{S}}^{\text{pure}}$. Let us assume that $\mu \perp \nu$, $\phi$. We have

$$
(\exists \chi \in (\tilde{\mathfrak{S}} \setminus \overline{\mathfrak{S}}) | ((\nu \sqcap_{\tilde{\mathfrak{S}}} \phi) \sqsubseteq_{\tilde{\mathfrak{S}}'} \chi \text{ and } \chi \perp \mu)) \Leftrightarrow \qquad (6.30)
$$

$$
(\exists I \subseteq \{1, \dots, N\}, \mathrm{Card}(I) = N-2 | \forall i \in I, \forall \sigma, \kappa \in \overline{\mathfrak{S}}^{\text{pure}}, ((\sigma \sqcap_{\overline{\mathfrak{S}}'} \chi) \sqsubseteq_{\overline{\mathfrak{S}}'} \sigma, \chi, (\kappa \sqcap_{\overline{\mathfrak{S}}'} \chi) \sqsubseteq_{\overline{\mathfrak{S}}'} \kappa, \chi) \Rightarrow \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\sigma) = \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\kappa)).
$$

In particular, $\exists \chi \in \tilde{\mathfrak{S}} | ((\nu \sqcap_{\tilde{\mathfrak{S}}} \phi) \sqsubseteq_{\tilde{\mathfrak{S}}} \chi \text{ and } \chi \perp \mu))$ implies $\forall i \in I$, $\zeta_{(i)}^{\tilde{\mathfrak{S}}_{A_1} \cdots \tilde{\mathfrak{S}}_{A_N}}(\nu) = \zeta_{(i)}^{\tilde{\mathfrak{S}}_{A_1} \cdots \tilde{\mathfrak{S}}_{A_N}}(\phi)$.
Let us denote $j,k \in \{1, \dots, N\}$ with $j < k$ such that for any $i \in \{1, \dots, N\} \setminus \{j,k\}$ we have $\zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(v) = \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\phi)$. Endly, let us denote $\alpha := \zeta_{(j)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\mu)$,
$\beta := \zeta_{(k)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\mu)$, $\alpha' := \zeta_{(j)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(v)$, $\beta' := \zeta_{(k)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(v)$, $\alpha'' := \zeta_{(j)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\phi)$,
$\beta'' := \zeta_{(k)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\phi)$.

As long as $\lambda := \mu \sqcup_{\mathfrak{S}} (v \sqcap_{\mathfrak{S}} \phi)$ exists in $\mathfrak{S} \setminus \bar{\mathfrak{S}}$, we have (see (6.13))

$$ \Theta \bar{\mathfrak{S}}(\lambda) = \left\{ \Delta_{\alpha'' \bar{\beta}''}^{(j,k);v} \sqcap_{\mathfrak{S}} \Delta_{\alpha' \bar{\beta}'}^{(j,k);v} \sqcap_{\bar{\mathfrak{S}}} \Delta_{\alpha^* \bar{\beta}^*}^{(j,k);v} \sqcap_{\bar{\mathfrak{S}}} \Delta_{\alpha^* \bar{\beta}''}^{(j,k);v} \right\} \quad (6.31) $$

*Proof.* Once again we will fix $N=3$. It suffices to remark, using the expression (6.24), the expression of the infimum (3.21), the property (6.2) and the expansion formula (4.18), that the relations $(\sigma \sqcap_{\mathfrak{S}} \chi) \bar{\sqcap}_{\mathfrak{S}} \sigma, \chi$ and $(\kappa \sqcap_{\bar{\mathfrak{S}}} \chi) \bar{\sqcap}_{\bar{\mathfrak{S}}} \kappa, \chi$ impose necessarily $\sigma$ and $\kappa$ to be both elements of the following subset of pure states

$$ \{ (\sigma'_{A_1} \bar{\otimes} \sigma''_{A_2} \bar{\otimes} \sigma''_{A_3}), (\sigma'_{A_1} \bar{\otimes} \sigma'_{A_2} \bar{\otimes} \sigma'_{A_3}), (\sigma'_{A_1} \bar{\otimes} \sigma^*_{A_2} \bar{\otimes} \sigma''_{A_3}), (\sigma'_{A_1} \bar{\otimes} \sigma'_{A_2} \bar{\otimes} \sigma^*_{A_3}), (\sigma'_{A_1} \bar{\otimes} \sigma''_{A_2} \bar{\otimes} \sigma^*_{A_3}), (\sigma'_{A_1} \bar{\otimes} \sigma''_{A_2} \bar{\otimes} \sigma'_{A_3}) \} \quad (6.32) $$

This concludes the proof, which can be straightforwardly generalized to generic $N$. $\square$

We end this subsection by giving some basic properties that will be fruitful in the sequel.

**Theorem 6.5.**

$$ \forall \alpha, \beta \in \bar{\mathfrak{S}}^{\text{pure}}, \quad \beta \neq \alpha \Rightarrow (\alpha \sqcap_{\mathfrak{S}} \beta) \bar{\sqcap}_{\mathfrak{S}} \alpha, \beta. \quad (6.33) $$

*Proof.* The property (6.33) is obtained by a trivial recursion proof (on the number $N$) using Theorem 4.29. $\square$

**Theorem 6.6.** *We have the following basic property*

$$ \begin{gathered} \forall \lambda, \alpha, \beta, \gamma, \delta \in \bar{\mathfrak{S}}^{\text{pure}} \text{ distinct,} \\ ((\alpha \sqcap_{\mathfrak{S}} \beta) \neq (\gamma \sqcap_{\mathfrak{S}} \delta) \text{ and } (\alpha \sqcap_{\bar{\mathfrak{S}}} \beta), (\gamma \sqcap_{\bar{\mathfrak{S}}} \delta) \bar{\sqcap}_{\bar{\mathfrak{S}}} \lambda) \Rightarrow \\ (\alpha \sqcap_{\mathfrak{S}} \beta \sqcap_{\bar{\mathfrak{S}}} \gamma \sqcap_{\bar{\mathfrak{S}}} \delta) \bar{\sqcap}_{\bar{\mathfrak{S}}} (\alpha \sqcap_{\bar{\mathfrak{S}}} \beta), (\gamma \sqcap_{\bar{\mathfrak{S}}} \delta). \end{gathered} \quad (6.34) $$

*Proof.* The property (6.34) is obtained by a trivial recursion proof (on the number $N$) using Theorem 4.30. $\square$

**Theorem 6.7.**

$$ \forall \lambda, \alpha, \beta, \gamma, \delta \in \bar{\mathfrak{S}}^{\text{pure}} \text{ distinct}, ((\alpha \sqcap_{\mathfrak{S}} \beta) \neq (\gamma \sqcap_{\mathfrak{S}} \delta) \text{ and } (\alpha \sqcap_{\bar{\mathfrak{S}}} \beta), (\gamma \sqcap_{\bar{\mathfrak{S}}} \delta) \bar{\sqcap}_{\bar{\mathfrak{S}}} \lambda) \Rightarrow \quad (6.35) $$

$$ \exists I \subseteq \{1, \dots, N\}, \text{Card}(I) = N-2 \mid \forall i \in I, \forall \sigma, \kappa \in \{\lambda, \alpha, \beta, \gamma, \delta\}, \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\sigma) = \zeta_{(i)}^{\mathfrak{S}_{A_1} \cdots \mathfrak{S}_{A_N}}(\kappa). $$

*Proof.* Direct consequence of the expansion formula (4.18). Let us consider $\lambda := \sigma_{A_1}^{\lambda} \bar{\otimes} \cdots \bar{\otimes} \sigma_{A_N}^{\lambda}$, $\alpha := \sigma_{A_1}^{\alpha} \bar{\otimes} \cdots \bar{\otimes} \sigma_{A_N}^{\alpha}$, $\beta := \sigma_{A_1}^{\beta} \bar{\otimes} \cdots \bar{\otimes} \sigma_{A_N}^{\beta}$. The property $(\alpha \sqcap_{\mathfrak{S}} \beta) \bar{\sqcap}_{\bar{\mathfrak{S}}} \lambda$ imposes

$$ \exists i \in \{1, \dots, N\}, \mid (\forall j \in \{1, \dots, N\} \setminus \{i\}, \sigma_{A_j}^{\alpha} = \sigma_{A_j}^{\beta} = \sigma_{A_j}^{\lambda} \text{ and } (\sigma_{A_i}^{\alpha} \sqcap_{\bar{\mathfrak{S}}_{A_i}} \sigma_{A_i}^{\beta}) \bar{\sqcap}_{\bar{\mathfrak{S}}_{A_i}} \sigma_{A_i}^{\lambda}). \quad (6.36) $$

This leads directly to the announced result. $\square$
## 6.2. Towards Hilbert geometry

Let us now focus on projectivity and orthogonality properties relative to the multi-dimensional linear indeterministic space of states.

**Definition 6.8.** We will define the following subset of $\mathfrak{G}$:

$$ \check{\mathfrak{G}} := \bar{\mathfrak{G}} \cup \{ \chi \in \mathfrak{G} \setminus \bar{\mathfrak{G}} \mid \exists \mu, \nu, \phi \in \bar{\mathfrak{G}}^{\text{pure}}, \gamma := \nu \sqcap_{\bar{\mathfrak{G}}} \phi \bar{\sqcap}_{\bar{\mathfrak{G}}} \nu, \phi, \gamma \sqsupseteq_{\bar{\mathfrak{G}}} \mu^*, \chi = \mu^* \sqcup_{\bar{\mathfrak{G}}} \gamma \} \quad (6.37) $$

$$ \check{\mathfrak{G}} := \bar{\mathfrak{G}} \cup \{ \chi \in \mathfrak{G} \setminus \bar{\mathfrak{G}} \mid \exists \mu, \phi \in \bar{\mathfrak{G}}^{\text{pure}}, \mu \sqcap_{\bar{\mathfrak{G}}} \phi \bar{\sqcap}_{\bar{\mathfrak{G}}} \mu, \chi = \mu^* \sqcup_{\bar{\mathfrak{G}}} (\mu \sqcap_{\bar{\mathfrak{G}}} \phi) \} \quad (6.38) $$

and

$$ \mathfrak{G}^{\check{\mathfrak{G}}} := \bar{\mathfrak{G}}^{\text{pure}} \cup (\check{\mathfrak{G}} \setminus \bar{\mathfrak{G}}), \quad (6.39) $$

$$ \mathfrak{G}^{\widetilde{\mathfrak{G}}} := \bar{\mathfrak{G}}^{\text{pure}} \cup (\check{\mathfrak{G}} \setminus \bar{\mathfrak{G}}). \quad (6.40) $$

We note obviously that $\check{\mathfrak{G}} \supseteq \widetilde{\mathfrak{G}}$ and $\mathfrak{G}^{\check{\mathfrak{G}}} \supseteq \mathfrak{G}^{\widetilde{\mathfrak{G}}}$.

**Definition 6.9.** On $\mathfrak{G}^{\check{\mathfrak{G}}}$, we will define the consistency relation, denoted $\asymp$, as follows (see Definition 6.2 for a definition of $\lozenge$):

$$ \forall \chi, \chi' \in \check{\mathfrak{G}} \setminus \bar{\mathfrak{G}}, \quad \chi \asymp \chi' \iff (\chi = \chi' \text{ or } \exists \varepsilon \in \Theta^{\bar{\mathfrak{G}}}(\chi) \cap \Theta^{\check{\mathfrak{G}}}(\chi'), \varepsilon = (\chi \sqcap_{\bar{\mathfrak{G}}} \chi')) \quad (6.41) $$

$$ \forall \chi \in \check{\mathfrak{G}} \setminus \bar{\mathfrak{G}}, \forall \sigma \in \bar{\mathfrak{G}}^{\text{pure}}, \quad \chi \asymp \sigma \iff \sigma \asymp \chi \iff (\exists \varepsilon \in \Theta^{\bar{\mathfrak{G}}}(\chi) \mid \varepsilon \sqsupseteq_{\bar{\mathfrak{G}}} \sigma) \quad (6.42) $$

$$ \forall \sigma, \sigma' \in \bar{\mathfrak{G}}^{\text{pure}}, \quad \sigma \asymp \sigma' \iff \sigma \lozenge \sigma' \quad (6.43) $$

Note that, due to the property (6.33) we have $\forall \sigma, \sigma' \in \bar{\mathfrak{G}}^{\text{pure}}, (\sigma \sqcap_{\bar{\mathfrak{G}}} \sigma') \bar{\sqcap}_{\bar{\mathfrak{G}}} \sigma, \sigma'$ or $\sigma = \sigma'$.

**Remark 6.10.** Note that, using (6.2), we obtain easily

$$ \forall \chi, \chi' \in \check{\mathfrak{G}} \setminus \bar{\mathfrak{G}}, \quad (\chi \asymp \chi' \text{ and } \chi \neq \chi') \Rightarrow \text{Card}(\Theta^{\bar{\mathfrak{G}}}(\chi) \cap \Theta^{\check{\mathfrak{G}}}(\chi')) = 1. \quad (6.44) $$

**Remark 6.11.** Note that, using (6.2), we obtain easily

$$ \forall \chi \in \check{\mathfrak{G}} \setminus \bar{\mathfrak{G}}, \forall \sigma \in \bar{\mathfrak{G}}^{\text{pure}}, \quad (\chi \asymp \sigma) \Rightarrow \text{Card}(\Theta^{\bar{\mathfrak{G}}}(\chi) \cap (\lozenge_{\bar{\mathfrak{G}}} \sigma)) = 1. \quad (6.45) $$

**Definition 6.12.** A consistent subset of states is defined to be a subset of $\mathfrak{G}^{\check{\mathfrak{G}}}$ such that all its elements are pairwise consistent.

**Definition 6.13.** The consistency cover of $\mathfrak{G}^{\check{\mathfrak{G}}}$, denoted $\mathfrak{C}^{\check{\mathfrak{G}}}$, is defined as a family of maximal consistent subsets covering the whole set of elements of $\mathfrak{G}^{\check{\mathfrak{G}}}$:

$$ \left\{ \begin{array}{l} (\bigcup_{U \in \mathfrak{C}^{\check{\mathfrak{G}}}} U) = \mathfrak{G}^{\check{\mathfrak{G}}} \\ \forall U, U' \in \mathfrak{C}^{\check{\mathfrak{G}}}, \quad U \subseteq U' \Rightarrow U = U' \\ \forall U \in \mathfrak{C}^{\widetilde{\mathfrak{G}}}, \forall \sigma, \sigma' \in U, \sigma \asymp \sigma'. \end{array} \right. \quad (6.46) $$

**Definition 6.14.** Let us consider any consistent subset $U \in \mathfrak{C}^{\widetilde{\mathfrak{G}}}$.

We define a ternary relation $\mathfrak{t}^U$ on $U$ (it will be called colinearity) as follows

$$ \forall \sigma_1, \sigma_2, \sigma_3 \in U, \quad \mathfrak{t}^U(\sigma_1, \sigma_2, \sigma_3) \iff (\sigma_2 = \sigma_3 \text{ or } (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_3) \bar{\sqcap}_{\bar{\mathfrak{G}}} \sigma_1). \quad (6.47) $$
**Lemma 6.15.** We note the following sheaf condition

$$ \forall \sigma_1, \sigma_2, \sigma_3 \in \mathfrak{B}^{\mathfrak{G}}, \forall U, U' \in \mathfrak{C}^{\mathfrak{G}}, \quad \sigma_1, \sigma_2, \sigma_3 \in U \cap U' \Rightarrow \mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_3) = \mathfrak{r}^{U'}(\sigma_1, \sigma_2, \sigma_3). \quad (6.48) $$

Let us begin by checking a basic symmetry result of the colinearity relation.

**Lemma 6.16.** The ternary relation $\mathfrak{r}^U$ defined on the consistent subset $U$ is symmetric under any permutation of its arguments.

*Proof.* It suffices to prove that it is symmetric under the exchange of the two first arguments (because it is already symmetric under the exchange of its two last arguments). This is a simple computation. We reproduce only one of its cases.

Let us then consider $\sigma_1, \sigma_2, \sigma_3 \in \mathfrak{B}^{\mathfrak{G}}$ such that $\sigma_1, \sigma_2, \sigma_3$ are pairwise compatible. We will choose for example $\sigma_2, \sigma_3 \in \mathfrak{G} \setminus \bar{\mathfrak{G}}$.

Let us first consider that $\sigma_2 \neq \sigma_3$. Then, $\mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_3)$ is equivalent to $\sigma_1 \bar{\cap}_\mathfrak{G} (\sigma_2 \cap_\mathfrak{G} \sigma_3)$, which means that there exists $\varepsilon \bar{\cap}_\mathfrak{G} \sigma_1$ where $\{\varepsilon\} = \Theta^{\bar{\mathfrak{G}}}(\sigma_2) \cap \Theta^{\bar{\mathfrak{G}}}(\sigma_3)$ (here we use property (6.44) or (6.45)). We then deduce that $(\sigma_1 \cap_\mathfrak{G} \sigma_3) = \varepsilon$ and using $\varepsilon \in \Theta^{\bar{\mathfrak{G}}}(\sigma_2)$ we obtain $\sigma_2 \bar{\cap}_\mathfrak{G} (\sigma_1 \cap_\mathfrak{G} \sigma_3)$ from (6.2). This proves $\mathfrak{r}^U(\sigma_2, \sigma_1, \sigma_3)$.

Let us now consider that $\sigma_2 = \sigma_3$. Then, we have immediately $\mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_3)$. We have also trivially $\sigma_2 \bar{\cap}_\mathfrak{G} (\sigma_1 \cap_\mathfrak{G} \sigma_2)$ which means that we have also $\mathfrak{r}^U(\sigma_2, \sigma_1, \sigma_3)$. $\square$

<figure><img src="image_9.png" /></figure>

**Lemma 6.17.** [Veblen-Young's first geometric property]

$$ \forall U \in \mathfrak{C}^{\mathfrak{G}}, \forall \sigma_1, \sigma_2 \in U, \quad \mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_2). \quad (6.49) $$

*Proof.* Trivial. $\square$

<figure><img src="image_10.png" /></figure>

Before studying the different properties of the colinearity relation, we begin to prove an important lemma (Lemma 6.19) describing the structure of the consistent subsets. But before that, we will check a small technical result.

**Lemma 6.18.** Let us consider $\chi \in (\mathfrak{G} \setminus \bar{\mathfrak{G}})$. We have

$$ \forall \sigma_1, \sigma_2 \in \Theta^{\bar{\mathfrak{G}}}(\chi), \sigma_1 \neq \sigma_2, \quad \sigma_1 \sqcup_{\mathfrak{G}} \sigma_2 = \chi. \quad (6.50) $$

*Proof.* From property (6.2) we know that $\forall \sigma \in \Theta^{\bar{\mathfrak{G}}}(\chi)$, $\sigma \bar{\cap}_\mathfrak{G} \chi$. Then, if we consider $\sigma_1, \sigma_2 \in \Theta^{\bar{\mathfrak{G}}}(\chi)$ with $\sigma_1 \neq \sigma_2$, we have $\sigma_1 \sqcup_\mathfrak{G} (\sigma_1 \sqcup_\mathfrak{G} \sigma_2) \sqsubseteq_\mathfrak{G} \chi$ and $\sigma_1 \bar{\cap}_\mathfrak{G} \chi$. Then, we conclude that $(\sigma_1 \sqcup_\mathfrak{G} \sigma_2) = \chi$. $\square$

<figure><img src="image_11.png" /></figure>

**Lemma 6.19.** Let $U \subseteq \mathfrak{B}^{\mathfrak{G}}$ be a consistent subset of states, such that there exists $\sigma_1, \sigma_2, \sigma_3, \sigma_4, \lambda \in U$ satisfying

$$ \forall i, j \in \{1, 2, 3, 4\}, i \neq j, \quad \sigma_i \neq \sigma_j \quad (6.51) $$

$$ \forall i \in \{1, 2, 3, 4\}, \quad \sigma_i \neq \lambda \quad (6.52) $$

$$ \forall i, j, k, l \in \{1, 2, 3, 4\}, \{i, j\} \neq \{k, l\}, \quad (\sigma_i \cap_{\mathfrak{G}} \sigma_j) \neq (\sigma_k \cap_{\mathfrak{G}} \sigma_l), \quad (6.53) $$

$$ \lambda \bar{\cap}_{\mathfrak{G}} (\sigma_1 \cap_{\mathfrak{G}} \sigma_2), \quad \lambda \bar{\cap}_{\mathfrak{G}} (\sigma_3 \cap_{\mathfrak{G}} \sigma_4), \quad (6.54) $$

then we have necessarily

$$ \forall i \in \{1, 2, 3, 4\}, \sigma_i \in \bar{\mathfrak{G}}^{\text{pure}}. \quad (6.55) $$
*Proof.* The proof is in the Appendix. □

**Lemma 6.20.** Let us consider $U \in \mathfrak{C}^{\mathfrak{G}}$ such that there exists $\chi \in (\mathfrak{G} \diagup \bar{\mathfrak{G}})$ with $\chi \in U$. The subset $U$ is necessarily of one of the two following types

* (Consistent Type 1) $\exists! \alpha \in \Theta^{\bar{\mathfrak{G}}}(\chi)$ such that $\forall \sigma \in U$ we have $(\sigma \sqcap_{\bar{\mathfrak{G}}} \chi) = \alpha$.

* (Consistent Type 2) $\exists \alpha, \beta \in \Theta^{\bar{\mathfrak{G}}}(\chi)$, $\alpha \neq \beta$ and $\exists \sigma, \sigma' \in U \cap \bar{\mathfrak{G}}^{\text{pure}}$ such that $(\sigma \sqcap_{\bar{\mathfrak{G}}} \chi) = \alpha$ and $(\sigma' \sqcap_{\bar{\mathfrak{G}}} \chi) = \beta$.

*Proof.* Direct consequence of Lemma 6.19 and the assumption of maximality on $U$. □

**Lemma 6.21.** Let us consider a maximal consistent subset of states $U \in \mathfrak{C}^{\mathfrak{G}}$ and let us assume that there exists $\lambda \in (\mathfrak{G} \diagup \bar{\mathfrak{G}})$ and $\sigma_1, \sigma_2, \sigma_3, \sigma_4 \in \bar{\mathfrak{G}}^{\text{pure}}$ such that $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U$ and such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \mathfrak{r}^U(\mu, \nu, \rho)$. Then, the consistent subset $U$ is necessarily of consistent type 2.

*Proof.* As long as $\lambda \in (\mathfrak{G} \diagup \bar{\mathfrak{G}})$ $\Theta^{\bar{\mathfrak{G}}}(\lambda)$ is given as in Theorem 6.4. The conditions $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \mathfrak{r}^U(\mu, \nu, \rho)$ impose the existence of $\varepsilon, \delta \in \Theta^{\bar{\mathfrak{G}}}(\lambda)$ such that $\varepsilon \neq \delta$ and $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_2) = \varepsilon$ and $(\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \sigma_4) = \delta$. We note that these conditions impose also the distinctness of the elements $\sigma_i$, i.e. $\alpha' \neq \alpha''$ and $\beta' \neq \beta''$. □

Let us now come back to the properties of the colinearity relation.

**Lemma 6.22.** [Veblen-Young's second geometric property]

$$ \forall U \in \mathfrak{C}^{\mathfrak{G}}, \forall \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U, \quad (\mathfrak{r}^U(\sigma_1, \sigma_3, \sigma_4) \text{ and } \mathfrak{r}^U(\sigma_2, \sigma_3, \sigma_4)) \Rightarrow \mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_3). \quad (6.56) $$

*Proof.* This is a straightforward check. We reproduce one of its cases. Let us then consider for example $\sigma_3, \sigma_4 \in (\mathfrak{G} \diagup \bar{\mathfrak{G}})$ with $\sigma_3 \neq \sigma_4$.

Then, $\mathfrak{r}^U(\sigma_1, \sigma_3, \sigma_4)$ is equivalent to $\sigma_1 \bar{\sqcap}_{\bar{\mathfrak{G}}} (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ or, using property (6.44), $\varepsilon \bar{\sqcap}_{\bar{\mathfrak{G}}} \sigma_1$ with $\{\varepsilon\} = \Theta^{\bar{\mathfrak{G}}}(\sigma_3) \cap \Theta^{\bar{\mathfrak{G}}}(\sigma_4)$. Analogously, $\mathfrak{r}^U(\sigma_2, \sigma_3, \sigma_4)$ is equivalent to $\sigma_2 \bar{\sqcap}_{\bar{\mathfrak{G}}} (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ or, using property (6.44), $\mu \bar{\sqcap}_{\bar{\mathfrak{G}}} \sigma_2$ with $\{\mu\} = \Theta^{\bar{\mathfrak{G}}}(\sigma_3) \cap \Theta^{\bar{\mathfrak{G}}}(\sigma_4)$. Using property (6.44), we also know that $\mu = \varepsilon$. We then note that $\mu = (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_3)$ and then $\sigma_1 \bar{\sqcap}_{\bar{\mathfrak{G}}} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_3)$, i.e. $\mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_3)$. □

**Lemma 6.23.** [Veblen-Young's third geometric property]

We will consider $\sigma_1, \sigma_2, \sigma_3, \sigma_4 \in \mathfrak{C}^{\mathfrak{G}}$. We have the following result

$$ \begin{align} & (\exists U \in \mathfrak{C}^{\mathfrak{G}}, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U, \exists \lambda \in U \mid \mathfrak{r}^U(\lambda, \sigma_1, \sigma_2) \text{ and } \mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)) \Rightarrow \nonumber \\ & ((\exists U', U'' \in \mathfrak{C}^{\mathfrak{G}} \mid \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U', U'') \text{ and } \exists \mu \in U' \mid \mathfrak{r}^{U'}(\mu, \sigma_1, \sigma_3) \text{ and } \mathfrak{r}^{U'}(\mu, \sigma_2, \sigma_4) \nonumber \\ & \phantom{((\exists U', U'' \in \mathfrak{C}^{\mathfrak{G}} \mid \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U', U'') \text{ and }} \text{ and } \exists \chi \in U'' \mid \mathfrak{r}^{U''}(\chi, \sigma_2, \sigma_3) \text{ and } \mathfrak{r}^{U''}(\chi, \sigma_1, \sigma_4)). \tag{6.57} \end{align} $$

*Proof.* Let us first note that if $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_2) = (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ then we have also $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_3) = (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ and also $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_4) = (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_3)$ and then the proof of (6.57) is trivial. We will then assume $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_2) \neq (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ and $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_3) \neq (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ and $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_4) \neq (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_3)$.
We will also assume $\lambda \neq \sigma_1$, $\lambda \neq \sigma_2$, $\lambda \neq \sigma_3$, $\lambda \neq \sigma_4$. Indeed, if we have for example $\lambda = \sigma_1$ then it suffices to take $\mu := \sigma_4$ and $\chi := \sigma_3$ to satisfy the announced properties.

We will then assume the existence of $\lambda \in U \in \mathfrak{C}^{\mathfrak{G}}$ such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$.
The assumption $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$ is equivalent to

$$ (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_2) \sqsubseteq_{\mathfrak{G}} \lambda, \quad (\sigma_3 \sqcap_{\mathfrak{G}} \sigma_4) \sqsubseteq_{\mathfrak{G}} \lambda. \qquad (6.58) $$

We have then, using (6.3) (if $\lambda \in (\mathfrak{G} \setminus \mathfrak{G})$) or using (6.34) (if $\lambda \in \mathfrak{G}^{\text{pure}}$), the properties

$$ (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_2 \sqcap_{\mathfrak{G}} \sigma_3 \sqcap_{\mathfrak{G}} \sigma_4) \sqsubseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_2), (\sigma_3 \sqcap_{\mathfrak{G}} \sigma_4) \qquad (6.59) $$

We know from Theorem 6.7 and Theorem 6.4 that

$$ \exists I \subseteq \{1, \dots, N\}, \text{Card}(I) = N-2 \mid \forall i \in I, \forall \alpha, \beta \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \quad \zeta_{(i)}^{\mathfrak{G}A_1 \cdots \mathfrak{G}A_N}(\alpha) = \zeta_{(i)}^{\mathfrak{G}A_1 \cdots \mathfrak{G}A_N}(\beta). \qquad (6.60) $$

Having noted the property $(\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3) \neq (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4)$, we will consider any $\xi \in \mathfrak{G}^{\text{pure}}$ such that

$$ \xi \sqsupseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3)^*, \quad \xi \sqsupsetneq_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4)^*, \quad \forall i \in I, \zeta_{(i)}^{\mathfrak{G}A_1 \cdots \mathfrak{G}A_N}(\xi) = \zeta_{(i)}^{\mathfrak{G}A_1 \cdots \mathfrak{G}A_N}(\sigma_1) \quad (6.61) $$
(we note that $\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3$ and $\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4$ are different from $\perp_{\mathfrak{G}}$).

From (6.1), and using $\xi \sqsupsetneq_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4)^*$ and property (6.60), we know that there exists $v \in \mathfrak{G}^{\mathfrak{G}}$ such that

$$ v := \xi^* \sqcup_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4) \qquad (6.62) $$

From (6.2), we know that $v \sqsupseteq_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4)$ and then $\mathfrak{r}^U(v, \sigma_2, \sigma_4)$. We intent to show that we have also $\mathfrak{r}^U(v, \sigma_1, \sigma_3)$.
We then consider any $\omega \in \mathfrak{G}^{\text{pure}}$ such that

$$ \omega \sqsupseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3)^*, \quad \omega \neq \xi, \quad \forall i \in I, \zeta_{(i)}^{\mathfrak{G}A_1 \cdots \mathfrak{G}A_N}(\omega) = \zeta_{(i)}^{\mathfrak{G}A_1 \cdots \mathfrak{G}A_N}(\xi) \quad (6.63) $$

and we intent to show that $\omega^* \sqsubseteq_{\mathfrak{G}} v$ (we note that $(\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3)^*$ is not in $\mathfrak{G}^{\text{pure}}$ and then, such an $\omega$ different from $\xi$ always exists).

We first note that $(\xi \sqcap_{\mathfrak{G}} \omega)^* \sqsupseteq_{\mathfrak{G}} \sigma_2$. Indeed, if we had $(\xi \sqcap_{\mathfrak{G}} \omega)^* \sqsubseteq_{\mathfrak{G}} \sigma_2$, and using the fact that $(\xi \sqcap_{\mathfrak{G}} \omega)^* \sqsubseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3)$, we would have $(\xi \sqcap_{\mathfrak{G}} \omega)^* \sqsubseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3 \sqcap_{\mathfrak{G}} \sigma_2)$. But now, using (6.59), we have in particular $\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3 \sqcap_{\mathfrak{G}} \sigma_2 = (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_2 \sqcap_{\mathfrak{G}} \sigma_3 \sqcap_{\mathfrak{G}} \sigma_4)$. As a consequence, we would have $(\xi \sqcap_{\mathfrak{G}} \omega)^* \sqsubseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_2 \sqcap_{\mathfrak{G}} \sigma_3 \sqcap_{\mathfrak{G}} \sigma_4) \sqsubseteq_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4)$, and then $\xi^* \sqsubseteq_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_4)$ which is false by assumption (6.61).
Then, from (6.1), and using the fact that $\sigma_2$ is in $\mathfrak{G}^{\text{pure}}$ (consequence of Lemma 6.19) and the fact that $(\xi \sqcap_{\mathfrak{G}} \omega)^* \sqsupseteq_{\mathfrak{G}} \sigma_2$ and also the property (6.63), we know that there exists $v \in \mathfrak{G}^{\mathfrak{G}}$ such that

$$ v := \sigma_2^* \sqcup_{\mathfrak{G}} (\xi \sqcap_{\mathfrak{G}} \omega). \qquad (6.64) $$

Using $(\xi \sqcap_{\mathfrak{G}} \omega) \sqsupseteq_{\mathfrak{G}} (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_3)^*$, we deduce as before

$$ v \sqsupseteq_{\mathfrak{G}} (\sigma_2 \sqcap_{\mathfrak{G}} \sigma_1 \sqcap_{\mathfrak{G}} \sigma_3)^* = (\sigma_1 \sqcap_{\mathfrak{G}} \sigma_2 \sqcap_{\mathfrak{G}} \sigma_3 \sqcap_{\mathfrak{G}} \sigma_4)^* \qquad (6.65) $$
(we note using, for example, the property (6.3) and (6.58) that $(\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_3 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ is different from $\perp_{\bar{\mathfrak{G}}}$).

We then deduce

$$ v \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)^*. \qquad (6.66) $$

From this inequality we deduce that

$$ \exists \alpha \in \Theta^{\overline{\mathfrak{G}}}(v) \mid \alpha \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)^*. \qquad (6.67) $$

Let us fix one of these $\alpha$.

We recall that, as long as $\xi \in \overline{\mathfrak{G}}^{\text{pure}}$ we have $(\xi \sqcap_{\bar{\mathfrak{G}}} v) = \text{Max}(\{\xi \sqcap_{\bar{\mathfrak{G}}} \kappa \mid \kappa \in \Theta^{\overline{\mathfrak{G}}}(v)\})$.

Moreover, we have $(\xi \sqcap_{\bar{\mathfrak{G}}} \omega) \in \Theta^{\overline{\mathfrak{G}}}(v)$. Then, using (6.2) we deduce as well

$$ \xi \sqcap_{\bar{\mathfrak{G}}} \omega = \xi \sqcap_{\bar{\mathfrak{G}}} v. \qquad (6.68) $$

As a result, we obtain for the chosen $\alpha$

$$ \xi \sqcap_{\bar{\mathfrak{G}}} \omega \sqsupseteq_{\bar{\mathfrak{G}}} \xi \sqcap_{\bar{\mathfrak{G}}} \alpha. \qquad (6.69) $$

Now, since $v \sqsupseteq_{\bar{\mathfrak{G}}} \xi^*$ and $v \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)$ and $\alpha \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \sigma_4)^*$, we deduce that

$$ v \sqsupseteq_{\bar{\mathfrak{G}}} (\xi^* \sqcup_{\bar{\mathfrak{G}}} \alpha^*). \qquad (6.70) $$

and then using (6.69)

$$ v \sqsupseteq_{\bar{\mathfrak{G}}} (\xi^* \sqcup_{\bar{\mathfrak{G}}} \omega^*) \sqsupseteq_{\bar{\mathfrak{G}}} \omega^* \qquad (6.71) $$

We have then obtained the announced intermediary result : for any $\omega \in \overline{\mathfrak{G}}^{\text{pure}}$ such that the conditions (6.63) are satisfied, we have $\omega^* \sqsubseteq_{\bar{\mathfrak{G}}} v$. We note that we have also by assumption $\xi \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_3)^*$ and $\xi^* \sqsubseteq_{\bar{\mathfrak{G}}} v$.

Hence, we can conclude

$$ v \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_3) \qquad (6.72) $$

i.e. explicitly $\tau^{U'}(v, \sigma_1, \sigma_3)$. Obviously, we have $v \asymp \sigma_1, v \asymp \sigma_2, v \asymp \sigma_3, v \asymp \sigma_4$ and then the existence of the consistent subset $U'$ does not pose any problem.

As a final conclusion, we obtain the existence of $\mu := v \in \mathfrak{G}^{\mathfrak{G}}$ such that $\tau^{U'}(\mu, \sigma_1, \sigma_3)$ and $\tau^{U'}(\mu, \sigma_2, \sigma_4)$.

By exchanging the roles of $\sigma_3$ and $\sigma_4$ in our construction, we obtain also the existence of $\chi \in \mathfrak{G}^{\mathfrak{G}}$ such that $\tau^{U''}(\chi, \sigma_1, \sigma_4)$ and $\tau^{U''}(\chi, \sigma_2, \sigma_3)$. $\square$

Let us now summarize what we have found so far.

**Theorem 6.24.** $\mathfrak{G}^{\mathfrak{G}}$ is equipped with a structure of projective sheaf. In other words, apart from the sheaf condition (6.48), we have proved the Veblen-Young's geometric axioms of projective spaces (6.49)(6.56)(6.57).

*Proof.* This result summarizes Lemma 6.15, Lemma 6.17, Lemma 6.22, Lemma 6.23. $\square$

**Definition 6.25.** For any pair $\sigma_1, \sigma_2 \in \mathfrak{G}^{\mathfrak{G}}$, we define the line $\ell_{(\sigma_1 \sigma_2)}$ to be the set of elements $\sigma_3$ in $\mathfrak{G}^{\mathfrak{G}}$ such that $\sigma_3 = \sigma_1$ or $\sigma_3 = \sigma_2$ or $\sigma_3 \sqsupseteq_{\bar{\mathfrak{G}}} (\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_2)$.
**Definition 6.26.** For any triple $\sigma_1, \sigma_2, \sigma_3 \in \mathfrak{G}^{\mathfrak{G}}$ distinct, such that $\sigma_i \notin \ell_{(\sigma_j \sigma_k)}$ for any $\{i, j, k\} = \{1, 2, 3\}$, we define the plane $\mathcal{P}_{(\sigma_1, \sigma_2, \sigma_3)}$ to be the set of elements $\sigma_4$ in $\mathfrak{G}^{\mathfrak{G}}$ such that there exists $i, j, k$ with $\{i, j, k\} = \{1, 2, 3\}$ and $\lambda \in \mathfrak{G}^{\mathfrak{G}}$ and $\lambda \in \ell_{(\sigma_i \sigma_j)}$ and $\lambda \in \ell_{(\sigma_j \sigma_k)}$.

The set of planes on $\mathfrak{G}^{\mathfrak{G}}$ will be denoted $\wp^{\mathfrak{G}}$.

Any element of $\wp^{\mathfrak{G}}$ has a structure of a projective sheaf.

**Remark 6.27.** Any plane is stable under the application of the third Veblen-Young's property. In other words, if there exists $U \in \mathfrak{G}^{\mathfrak{G}}$ and $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in \mathfrak{G}^{\mathfrak{G}}$ such that $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U$ and such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \mathfrak{r}^U(\mu, \nu, \rho)$. Then, $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4$ are in a same plane $\wp$. Moreover, if we introduce $U' \in \mathfrak{G}^{\mathfrak{G}}$ and the element $\mu \in U'$ such that $\mathfrak{r}^{U'}(\mu, \sigma_1, \sigma_3)$ and $\mathfrak{r}^{U'}(\mu, \sigma_2, \sigma_4)$, and if we introduce $U'' \in \mathfrak{G}^{\mathfrak{G}}$ and $\chi \in U''$ such that $\mathfrak{r}^{U''}(\chi, \sigma_2, \sigma_3)$ and $\mathfrak{r}^{U''}(\chi, \sigma_1, \sigma_4))$ we have necessarily $\mu, \chi \in \wp$.

Let us now consider the properties of the orthogonality relation induced on $\mathfrak{G}^{\mathfrak{G}}$ from the orthogonality relation defined on $\mathfrak{G}$ (see subsection 2.4).

Even though we have found a very interesting projective structure on $\mathfrak{G}^{\mathfrak{G}}$, we must now proceed with caution. Indeed, the problem with $\mathfrak{G}^{\mathfrak{G}}$ is due to the following simple fact (see Remark 6.34)

$$ \forall U \in \mathfrak{G}^{\mathfrak{G}}, \forall \chi \in \mathfrak{G}^{\mathfrak{G}} \cap U, \forall \sigma \in \bar{\mathfrak{G}}^{\text{pure}} \cap U, \quad (\exists \lambda \in U \mid \mathfrak{r}^U(\chi, \sigma, \lambda) \text{ and } \chi \perp \lambda) \Rightarrow (\chi \in \mathfrak{G}^{\mathfrak{G}}) \quad (6.73) $$

The following sequence of results is the consequence of our attempt to deal with this annoying problem and with the necessity to restrict ourselves to $\mathfrak{G}^{\mathfrak{G}}$.

**Definition 6.28.** A consistent subset of states $U \subseteq \mathfrak{G}^{\mathfrak{G}}$ is said to be orthogonally complete iff

$$ \forall \sigma_1, \sigma_2, \sigma_3 \in U, \text{ distinct, } \mathfrak{r}^U(\sigma_1, \sigma_2, \sigma_3) \Rightarrow (\sigma_1 \perp \sigma_2 \text{ or } \sigma_1 \perp \sigma_3 \text{ or } \sigma_2 \perp \sigma_3). \quad (6.74) $$

$$ \forall \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U \mid (\exists \lambda \in U, \mathfrak{r}^U(\lambda, \sigma_1, \sigma_2), \mathfrak{r}^U(\lambda, \sigma_3, \sigma_4) \text{ and } \forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \mathfrak{r}^U(\mu, \nu, \rho)) $$

$$ \exists \alpha, \beta, \gamma \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \alpha \perp \beta \text{ and } \alpha \perp \gamma. \quad (6.75) $$

**Definition 6.29.** For any $\sigma_1, \sigma_2, \sigma_3 \in \bar{\mathfrak{G}}^{\text{pure}}$ satisfying the following conditions

(1) there exists $I \subseteq \{1, \dots, N\}, \text{Card}(I) = N-2 \mid \forall m \in I, \zeta_{(m)}^{\mathfrak{G}_{A_1} \cdots \mathfrak{G}_{A_N}}(\sigma_1) = \zeta_{(m)}^{\mathfrak{G}_{A_1} \cdots \mathfrak{G}_{A_N}}(\sigma_2) = \zeta_{(m)}^{\mathfrak{G}_{A_1} \cdots \mathfrak{G}_{A_N}}(\sigma_3)$,

(2) if we denote $j, k \in \{1, \dots, N\} \setminus I$ with $j < k$, there exists $\alpha \in \bar{\mathfrak{G}}_{A_j}^{\text{pure}}$ and $\beta \in \bar{\mathfrak{G}}_{A_k}^{\text{pure}}$ such that $\mathfrak{r}_{\sigma_1}^{(j,k)} = \alpha \otimes \beta, \mathfrak{r}_{\sigma_2}^{(j,k)} = \alpha^* \otimes \beta, \mathfrak{r}_{\sigma_3}^{(j,k)} = \alpha \otimes \beta^*$.

we say that the plane $\wp_{(\sigma_1, \sigma_2, \sigma_3)}$ is a starred plane. The set of starred planes on $\mathfrak{G}^{\mathfrak{G}}$ will be denoted $\wp_{\mathfrak{G}}^*.$

Let us now analyze the structure of orthogonally complete consistent subsets in the following lemmas.
**Lemma 6.30.** Let us consider a consistent subset of states $U \in \mathfrak{C}^{\mathfrak{G}}$ and let us assume that there exists $\chi \in (\mathfrak{G} \setminus \overline{\mathfrak{G}})$ and that the consistent subset $U$ is of consistent type 2.

Let us now assume that $U$ is orthogonally complete. Then, we have necessarily
$\chi \in (\mathfrak{G} \setminus \overline{\mathfrak{G}})$.

Conversely, let us assume that $\chi \in (\mathfrak{G} \setminus \overline{\mathfrak{G}})$ and that $U$ is maximal, then we have

$$
U = \{\chi\} \cup \{\varphi_{\alpha} \mid \alpha \in \Theta^{\overline{\mathfrak{G}}}(\chi)\} \cup \{\psi_{\alpha} \mid \alpha \in \Theta^{\overline{\mathfrak{G}}}(\chi)\} \quad (6.76)
$$

$$
\forall \alpha \in \Theta^{\overline{\mathfrak{G}}}(\chi), \varphi_{\alpha}, \psi_{\alpha} \in \overline{\mathfrak{G}}^{\text{pure}} \text{ and } \varphi_{\alpha}, \psi_{\alpha} \bar{=} \overline{\mathfrak{G}} \alpha \text{ and } \varphi_{\alpha} \neq \psi_{\alpha} \quad (6.77)
$$

$$
\exists! \gamma \in \Theta^{\overline{\mathfrak{G}}}(\chi) \mid \varphi_{\gamma} \perp \chi, \quad \varphi_{\gamma} \perp \underline{\psi}_{\gamma}, \quad \forall \alpha \in \Theta^{\overline{\mathfrak{G}}}(\chi) \setminus \{\gamma\}, \quad \varphi_{\gamma} \perp \varphi_{\alpha}, \quad \varphi_{\gamma} \perp \psi_{\alpha}, \quad (6.78)
$$

$$
\forall \alpha \in \Theta^{\overline{\mathfrak{G}}}(\chi) \setminus \{\gamma\}, \varphi_{\alpha} \perp \psi_{\alpha}, \varphi_{\alpha} \perp \chi, \psi_{\alpha} \perp \chi, \quad (6.79)
$$

$$
\forall \alpha, \beta \in \Theta^{\overline{\mathfrak{G}}}(\chi) \setminus \{\gamma\}, \alpha \neq \beta, \varphi_{\alpha} \perp \varphi_{\beta}, \varphi_{\alpha} \perp \psi_{\beta}, \quad (6.80)
$$

$$
\forall \alpha, \beta \in \Theta^{\overline{\mathfrak{G}}}(\chi) \setminus \{\gamma\}, \alpha \neq \beta, \exists \lambda_{\alpha\beta} \in \overline{\mathfrak{G}}^{\text{pure}} \mid (\varphi_{\beta} \sqcap_{\overline{\mathfrak{G}}} \psi_{\alpha}), (\psi_{\beta} \sqcap_{\overline{\mathfrak{G}}} \varphi_{\alpha}) \bar{=} \overline{\mathfrak{G}} \lambda_{\alpha\beta} \\
\text{and } \lambda_{\alpha\beta} \perp \varphi_{\alpha}, \lambda_{\alpha\beta} \perp \varphi_{\beta}, \qquad (6.81)
$$

$$
\forall \alpha \in \Theta^{\overline{\mathfrak{G}}}(\chi) \setminus \{\gamma\}, \exists \mu_{\alpha\gamma} \in \overline{\mathfrak{G}}^{\text{pure}} \mid (\varphi_{\gamma} \sqcap_{\overline{\mathfrak{G}}} \varphi_{\alpha}), (\psi_{\gamma} \sqcap_{\overline{\mathfrak{G}}} \psi_{\alpha}) \bar{=} \overline{\mathfrak{G}} \mu_{\alpha\gamma} \\
\text{and } \mu_{\alpha\gamma} \perp \psi_{\alpha}, \qquad (6.82)
$$

and then necessarily $U$ is orthogonally complete.

*Proof.* As long as $\lambda \in (\mathfrak{G} \setminus \overline{\mathfrak{G}})$, $\Theta^{\overline{\mathfrak{G}}}(\lambda)$ is given as in Theorem 6.4.

Now we use the condition of orthogonal completeness to conclude that necessarily
($\alpha'' = \alpha, \beta'' = \beta$) or ($\alpha' = \alpha, \beta' = \beta$). This concludes the proof of the first asser-
tion, i.e. $\lambda \in (\mathfrak{G} \setminus \overline{\mathfrak{G}})$.

Then, in (6.76)–(6.82), we adopt the same notation as in (6.5) (6.6) (6.7) (6.8) for
$\varphi_\alpha$ and $\psi_\alpha$ with $\alpha$ any element of $\Theta^{\overline{\mathfrak{G}}}(\chi)$. With these formulas, the last assertion is
trivial to check.
$\square$

**Lemma 6.31.** Let us consider an orthogonally complete consistent subset such that there exists $\chi \in (\mathfrak{G} \setminus \overline{\mathfrak{G}})$ with $\chi \in U$. We adopt the same notation as in (6.5) (6.6) (6.7) (6.8) for $\gamma$ and for $\varphi_\alpha$ and $\psi_\alpha$ with $\alpha$ any element of $\Theta^{\overline{\mathfrak{G}}}(\chi)$.

Let us now assume that $U$ is of consistent type 1, i.e. $\exists! \delta \in \Theta^{\overline{\mathfrak{G}}}(\chi)$ such that $\forall \sigma \in U$
we have $(\sigma \sqcap_{\overline{\mathfrak{G}}} \chi) = \delta$. Let us fix this element $\delta$.

Then, we are in one of the two following cases

(1) $\delta \in \Theta^{\overline{\mathfrak{G}}}(\chi) \setminus \{\gamma\}$. Then, there exists $\chi' \in \mathfrak{G} \setminus \overline{\mathfrak{G}}$ such that

$$
U = \{\chi, \varphi_{\delta}, \psi_{\delta}, \chi'\} \tag{6.83}
$$

$$
\varphi_{\delta}, \psi_{\delta} \in \overline{\mathfrak{G}}^{\text{pure}} \text{ and } \varphi_{\delta}, \psi_{\delta} \bar{=} \overline{\mathfrak{G}} \delta \text{ and } \varphi_{\delta} \neq \psi_{\delta} \quad (6.84)
$$

$$
\varphi_{\delta} \perp \psi_{\delta}, \quad \varphi_{\delta} \perp \chi, \quad \psi_{\delta} \perp \chi, \quad \varphi_{\delta} \perp \chi', \quad \psi_{\delta} \perp \chi', \quad \chi \perp \chi' \tag{6.85}
$$
(2) $\delta = \gamma$. Then, there exists $\chi' \in \tilde{\mathfrak{S}} \setminus \bar{\mathfrak{S}}$ such that

$$ U = \{\chi, \phi_{\gamma}, \psi_{\gamma}, \chi'\} \qquad (6.86) $$

$$ \phi_{\gamma}, \psi_{\gamma} \in \bar{\mathfrak{S}}^{\text{pure}} \quad \text{and} \quad \phi_{\gamma}, \psi_{\gamma} \equiv_{\bar{\mathfrak{S}}} \gamma \quad \text{and} \quad \phi_{\gamma} \neq \psi_{\gamma} \qquad (6.87) $$

$$ \phi_{\gamma} \perp \psi_{\gamma}, \phi_{\gamma} \perp \chi, \psi_{\gamma} \perp \chi, \phi_{\gamma} \perp \chi', \psi_{\gamma} \perp \chi', \chi \perp \chi' \qquad (6.88) $$

*Proof.* We also adopt the same notation as in (6.5) (6.6) (6.7). Without the constraint imposed by the Lemma 6.19, and for the maximality of $U$, we can consider the existence of an element $\chi' \in \tilde{\mathfrak{S}} \setminus \bar{\mathfrak{S}}$ distinct from $\chi$ such that $\chi' \in U$. In order to guaranty that $U$ is orthogonality complete, we have to require new orthogonality conditions.

In the first configuration, i.e. $\delta \in \Theta_{\bar{\mathfrak{S}}}(\chi) \setminus \{\gamma\}$. We choose $\chi' := \delta \sqcup_{\bar{\mathfrak{S}}} \gamma^*$. Such an element does effectively exist in $\tilde{\mathfrak{S}} \setminus \bar{\mathfrak{S}}$ because of property (6.11). The announced orthogonality relations are easy to check.

In the second configuration, i.e. $\delta = \gamma$. We choose $\chi' := \psi_{\gamma}^* \sqcup_{\bar{\mathfrak{S}}} \gamma$. Such an element does effectively exist in $\tilde{\mathfrak{S}} \setminus \bar{\mathfrak{S}}$. The announced orthogonality relations are easy to check. $\square$

**Definition 6.32.** We will denote by $\mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ the following consistency cover of $\mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ (Note that we have a cover of $\mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ and not $\mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ because of Lemma 6.30) :

$$ \mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}} := \{ U \mid \exists V \in \mathfrak{e}_{\tilde{\mathfrak{S}}}^*, U \subseteq V, \text{ and } U \text{ is orthogonally complete} \}. \quad (6.89) $$

**Lemma 6.33.** Let us consider $U \in \mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ such that $U \subseteq \bar{\mathfrak{S}}^{\text{pure}}$. Let us moreover consider $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U$, all distinct, such that $\tau^U(\lambda, \sigma_1, \sigma_2)$ and $\tau^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \tau^U(\mu, \nu, \rho)$. Endly, we will suppose that the plane to which $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4$ belong is not a starred plane.

Then, there exists $\mu \in \mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}} \setminus \bar{\mathfrak{S}}$ and there exists $U' \in \mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\tau^{U'}(\mu, \sigma_1, \sigma_3)$ and $\tau^{U'}(\mu, \sigma_2, \sigma_4)$, and there exists $\nu \in \mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}} \setminus \bar{\mathfrak{S}}$ and there exists $U' \in \mathfrak{e}_{\perp}^{\tilde{\mathfrak{S}}}$ with $\nu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\tau^{U'}(\nu, \sigma_2, \sigma_3)$ and $\tau^{U'}(\nu, \sigma_1, \sigma_4)$.

*Proof.* From Theorem 6.7, we know that the conditions $\tau^U(\lambda, \sigma_1, \sigma_2)$ and $\tau^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \tau^U(\mu, \nu, \rho)$ with $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U \subseteq \bar{\mathfrak{S}}^{\text{pure}}$ imply $\exists I \subseteq \{1, \dots, N\}, \text{Card}(I) = N-2 \mid \forall i \in I, \forall \sigma, \kappa \in \{\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \zeta_{(i)}^{\bar{\mathfrak{S}}_{A_1} \cdots \bar{\mathfrak{S}}_{A_N}}(\sigma) = \zeta_{(i)}^{\bar{\mathfrak{S}}_{A_1} \cdots \bar{\mathfrak{S}}_{A_N}}(\kappa)$. Let us now denote $j$ and $k$ such that $\{j, k\} = \{1, \dots, N\} \setminus I$ and $j < k$ and let us now adopt the notations of Definition 6.3.

The condition (6.74) imposed on $U$ by the requirement of orthogonal completeness ensures that $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4$ can be written in terms of four elements $\alpha, \alpha' \in$
$\overline{\mathfrak{G}}_A^{\text{pure}}$, $\beta, \beta' \in \overline{\mathfrak{G}}_B^{\text{pure}}$ with $\alpha' \neq \alpha, \alpha^*$ and $\beta' \neq \beta, \beta^*$, in one of the following configurations

$$ (i) \quad \Upsilon_{\lambda}^{(j,k)} := \alpha^{\otimes} \beta, \quad \Upsilon_{\sigma_1}^{(j,k)} := \alpha'^{\otimes} \beta, \quad \Upsilon_{\sigma_2}^{(j,k)} := \alpha^{*\otimes} \beta, \quad \Upsilon_{\sigma_3}^{(j,k)} := \alpha^{\otimes} \beta', \quad \Upsilon_{\sigma_4} := \alpha^{\otimes} \beta^* \qquad (6.90) $$

$$ (ii) \quad \Upsilon_{\lambda}^{(j,k)} := \alpha' \tilde{\otimes} \beta, \quad \Upsilon_{\sigma_1}^{(j,k)} := \alpha \tilde{\otimes} \beta, \quad \Upsilon_{\sigma_2}^{(j,k)} := \alpha^* \tilde{\otimes} \beta, \quad \Upsilon_{\sigma_3}^{(j,k)} := \alpha' \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_4}^{(j,k)} := \alpha' \tilde{\otimes} \beta^* \qquad (6.91) $$

$$ (iii) \quad \Upsilon_{\lambda}^{(j,k)} := \alpha \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_1}^{(j,k)} := \alpha' \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_2}^{(j,k)} := \alpha^* \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_3}^{(j,k)} := \alpha \tilde{\otimes} \beta, \quad \Upsilon_{\sigma_4}^{(j,k)} := \alpha \tilde{\otimes} \beta^* \qquad (6.92) $$

$$ (iv) \quad \Upsilon_{\lambda} := \alpha' \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_1} := \alpha \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_2} := \alpha^* \tilde{\otimes} \beta', \quad \Upsilon_{\sigma_3} := \alpha' \tilde{\otimes} \beta, \quad \Upsilon_{\sigma_4} := \alpha' \tilde{\otimes} \beta^*. \qquad (6.93) $$

The condition (6.75) imposed on $U$ by the requirement of orthogonal completeness ensures that the configuration (iv) is excluded.

Endly, the condition requiring that the plane to which $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4$ belong is not a starred plane ensures that the configuration (i) is excluded.

On another part we note the following properties.

In the configuration (ii)

$$ (\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_3) \sqcup_{\overline{\mathfrak{G}}} (\sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) = (\Delta_{\alpha \tilde{\otimes} \beta}^{(j,k); \sigma_1})^* \sqcup_{\overline{\mathfrak{G}}} (\Delta_{\alpha \tilde{\otimes} \beta}^{(j,k); \sigma_1} \sqcap_{\overline{\mathfrak{G}}} \Delta_{\alpha' \tilde{\otimes} \beta'}^{(j,k); \sigma_1}) \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}} \qquad (6.94) $$

$$ (\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) \sqcup_{\overline{\mathfrak{G}}} (\sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_3) = (\Delta_{\alpha^* \tilde{\otimes} \beta}^{(j,k); \sigma_1})^* \sqcup_{\overline{\mathfrak{G}}} (\Delta_{\alpha^* \tilde{\otimes} \beta}^{(j,k); \sigma_1} \sqcap_{\overline{\mathfrak{G}}} \Delta_{\alpha' \tilde{\otimes} \beta'}^{(j,k); \sigma_1}) \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}}. \qquad (6.95) $$

In the configuration (iii)

$$ (\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_3) \sqcup_{\overline{\mathfrak{G}}} (\sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) = (\Delta_{\alpha \tilde{\otimes} \beta}^{(j,k); \sigma_1})^* \sqcup_{\overline{\mathfrak{G}}} (\Delta_{\alpha \tilde{\otimes} \beta}^{(j,k); \sigma_1} \sqcap_{\overline{\mathfrak{G}}} \Delta_{\alpha' \tilde{\otimes} \beta'}^{(j,k); \sigma_1}) \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}} \qquad (6.96) $$

$$ (\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) \sqcup_{\overline{\mathfrak{G}}} (\sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_3) = (\Delta_{\alpha \tilde{\otimes} \beta^*}^{(j,k); \sigma_1})^* \sqcup_{\overline{\mathfrak{G}}} (\Delta_{\alpha \tilde{\otimes} \beta^*}^{(j,k); \sigma_1} \sqcap_{\overline{\mathfrak{G}}} \Delta_{\alpha' \tilde{\otimes} \beta'}^{(j,k); \sigma_1}) \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}}. \qquad (6.97) $$

This concludes the proof. $\square$

**Remark 6.34.** Let us formulate a simple remark justifying the introduction of starred planes.

In the configuration (i) we have the following problem

$$ \chi := (\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_3) \sqcup_{\overline{\mathfrak{G}}} (\sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) = (\Delta_{\alpha^* \tilde{\otimes} \beta}^{(j,k); \sigma_1})^* \sqcup_{\overline{\mathfrak{G}}} (\Delta_{\alpha \tilde{\otimes} \beta}^{(j,k); \sigma_1} \sqcap_{\overline{\mathfrak{G}}} \Delta_{\alpha' \tilde{\otimes} \beta}^{(j,k); \sigma_1}) \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}}. \qquad (6.98) $$

We could still hope to have a property of the type $\forall \sigma \in \overline{\mathfrak{G}}^{\text{pure}} | \sigma \asymp \chi, \exists \lambda \in \mathfrak{G} | \chi \sqcap_{\mathfrak{G}} \sigma \sqsubseteq_{\mathfrak{G}} \lambda$ and $\chi \perp \lambda$. In other words, we could still hope to have $\forall \mu \in \Theta^{\overline{\mathfrak{G}}}(\chi), \exists \nu \in \Theta^{\overline{\mathfrak{G}}}(\chi)$ such that $\mu^* \sqcup_{\mathfrak{G}} \nu \in \mathfrak{G}$. But this point is false. Indeed, we have for

$$ \Theta^{\overline{\mathfrak{G}}}(\chi) = \{ (\alpha \tilde{\otimes} \beta') \sqcap_{\overline{\mathfrak{G}}} (\alpha' \tilde{\otimes} \beta), (\alpha \tilde{\otimes} \beta^*) \sqcap_{\overline{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta), (\alpha' \tilde{\otimes} \beta^*) \sqcap_{\overline{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta') \} \qquad (6.99) $$

the following result

$$ \begin{aligned} cl_c^{\overline{\mathfrak{G}}} & \{ ((\alpha \tilde{\otimes} \beta') \sqcap_{\overline{\mathfrak{G}}} (\alpha' \tilde{\otimes} \beta))^*, ((\alpha \tilde{\otimes} \beta^*) \sqcap_{\overline{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta)) \} = \\ &= cl_c^{\overline{\mathfrak{G}}} \{ ((\alpha^* \tilde{\otimes} \perp) \sqcap_{\overline{\mathfrak{G}}} (\perp \tilde{\otimes} \beta'^*)), ((\perp \tilde{\otimes} \beta^*) \sqcap_{\overline{\mathfrak{G}}} (\alpha'^* \tilde{\otimes} \perp)), ((\alpha \tilde{\otimes} \beta^*) \sqcap_{\overline{\mathfrak{G}}} (\alpha^* \tilde{\otimes} \beta)) \} = \\ &= cl_c^{\overline{\mathfrak{G}}} \{ ((\alpha^* \tilde{\otimes} \perp) \sqcap_{\overline{\mathfrak{G}}} (\perp \tilde{\otimes} \beta'^*)), (\alpha \tilde{\otimes} \beta^*) \} \end{aligned} \qquad (6.100) $$

but we have also easily

$$ \{ (\alpha^* \tilde{\otimes} \beta^*), (\alpha \tilde{\otimes} \beta^*) \} \sqsubseteq cl_c^{\overline{\mathfrak{G}}} \{ ((\alpha^* \tilde{\otimes} \perp) \sqcap_{\overline{\mathfrak{G}}} (\perp \tilde{\otimes} \beta'^*)), (\alpha \tilde{\otimes} \beta^*) \} \qquad (6.101) $$
and then, as a consequence,

$$cl_c^{\mathcal{G}} \left\{ \left( (\alpha \otimes \beta') \cap_{\mathcal{G}} (\alpha' \otimes \beta) \right)^*, \left( (\alpha \otimes \beta^*) \cap_{\mathcal{G}} (\alpha^* \otimes \beta) \right) \right\} \notin \mathcal{Q}_c(\bar{\mathcal{G}}). \quad (6.102)$$

**Lemma 6.35.** We will consider $\sigma_1, \sigma_2, \sigma_3, \sigma_4 \in \bar{\mathcal{G}}^{\text{pure}}$ and $\lambda \in \mathfrak{G} \setminus \bar{\mathcal{G}}$ and we will assume the existence of $U \in \mathfrak{G}^{\mathcal{S}}$ with $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U$ and such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \omega, v, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \mathfrak{r}^U(\omega, v, \rho)$. Endly, we will assume that the plane to which $\sigma_1, \sigma_2, \sigma_3, \sigma_4$ belong is not a starred plane.

Then, there exists $\mu \in \bar{\mathcal{G}}^{\text{pure}}$ and there exists $U' \in \mathfrak{G}^{\mathcal{S}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\mathfrak{r}^U(\mu, \sigma_1, \sigma_3)$ and $\mathfrak{r}^U(\mu, \sigma_2, \sigma_4)$ and there exists $\chi \in \mathfrak{G}^{\mathcal{S}} \setminus \bar{\mathcal{G}}$ and there exists $U'' \in \mathfrak{G}^{\mathcal{S}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\mathfrak{r}^{U''}(\chi, \sigma_2, \sigma_3)$ and $\mathfrak{r}^{U''}(\chi, \sigma_1, \sigma_4)$, or there exists $\mu \in \bar{\mathcal{G}}^{\text{pure}}$ and there exists $U' \in \mathfrak{G}^{\mathcal{S}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\mathfrak{r}^U(\mu, \sigma_2, \sigma_3)$ and $\mathfrak{r}^U(\mu, \sigma_1, \sigma_4)$ and there exists $\chi \in \mathfrak{G}^{\mathcal{S}} \setminus \bar{\mathcal{G}}$ and there exists $U'' \in \mathfrak{G}^{\mathcal{S}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\mathfrak{r}^{U''}(\chi, \sigma_1, \sigma_3)$ and $\mathfrak{r}^{U''}(\chi, \sigma_2, \sigma_4)$.

*Proof.* As long as $\lambda \in (\mathfrak{G} \setminus \bar{\mathcal{G}})$, we have the results of Lemma 6.30 and we can adopt the notations of this lemma.

We note that the conditions $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \omega, v, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \mathfrak{r}^U(\omega, v, \rho)$ impose the existence of $\delta, \varepsilon \in \Theta^{\bar{\mathcal{G}}}(\lambda)$ such that $\varepsilon \neq \delta$ and $(\sigma_1 \cap_{\bar{\mathcal{G}}} \sigma_2) = \varepsilon$ and $(\sigma_3 \cap_{\bar{\mathcal{G}}} \sigma_4) = \delta$.

Let us now distinguish the different cases exhibited from the condition of orthogonal completeness. We denote as usual $\gamma$ the unique element of $\Theta^{\bar{\mathcal{G}}}(\lambda)$ such that $\lambda = \varphi_{\gamma}^* \sqcup \gamma$ and $\gamma := (\varphi_{\gamma} \sqcap_{\bar{\mathcal{G}}} \psi_{\gamma}) \bar{\sqcup}_{\bar{\mathcal{G}}} \varphi_{\gamma}, \psi_{\gamma}$.

If $\sigma_2 \bar{\sqcup}_{\bar{\mathcal{G}}} (\sigma_1 \cap_{\bar{\mathcal{G}}} \sigma_3)^*$ we are then in one of the two following configurations

(i) $\gamma = \delta$ and then $\varphi_{\gamma} = \sigma_3, \psi_{\gamma} = \sigma_4$ and $\{\varphi_{\varepsilon}, \psi_{\varepsilon}\} = \{\sigma_1, \sigma_2\}$,

(ii) $\gamma \neq \delta, \gamma \neq \varepsilon$ and then $\varphi_{\delta} = \sigma_3, \psi_{\delta} = \sigma_4, \varphi_{\varepsilon} = \sigma_2, \psi_{\varepsilon} = \sigma_1$.

In the configuration (i) and if $\sigma_1 = \varphi_{\varepsilon}$, using (6.82), we deduce that there exists $\mu := \mu_{\varepsilon\gamma} \in \bar{\mathcal{G}}^{\text{pure}}$ such that $(\sigma_1 \cap_{\bar{\mathcal{G}}} \sigma_3) = (\varphi_{\gamma} \cap_{\bar{\mathcal{G}}} \varphi_{\varepsilon}) \bar{\sqcup}_{\bar{\mathcal{G}}} \mu$ and $(\sigma_2 \cap_{\bar{\mathcal{G}}} \sigma_4) = (\psi_{\gamma} \cap_{\bar{\mathcal{G}}} \psi_{\varepsilon}) \bar{\sqcup}_{\bar{\mathcal{G}}} \mu$. Moreover, we have $\mu_{\alpha\gamma} \perp \psi_{\varepsilon}$, i.e. $\mu \perp \sigma_2$, and $\varphi_{\gamma} \perp \varphi_{\varepsilon}$, i.e. $\sigma_1 \perp \sigma_3$. As a result, there exists $U' \in \mathfrak{G}^{\mathcal{S}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\mathfrak{r}^U(\mu, \sigma_1, \sigma_3)$ and $\mathfrak{r}^U(\mu, \sigma_2, \sigma_4)$. Now we can use Lemma 6.33 to deduce that there exists $U'' \in \mathfrak{G}^{\mathcal{S}}$ and $\chi \in \mathfrak{G}^{\mathcal{S}} \setminus \bar{\mathcal{G}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\mathfrak{r}^{U''}(\chi, \sigma_2, \sigma_3)$ and $\mathfrak{r}^{U''}(\chi, \sigma_1, \sigma_4)$.

In the configuration (i) and if $\sigma_1 = \psi_{\varepsilon}$, using (6.82), we deduce analogously that there exists $\mu \in \bar{\mathcal{G}}^{\text{pure}}$ and $U' \in \mathfrak{G}^{\mathcal{S}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\mathfrak{r}^U(\mu, \sigma_2, \sigma_3)$ and $\mathfrak{r}^U(\mu, \sigma_1, \sigma_4)$. Now we can use Lemma 6.33 to deduce that there exists $U'' \in \mathfrak{G}^{\mathcal{S}}$ and $\chi \in \mathfrak{G}^{\mathcal{S}} \setminus \bar{\mathcal{G}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\mathfrak{r}^{U''}(\chi, \sigma_1, \sigma_3)$ and $\mathfrak{r}^{U''}(\chi, \sigma_2, \sigma_4)$.

In the configuration (ii), using (6.81), we deduce that there exists $\mu := \lambda_{\varepsilon\delta} \in \bar{\mathcal{G}}^{\text{pure}}$ such that $(\sigma_1 \cap_{\bar{\mathcal{G}}} \sigma_3) = (\varphi_{\delta} \cap_{\bar{\mathcal{G}}} \psi_{\varepsilon}) \bar{\sqcup}_{\bar{\mathcal{G}}} \mu$ and $(\sigma_2 \cap_{\bar{\mathcal{G}}} \sigma_4) = (\psi_{\delta} \cap_{\bar{\mathcal{G}}} \varphi_{\varepsilon}) \bar{\sqcup}_{\bar{\mathcal{G}}} \mu$.
Moreover, we have $\mu = \lambda_{\varepsilon\delta}\perp\varphi_{\varepsilon} = \sigma_2$ and $\mu = \lambda_{\varepsilon\delta}\perp\varphi_{\delta} = \sigma_3$. In other words, there exists $U' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\tau^{U'}(\mu, \sigma_1, \sigma_3)$ and $\tau^{U'}(\mu, \sigma_2, \sigma_4)$. Now we can use Lemma 6.33 to deduce that there exists $U'' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ and $\chi \in \mathfrak{B}_{\tilde{\mathfrak{G}}} \setminus \overline{\mathfrak{G}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\tau^{U''}(\chi, \sigma_2, \sigma_3)$ and $\tau^{U''}(\chi, \sigma_1, \sigma_4)$.

If we have $\sigma_3 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_2 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_4)^*$ or $\sigma_1 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_2 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_4)^*$ or $\sigma_4 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_1 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_3)^*$, and using the natural symmetries of the assumption $(\sigma_1 \leftrightarrow \sigma_2, \sigma_3 \leftrightarrow \sigma_4, (\sigma_1, \sigma_2) \leftrightarrow (\sigma_3, \sigma_4))$, we obtain the same result.

If we have $\sigma_1 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_3 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_4)^*$, the situation is even simpler. Indeed, we have necessarily $\gamma = \delta$ and $\varphi_{\gamma} = \sigma_1, \psi_{\gamma} = \sigma_2, \{\varphi_{\varepsilon}, \psi_{\varepsilon}\} = \{\sigma_3, \sigma_4\}$.

If $\sigma_3 = \varphi_{\varepsilon}$, using (6.82), we deduce that there exists $\mu := \mu_{\varepsilon\gamma} \in \overline{\mathfrak{G}}^{\text{pure}}$ such that $(\sigma_1 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_3) = (\varphi_{\gamma} \sqsubseteq_{\overline{\mathfrak{G}}} \varphi_{\varepsilon}) \sqsubseteq_{\overline{\mathfrak{G}}} \mu$ and $(\sigma_2 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_4) = (\psi_{\gamma} \sqsubseteq_{\overline{\mathfrak{G}}} \psi_{\varepsilon}) \sqsubseteq_{\overline{\mathfrak{G}}} \mu$. Moreover, we have $\mu_{\alpha\gamma}\perp\psi_{\varepsilon}$, i.e. $\mu\perp\sigma_4$, and $\varphi_{\gamma}\perp\varphi_{\varepsilon}$, i.e. $\sigma_1\perp\sigma_3$. As a result, there exists $U' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\tau^{U'}(\mu, \sigma_1, \sigma_3)$ and $\tau^{U'}(\mu, \sigma_2, \sigma_4)$. Now we can use Lemma 6.33 to deduce that there exists $U'' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ and $\chi \in \mathfrak{B}_{\tilde{\mathfrak{G}}} \setminus \overline{\mathfrak{G}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\tau^{U''}(\chi, \sigma_2, \sigma_3)$ and $\tau^{U''}(\chi, \sigma_1, \sigma_4)$.

If $\sigma_4 = \varphi_{\varepsilon}$, using (6.82), we deduce analogously that there exists $\mu \in \overline{\mathfrak{G}}^{\text{pure}}$ and $U' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ with $\mu, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ such that $\tau^{U'}(\mu, \sigma_2, \sigma_3)$ and $\tau^{U'}(\mu, \sigma_1, \sigma_4)$. Now we can use Lemma 6.33 to deduce that there exists $U'' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ and $\chi \in \mathfrak{B}_{\tilde{\mathfrak{G}}} \setminus \overline{\mathfrak{G}}$ with $\chi, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ such that $\tau^{U''}(\chi, \sigma_1, \sigma_3)$ and $\tau^{U''}(\chi, \sigma_2, \sigma_4)$.

If we have $\sigma_2 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_3 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_4)^*$ or $\sigma_3 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_1 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_2)^*$ or $\sigma_4 \sqsubseteq_{\overline{\mathfrak{G}}} (\sigma_1 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_2)^*$, and using the natural symmetries of the assumption $(\sigma_1 \leftrightarrow \sigma_2, \sigma_3 \leftrightarrow \sigma_4, (\sigma_1, \sigma_2) \leftrightarrow (\sigma_3, \sigma_4))$, we obtain the same result. $\square$

Let us now summarize the main result of previous lemmas in the following theorem.

**Theorem 6.36.** [restricted Veblen-Young's third geometric property]

Let us consider $U \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ and $\lambda, \sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U$ such that $\tau^U(\lambda, \sigma_1, \sigma_2)$ and $\tau^U(\lambda, \sigma_3, \sigma_4)$ and $\forall \mu, \nu, \rho \in \{\sigma_1, \sigma_2, \sigma_3, \sigma_4\}, \neg \tau^U(\mu, \nu, \rho)$. Endly, we will assume that the plane to which $\sigma_1, \sigma_2, \sigma_3, \sigma_4$ belong is not a starred plane.

From Lemma 6.19, we know that $\sigma_1, \sigma_2, \sigma_3, \sigma_4 \in \overline{\mathfrak{G}}^{\text{pure}}$. From Lemma 6.30, we know that, as long as $U$ is orthogonally complete, we have $\lambda \in \mathfrak{B}_{\tilde{\mathfrak{G}}}$. Endly, from Lemma 6.35 and Lemma 6.33, we know that there exists $U' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$ with $\sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U'$ and $\lambda' \in U' \cap \mathfrak{B}_{\tilde{\mathfrak{G}}}$ satisfying $\tau^{U'}(\lambda', \sigma_1, \sigma_3)$ and $\tau^{U'}(\lambda', \sigma_2, \sigma_4)$, and there exists $U'' \in \mathfrak{C}_{\perp}^{\tilde{\mathfrak{G}}}$, $\sigma_1, \sigma_2, \sigma_3, \sigma_4 \in U''$ and $\lambda'' \in U'' \cap \mathfrak{B}_{\tilde{\mathfrak{G}}}$ satisfying $\tau^{U''}(\lambda'', \sigma_1, \sigma_4)$ and $\tau^{U''}(\lambda'', \sigma_2, \sigma_3)$.
**Theorem 6.37.** [Orthogonality properties]

The restriction to $\mathfrak{G}^{\tilde{\mathfrak{G}}}$ of the orthogonality relation $\perp$ satisfies the following properties.

$$ \forall \alpha, \beta \in \mathfrak{G}^{\tilde{\mathfrak{G}}}, \quad \alpha \perp \beta \Rightarrow \alpha \neq \beta, \qquad (6.103) $$

$$ \forall \alpha, \beta \in \mathfrak{G}^{\tilde{\mathfrak{G}}}, \quad \alpha \perp \beta \Rightarrow \beta \perp \alpha, \qquad (6.104) $$

$$ \forall U \in \mathfrak{G}_{\perp}^{\tilde{\mathfrak{G}}}, \forall \alpha, \beta, \varepsilon, \delta \in U, \quad (\alpha \neq \beta, \alpha \perp \varepsilon, \beta \perp \varepsilon, \mathfrak{r}^{U}(\delta, \alpha, \beta)) \Rightarrow (\varepsilon \perp \delta), \qquad (6.105) $$

$$ \forall \alpha, \beta \in \mathfrak{G}_{\perp}^{\tilde{\mathfrak{G}}}, \alpha \neq \beta, \alpha \asymp \beta, \quad \exists U \in \mathfrak{G}_{\perp}^{\tilde{\mathfrak{G}}}, \alpha, \beta \in U, \exists \varepsilon \in U \mid \mathfrak{r}^{U}(\varepsilon, \alpha, \beta) \text{ and } \varepsilon \perp \alpha. \qquad (6.106) $$

*Proof.* The properties (6.103) and (6.104) are trivial.

Let us now check the property (6.105). As long as $\alpha \perp \varepsilon$ and $\beta \perp \varepsilon$, we have $\varepsilon^* \sqsubseteq_{\mathfrak{G}} (\alpha \sqcap_{\mathfrak{G}} \beta)$, but we have also $(\alpha \sqcap_{\mathfrak{G}} \beta) \overline{\sqsubseteq}_{\mathfrak{G}} \delta$ because $\mathfrak{r}^U(\delta, \alpha, \beta)$. We then have $\varepsilon^* \sqsubseteq_{\mathfrak{G}} \delta$, which means $\varepsilon \perp \delta$.

Let us now check the property (6.106). We have to distinguish the different cases.

Let us begin with $\alpha, \beta \in \overline{\mathfrak{G}}^{\text{pure}}$ with $\alpha \neq \beta$ and $\alpha \asymp \beta$. From Lemma 6.4 and $\alpha \asymp \beta$, we know that $\varepsilon := \alpha^* \sqcup_{\mathfrak{G}} (\alpha \sqcap_{\mathfrak{G}} \beta)$ exists in $\mathfrak{G}^{\tilde{\mathfrak{G}}}$. Moreover, using (6.2), we have $(\alpha \sqcap_{\mathfrak{G}} \beta) \overline{\sqsubseteq}_{\mathfrak{G}} \varepsilon$, i.e. $\varepsilon \perp \alpha$.

Secondly, let us consider the case where $\alpha \in \overline{\mathfrak{G}}^{\text{pure}}$ and $\beta \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}^{\text{pure}}}$. The same analysis with the same expression for $\varepsilon$ gives the same result.

Thirdly, let us consider the case where $\beta \in \overline{\mathfrak{G}}^{\text{pure}}$ and $\alpha \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}^{\text{pure}}}$. By definition, there exists $\gamma \in \Theta \overline{\mathfrak{G}}(\alpha)$ and $\varphi_{\gamma}, \psi_{\gamma} \in \overline{\mathfrak{G}}^{\text{pure}}$ such that $\gamma = (\varphi_{\gamma} \sqcap_{\mathfrak{G}} \psi_{\gamma}) \overline{\sqsubseteq}_{\mathfrak{G}} \varphi_{\gamma}, \psi_{\gamma}$ and $\alpha = \varphi_{\gamma}^* \sqcup_{\mathfrak{G}} \gamma$. Let us denote by $\delta := \beta \sqcap_{\mathfrak{G}} \alpha \in \Theta \overline{\mathfrak{G}}(\alpha)$. We have to distinguish between two sub-cases :

(i) $\delta = \gamma$. In this case, it suffice to choose $\varepsilon := \varphi_{\gamma}$ and we check immediately $(\alpha \sqcap_{\mathfrak{G}} \beta) \overline{\sqsubseteq}_{\mathfrak{G}} \varepsilon$ and $\varepsilon^* \sqsubseteq_{\mathfrak{G}} \alpha$.

(ii) $\delta \neq \gamma$. In this case, it suffice to choose $\varepsilon := \delta \sqcup_{\mathfrak{G}} \gamma^*$ which is an element of $\mathfrak{G}^{\tilde{\mathfrak{G}}}$ because of property (6.11). We check immediately $(\alpha \sqcap_{\mathfrak{G}} \beta) \overline{\sqsubseteq}_{\mathfrak{G}} \varepsilon$ using property (6.2). We check also trivially $\gamma^* \sqsubseteq_{\mathfrak{G}} \varepsilon$ and $\gamma \sqsubseteq_{\mathfrak{G}} \alpha$, which means $\varepsilon \perp \alpha$.

Fourthly, let us consider the case where $\beta, \alpha \in \mathfrak{G}^{\tilde{\mathfrak{G}} \setminus \overline{\mathfrak{G}}^{\text{pure}}}$. The same construction as above leads to the result. $\square$

The last element to check is the fact that the tensor product respects the property of linearity defined in Definition 2.56.

**Theorem 6.38.** [Irreducibility]

The space of states $\mathfrak{G}$ equipped with its real structure $(\overline{\mathfrak{G}}, \star)$ satisfies the following irreducibility property

$$ \forall \sigma, \lambda \in \mathfrak{G}^{\tilde{\mathfrak{G}}}, \sigma \neq \lambda, \sigma \asymp \lambda, \exists U \in \mathfrak{G}_{\perp}^{\tilde{\mathfrak{G}}}, \sigma, \lambda \in U, \exists \kappa \in U \mid (\kappa \neq \sigma \text{ and } \kappa \neq \lambda \text{ and } \mathfrak{r}^{U}(\kappa, \lambda, \sigma)). \qquad (6.107) $$

*Proof.* First of all, we note that, if $\sigma \perp \lambda$ and $\lambda \neq \sigma$ and $\sigma \asymp \lambda$, using property (6.106), we deduce the existence of $U \in \mathfrak{G}^{\tilde{\mathfrak{G}}}$ such that $\sigma, \lambda \in U$ and the existence
of $\kappa \in U$ such that $\tau^U(\kappa, \lambda, \sigma)$ and $\sigma \perp \kappa$. As long as $\sigma \perp \kappa$ we deduce $\sigma \neq \kappa$, and as long as $\sigma \perp \lambda$ we deduce $\kappa \neq \lambda$. This proves the property (6.107) when $\sigma \perp \lambda$.

It remains to study the case $\sigma \perp \lambda$, $\sigma \asymp \lambda$. Let us distinguish the different cases.

(1) Let us first study the case $\sigma, \lambda \in \bar{\mathcal{G}}^{\text{pure}}$ with $\sigma \asymp \lambda$ and $\sigma \perp \lambda$. We will introduce $I \subseteq \{1, \dots, N\}$ such that $\text{Card}(I) = N-2$ and $\forall i \in I, \zeta_{(i)}^{\bar{\mathcal{G}}_{A_1} \cdots \bar{\mathcal{G}}_{A_N}}(\sigma) = \zeta_{(i)}^{\bar{\mathcal{G}}_{A_1} \cdots \bar{\mathcal{G}}_{A_N}}(\lambda)$. We will also denote $j$ and $k$ such that $\{j,k\} = \{1, \dots, N\} \setminus I$ and $j < k$. Then, we are in one of the two following sub-cases :

(1A) there exists $\alpha \in \bar{\mathcal{G}}_{A_j}^{\text{pure}}$ and $\beta, \beta' \in \bar{\mathcal{G}}_{A_k}^{\text{pure}}$ such that $\Upsilon_{\sigma}^{(j,k)} = \alpha \tilde{\otimes} \beta$ and $\Upsilon_{\lambda}^{(j,k)} = \alpha' \tilde{\otimes} \beta'$. If $\beta = \beta'$ then for any $\alpha' \in \bar{\mathcal{G}}_{A_j}^{\text{pure}}$ such that $\alpha' \neq \alpha, \alpha'$, we can define $\kappa := \alpha' \tilde{\otimes} \beta$ which satisfies the announced constraints. If $\beta \neq \beta'$, then, for any $\alpha' \in \bar{\mathcal{G}}_{A_j}^{\text{pure}}$ such that $\alpha' \neq \alpha, \alpha'$ we can define $\kappa := (\Delta_{\alpha \tilde{\otimes} \beta^*}^{(j,k);\sigma})^* \sqcup_{\mathcal{G}} (\Delta_{\alpha \tilde{\otimes} \beta^*}^{(j,k);\sigma} \sqcap_{\bar{\mathcal{G}}} \Delta_{\alpha' \tilde{\otimes} \beta'}^{(j,k);\sigma})$ which satisfies indeed $\kappa \neq \sigma$, $\kappa \neq \lambda$ and $(\sigma \sqcap_{\bar{\mathcal{G}}} \lambda) \in \Theta \bar{\mathcal{G}}(\kappa)$.

(1B) there exists $\alpha, \alpha' \in \bar{\mathcal{G}}_{A_j}^{\text{pure}}$ and $\beta \in \bar{\mathcal{G}}_{A_k}^{\text{pure}}$ such that $\Upsilon_{\sigma}^{(j,k)} = \alpha \tilde{\otimes} \beta$ and $\Upsilon_{\lambda}^{(j,k)} = \alpha' \tilde{\otimes} \beta^*$. Same treatment.

(2) Let us now study the case $\sigma \in \bar{\mathcal{G}}^{\text{pure}}$, $\lambda \in \mathfrak{B}^{\tilde{\mathcal{G}} \setminus \mathfrak{B}^{\text{pure}}}$ with $\sigma \asymp \lambda$ and $\sigma \perp \lambda$. We deduce that $\lambda = \varphi_{\gamma}^* \sqcup_{\mathcal{G}} (\varphi_{\gamma} \sqcap_{\bar{\mathcal{G}}} \psi_{\gamma})$ and $\sigma = \varphi_{\gamma}$ (we have adopted the same notation as in (6.5) (6.6) (6.7) (6.8)). It then suffices to choose $\kappa := \psi_{\gamma}$ to obtain $\kappa \neq \sigma$, $\kappa \neq \lambda$ and $(\sigma \sqcap_{\bar{\mathcal{G}}} \lambda) \in \Theta \bar{\mathcal{G}}(\kappa)$.

(3) Let us now study the case $\sigma, \lambda \in \mathfrak{B}^{\tilde{\mathcal{G}} \setminus \mathfrak{B}^{\text{pure}}}$ with $\sigma \asymp \lambda$ and $\sigma \perp \lambda$. In this case, it suffices to choose $\kappa := \psi_{\alpha}$ for $\{\alpha\} := \Theta \bar{\mathcal{G}}(\sigma) \cap \Theta \bar{\mathcal{G}}(\lambda)$.

This concludes the proof.

□

The result of this section appears to be outstanding in light of the traditional approach to quantum logic. We have reconstructed a structure on the tensor product that is as close as possible to irreducible Hilbert geometry, even though we did not impose anything of the sort *ab initio*. This is an absolutely non-trivial consequence of our construction of the tensor product and of the ontic completions.

## 7. Fundamental properties of the bipartite indeterministic experiments

This last section is devoted to the study of typical quantum-like properties: the absence of broadcasting and the existence of Bell non-local states. Recall that the core criticism of the quantum logic program [38][26][39] is the impossibility of constructing entangled states exhibiting Bell non-locality within the framework of a logical approach to quantum mechanics. The outcome of this section is then one of the main achievements of our approach.
**7.1. No-broadcasting theorem**

**Definition 7.1. [Broadcasting]**

We will say that a given space of states $\mathfrak{S}$ allows the broadcasting iff

$$ \exists \Psi \in \overline{\mathfrak{C}}(\mathfrak{S}, \mathfrak{S} \bar{\otimes} \mathfrak{S}) \mid \forall \sigma \in \overline{\mathfrak{S}}, (\zeta_{(1)}^{\mathfrak{S}\mathfrak{S}} \circ \Psi)(\sigma) = (\zeta_{(2)}^{\mathfrak{S}\mathfrak{S}} \circ \Psi)(\sigma) = \sigma. \quad (7.1) $$

**Lemma 7.2.** Let $m_1, m_2 \in \overline{\mathfrak{M}}_{\mathfrak{S}}$ be two real measurements on the space of states $\mathfrak{S}$ (they are real morphisms from $\mathfrak{S}$ to $\mathfrak{B}$).

If there exists a real morphism $\Psi \in \overline{\mathfrak{C}}(\mathfrak{S}, \mathfrak{S} \bar{\otimes} \mathfrak{S})$ such that $\forall \sigma \in \overline{\mathfrak{S}}, (\zeta_{(1)}^{\mathfrak{S}\mathfrak{S}} \circ \Psi)(\sigma) = (\zeta_{(2)}^{\mathfrak{S}\mathfrak{S}} \circ \Psi)(\sigma) = \sigma$, then there exists a real morphism $\Psi_{(m_1, m_2)} \in \overline{\mathfrak{C}}(\mathfrak{S}, \mathfrak{B} \bar{\otimes} \mathfrak{B})$ such that $\forall \sigma \in \overline{\mathfrak{S}}, (\zeta_{(1)}^{\mathfrak{B}\mathfrak{B}} \circ \Psi_{(m_1, m_2)})(\sigma) = m_1(\sigma)$ and $(\zeta_{(2)}^{\mathfrak{B}\mathfrak{B}} \circ \Psi_{(m_1, m_2)})(\sigma) = m_2(\sigma)$.

*Proof.* It suffices to choose $\Psi_{(m_1, m_2)} := (m_1 \bar{\otimes} m_2) \circ \Psi$. $\square$

<figure><img src="image_12.png" /></figure>

**Theorem 7.3. [No-broadcasting theorem]**

If the space of states $\mathfrak{S}$ (equipped with its real structure $(\overline{\mathfrak{S}}, \star)$) is not deterministic, it does not allow the broadcasting.

*Proof.* Let us consider the space of states $\mathfrak{S}$, and let us assume that it is NOT deterministic. We know, from the definition of indeterminism, that there exists $\sigma_1, \sigma_2 \in \overline{\mathfrak{S}}^{\text{pure}}$ such that $\sigma_2 \neq \sigma_1$ and $\sigma_2 \not\perp_{\overline{\mathfrak{S}}} \sigma_1^*$. Let us then define the following real measurements : $m_1 := m_{(\sigma_1, \sigma_1^*)}$ and $m_2 := m_{(\sigma_2, \sigma_2^*)}$. Let us now suppose that there exists a real morphism $\Psi_{(m_1, m_2)} \in \overline{\mathfrak{C}}(\mathfrak{S}, \mathfrak{B} \bar{\otimes} \mathfrak{B})$ such that $\zeta_{(1)}^{\mathfrak{B}\mathfrak{B}} \circ \Psi_{(m_1, m_2)} = m_1$ and $\zeta_{(2)}^{\mathfrak{B}\mathfrak{B}} \circ \Psi_{(m_1, m_2)} = m_2$, and let us exhibit a contradiction.

From $m_1(\sigma_1) = \mathbf{Y}$ and $m_2(\sigma_1) = \perp$, we deduce that $\Psi_{(m_1, m_2)}(\sigma_1) = \mathbf{Y} \bar{\otimes} \perp$.

From $m_1(\sigma_1^*) = \mathbf{N}$ and $m_2(\sigma_1^*) = \perp$, we deduce that $\Psi_{(m_1, m_2)}(\sigma_1^*) = \mathbf{N} \bar{\otimes} \perp$.

From $m_1(\sigma_2^*) = \perp$ and $m_2(\sigma_2^*) = \mathbf{N}$, we deduce that $\Psi_{(m_1, m_2)}(\sigma_2^*) = \perp \bar{\otimes} \mathbf{N}$.

From $\Psi_{(m_1, m_2)}(\sigma_1) = \mathbf{Y} \bar{\otimes} \perp$ and $\Psi_{(m_1, m_2)}(\sigma_1^*) = \mathbf{N} \bar{\otimes} \perp$ and the homomorphic properties satisfied by $\Psi_{(m_1, m_2)}$ we deduce that $\Psi_{(m_1, m_2)}(\perp_{\mathfrak{S}}) = \Psi_{(m_1, m_2)}(\sigma_1 \sqcap_{\mathfrak{S}} \sigma_1^*) = \mathbf{Y} \bar{\otimes} \perp \sqcap_{\mathfrak{B} \bar{\otimes} 2} \mathbf{N} \bar{\otimes} \perp = \perp \bar{\otimes} \perp$.

On another part, from $\Psi_{(m_1, m_2)}(\sigma_2^*) = \perp \bar{\otimes} \mathbf{N}$ and $\Psi_{(m_1, m_2)}(\sigma_1^*) = \mathbf{N} \bar{\otimes} \perp$ and the homomorphic properties satisfied by $\Psi_{(m_1, m_2)}$ we deduce that $\Psi_{(m_1, m_2)}(\perp_{\mathfrak{S}}) = \Psi_{(m_1, m_2)}(\sigma_1^* \sqcap_{\mathfrak{S}} \sigma_2^*) = \perp \bar{\otimes} \mathbf{N} \sqcap_{\mathfrak{B} \bar{\otimes} 2} \mathbf{N} \bar{\otimes} \perp$.

We have then obtained a contradiction.

As a conclusion, and using Lemma 7.2, we deduce that there cannot exist a real morphism $\Psi \in \overline{\mathfrak{C}}(\mathfrak{S}, \mathfrak{S} \bar{\otimes} \mathfrak{S})$ such that $\forall \sigma \in \overline{\mathfrak{S}}, (\zeta_{(1)}^{\mathfrak{S}\mathfrak{S}} \circ \Psi)(\sigma) = (\zeta_{(2)}^{\mathfrak{S}\mathfrak{S}} \circ \Psi)(\sigma) = \sigma$. Hence, there is no broadcasting. $\square$

<figure><img src="image_13.png" /></figure>

**7.2. Bell non-locality**

During this subsection, we will consider the two copies of one-dimensional indeterministic spaces of states $\overline{\mathfrak{G}}_A = \mathfrak{Z}'_{N_A}$ and $\overline{\mathfrak{G}}_B = \mathfrak{Z}'_{N_B}$ (with $N_A, N_B \ge 2$) and we consider the space of states $\overline{S}_{AB} := \overline{\mathfrak{G}}_A \bar{\otimes} \overline{\mathfrak{G}}_B$. We denote $\mathfrak{G}_A = \mathfrak{J}^{\mathfrak{c}}_{\overline{\mathfrak{G}}_A}$ and $\mathfrak{G}_B = \mathfrak{J}^{\mathfrak{c}}_{\overline{\mathfrak{G}}_B}$. The space of states (called two-dimensional indeterministic space of states) has been build, according to the subsection 4.7, as the ontic completion $\tilde{S}_{AB} := \mathfrak{G}_A \bar{\otimes} \mathfrak{G}_B = \mathfrak{J}^{\mathfrak{c}}_{\overline{\mathfrak{G}}_A \bar{\otimes} \overline{\mathfrak{G}}_B}$. If
necessary, in order to shorten our formulas, we will eventually replace the notation $\tilde{S}_{AB}$ by $\tilde{\mathfrak{S}}$ and the notation $\tilde{S}_{AB}$ by $\tilde{\mathfrak{S}}$.

First of all, let $\Sigma$ be an element of $\mathfrak{S}_A \hat{\otimes} \mathfrak{S}_B$. Let $\phi_1 \in \bar{\mathfrak{C}}(\mathfrak{S}_A, \mathfrak{B})$ et $\phi_2 \in \bar{\mathfrak{C}}(\mathfrak{S}_A, \mathfrak{B})$ be a pair of compatible real measurements, and $\rho_1 \in \bar{\mathfrak{C}}(\mathfrak{S}_B, \mathfrak{B})$ et $\rho_2 \in \bar{\mathfrak{C}}(\mathfrak{S}_B, \mathfrak{B})$ another pair of compatible real measurements. Then, there exists an element $\Lambda \in \mathfrak{B}^{\tilde{\otimes}4}$ such that

$$ \zeta_{(1)(3)}^{\tilde{\otimes}4}(\Lambda) = (\phi_1 \tilde{\otimes} \rho_1)(\Sigma) \qquad (7.2) $$

$$ \zeta_{(1)(4)}^{\tilde{\otimes}4}(\Lambda) = (\phi_1 \tilde{\otimes} \rho_2)(\Sigma) \qquad (7.3) $$

$$ \zeta_{(2)(3)}^{\tilde{\otimes}4}(\Lambda) = (\phi_2 \tilde{\otimes} \rho_1)(\Sigma) \qquad (7.4) $$

$$ \zeta_{(2)(4)}^{\tilde{\otimes}4}(\Lambda) = (\phi_2 \tilde{\otimes} \rho_2)(\Sigma) \qquad (7.5) $$

Indeed, it suffices to define

$$ \Lambda := (\Psi_{(\phi_1, \phi_2)} \tilde{\otimes} \Psi_{(\rho_1, \rho_2)})(\Sigma), $$

where $\Psi_{(\phi_1, \phi_2)} \in \bar{\mathfrak{C}}(\mathfrak{S}_A, \mathfrak{B} \tilde{\otimes} \mathfrak{B})$ is the joint channel associated to $\phi_1$ and $\phi_2$, and $\Psi_{(\rho_1, \rho_2)} \in \bar{\mathfrak{C}}(\mathfrak{S}_D, \mathfrak{B} \tilde{\otimes} \mathfrak{B})$ is the joint channel associated to $\rho_1$ and $\rho_2$.

Secondly, if $\Sigma$ is a real state (i.e. $\Sigma \in \tilde{S}_{AB} = \tilde{\mathfrak{S}}_A \tilde{\otimes} \tilde{\mathfrak{S}}_B$), then there exists an element $\Lambda \in \mathfrak{B}^{\tilde{\otimes}4}$ such that the four relations (7.2) (7.3) (7.4) (7.5) are simultaneously satisfied. Indeed, if $\Sigma = \bigcap_{i \in I}^{\mathfrak{S}_A \tilde{\otimes} \mathfrak{S}_B} \sigma_{i,A} \tilde{\otimes} \sigma_{i,B}$ with $\{(\sigma_{i,A}, \sigma_{i,B}) \mid i \in I\} \subseteq \bar{\mathfrak{E}}_A^{\text{pure}} \times \bar{\mathfrak{E}}_B^{\text{pure}}$, it suffices to define

$$ \Lambda = \bigcap_{i \in I}^{\mathfrak{B}^{\tilde{\otimes}4}} \phi_1(\sigma_{i,A}) \tilde{\otimes} \phi_2(\sigma_{i,A}) \tilde{\otimes} \rho_1(\sigma_{i,B}) \tilde{\otimes} \rho_2(\sigma_{i,B}). \qquad (7.6) $$

On the contrary, we can now introduce the notion of Bell non-locality in a perspective that is adapted to our purpose. Note that it has been shown by M. Plávala [37] that the standard notion of Bell non-locality given as in the CHSH scenario [16] is a particular case of the general framework on Bell non-locality adapted to Generalized Probabilistic Theories (GPT) (see the introduction by M. Plávala [36]).

**Lemma 7.4.** If there exists two pairs of real measurements $\phi_1 \in \bar{\mathfrak{C}}(\mathfrak{S}_A, \mathfrak{B})$ and $\phi_2 \in \bar{\mathfrak{C}}(\mathfrak{S}_A, \mathfrak{B})$, $\rho_1 \in \bar{\mathfrak{C}}(\mathfrak{S}_D, \mathfrak{B})$ and $\rho_2 \in \bar{\mathfrak{C}}(\mathfrak{S}_D, \mathfrak{B})$ such that there exists no state $\Lambda \in \mathfrak{B}^{\tilde{\otimes}4}$ satisfying simultaneously the four relations (7.2) (7.3) (7.4) (7.5), then the state $\Sigma$ is a Bell non-local state.

*Proof.* see [37]. □

We have the following fundamental result.

**Theorem 7.5.** Bell non-local states do exist in $\tilde{S}_{AB}$.

*Proof.* We will consider $\sigma_1, \sigma_2 \in \bar{\mathfrak{E}}_A^{\text{pure}}$ and $\tau_1, \tau_2 \in \bar{\mathfrak{E}}_B^{\text{pure}}$ such that $\sigma_1 \neq \sigma_2, \sigma_2^*$ and $\tau_1 \neq \tau_2, \tau_2^*$. Let us now introduce a particular hidden-state.
$$ \Sigma := (\sigma_1 \tilde{\otimes} \tau_1 \sqcap_{\bar{S}_{AB}} \sigma_2 \tilde{\otimes} \tau_2) \sqcup_{\bar{S}_{AB}} (\sigma_1^* \tilde{\otimes} \perp_{\bar{S}_B} \sqcap_{\bar{S}_{AB}} \perp_{\bar{S}_A} \tilde{\otimes} \tau_1^*). \quad (7.7) $$

This state do exist in the ontic completion $\hat{S}_{AB} := \mathfrak{J}_{\bar{S}_{AB}}^\zeta$ of the real space $\bar{S}_{AB} := \bar{\mathfrak{S}}_A \tilde{\otimes} \bar{\mathfrak{S}}_B$. More precisely, we have $\Sigma \in \hat{S}_{AB} \setminus \bar{S}_{AB}$.

Indeed, as shown in subsection 6.1, we have

$$ \begin{aligned} c l_{\bar{S}_{AB}}^{\bar{\mathfrak{S}}_{AB}} & \left( \left\{ (\sigma_1 \tilde{\otimes} \tau_1 \sqcap_{\bar{S}_{AB}} \sigma_2 \tilde{\otimes} \tau_2), (\sigma_1^* \tilde{\otimes} \perp_{\bar{S}_B} \sqcap_{\bar{S}_{AB}} \perp_{\bar{S}_A} \tilde{\otimes} \tau_1^* ) \right\} \right) = \\ & = \left\{ (\sigma_1 \tilde{\otimes} \tau_1 \sqcap_{\bar{S}_{AB}} \sigma_2 \tilde{\otimes} \tau_2), (\sigma_1 \tilde{\otimes} \tau_1^* \sqcap_{\bar{S}_{AB}} \sigma_1^* \tilde{\otimes} \tau_2), (\sigma_2 \tilde{\otimes} \tau_1^* \sqcap_{\bar{S}_{AB}} \sigma_1^* \tilde{\otimes} \tau_1) \right\} \in \mathcal{H}(\bar{S}_{AB}) \end{aligned} \quad (7.8) $$

Let us now consider the following real measurement operators

$$ \phi_1 := m_{l(\sigma_1, \sigma_1^*)}, \quad \phi_2 := m_{l(\sigma_2, \sigma_2^*)}, \quad (7.9) $$

$$ \rho_1 := m_{l(\tau_1, \tau_1^*)}, \quad \rho_2 := m_{l(\tau_2, \tau_2^*)}. \quad (7.10) $$

We now intent to prove that $\Sigma$ is a Bell non-local state with respect to the pairs of incompatible real measurements $(\phi_1, \phi_2)$ and $(\rho_1, \rho_2)$. In the following computation we use obviously the defining property (4.73).

The equality

$$ \Phi_{13} := (\phi_1 \tilde{\otimes} \rho_1)(\Sigma) = \mathbf{N} \tilde{\otimes} \perp \sqcap_{\mathfrak{B} \tilde{\otimes} 2} \mathbf{Y} \tilde{\otimes} \mathbf{N}. \quad (7.11) $$

and the property $\zeta_{(1)(3)}^{\mathfrak{B}\mathfrak{B}\mathfrak{B}}(\Lambda) = \Phi_{13}$ imply

$$ \Lambda = \sqcap_i^{\mathfrak{B}\tilde{\otimes} 4} \mathbf{N} \tilde{\otimes} \alpha_i \tilde{\otimes} \mathbf{N} \tilde{\otimes} \beta_i \sqcap_{\mathfrak{B}\tilde{\otimes} 4} \sqcap_j^{\mathfrak{B}\tilde{\otimes} 4} \mathbf{N} \tilde{\otimes} \gamma_j \tilde{\otimes} \mathbf{Y} \tilde{\otimes} \delta_j \sqcap_{\mathfrak{B}\tilde{\otimes} 4} \sqcap_k^{\mathfrak{B}\tilde{\otimes} 4} \mathbf{Y} \tilde{\otimes} \varepsilon_k \tilde{\otimes} \mathbf{N} \tilde{\otimes} \varphi_k \quad (7.12) $$

for certain collections $(\alpha_i, \beta_i)_i$, $(\gamma_j, \delta_j)_j$, $(\varepsilon_k, \varphi_k)_k$ of pairs of elements of $\{\mathbf{Y}, \mathbf{N}\}$.

The equality

$$ \Phi_{14} := (\phi_1 \tilde{\otimes} \rho_2)(\Sigma) = \mathbf{Y} \tilde{\otimes} \perp \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \mathbf{N} \tilde{\otimes} \mathbf{Y}. \quad (7.13) $$

and the property $\zeta_{(1)(4)}^{\mathfrak{B}\mathfrak{B}\mathfrak{B}}(\Lambda) = \Phi_{14}$ imply

$$ \sqcap_i^{\mathfrak{B}\tilde{\otimes} 2} \mathbf{N} \tilde{\otimes} \beta_i \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \sqcap_j^{\mathfrak{B}\tilde{\otimes} 2} \mathbf{N} \tilde{\otimes} \delta_j \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \sqcap_k^{\mathfrak{B}\tilde{\otimes} 2} \mathbf{Y} \tilde{\otimes} \varphi_k = \mathbf{Y} \tilde{\otimes} \perp \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \mathbf{N} \tilde{\otimes} \mathbf{Y}, \quad (7.14) $$

and then

$$ \bigwedge_i \beta_i \wedge \bigwedge_j \delta_j = \mathbf{Y}, \quad \bigwedge_k \varphi_k = \perp. \quad (7.15) $$

The equality

$$ \Phi_{23} := (\phi_2 \tilde{\otimes} \rho_1)(\Sigma) = \mathbf{Y} \tilde{\otimes} \mathbf{N} \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \perp \tilde{\otimes} \mathbf{Y}. \quad (7.16) $$

and $\zeta_{(2)(3)}^{\mathfrak{B}\mathfrak{B}\mathfrak{B}}(\Lambda) = \Phi_{23}$ imply

$$ \sqcap_i^{\mathfrak{B}\tilde{\otimes} 2} \alpha_i \tilde{\otimes} \mathbf{N} \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \sqcap_j^{\mathfrak{B}\tilde{\otimes} 2} \gamma_j \tilde{\otimes} \mathbf{Y} \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \sqcap_k^{\mathfrak{B}\tilde{\otimes} 2} \varepsilon_k \tilde{\otimes} \mathbf{N} = \mathbf{Y} \tilde{\otimes} \mathbf{N} \sqcap_{\mathfrak{B}\tilde{\otimes} 2} \perp \tilde{\otimes} \mathbf{Y}, \quad (7.17) $$

and then

$$ \bigwedge_i \alpha_i \wedge \bigwedge_k \varepsilon_k = \mathbf{Y}, \quad \bigwedge_j \gamma_j = \perp. \quad (7.18) $$
If we summarize what we have obtained, we have

$$ \Lambda = \mathbf{N} \tilde{\otimes} \mathbf{Y} \tilde{\otimes} \mathbf{N} \tilde{\otimes} \mathbf{Y} \sqcap_{\mathfrak{B} \tilde{\otimes} 4} \mathbf{N} \tilde{\otimes} \perp \tilde{\otimes} \mathbf{Y} \tilde{\otimes} \mathbf{Y} \sqcap_{\mathfrak{B} \tilde{\otimes} 4} \mathbf{Y} \tilde{\otimes} \mathbf{Y} \tilde{\otimes} \mathbf{N} \tilde{\otimes} \perp \quad (7.19) $$

Now we use the equality

$$ \Phi_{24} := (\phi_2 \tilde{\otimes} \rho_2)(\Sigma) = \perp \tilde{\otimes} \perp. \quad (7.20) $$

and $\zeta_{(2)(4)}^{\mathfrak{B}\mathfrak{B}\mathfrak{B}}(\Lambda) = \Phi_{24}$ to obtain the announced contradiction. Indeed, we have

$$ \mathbf{Y} \tilde{\otimes} \perp \sqcap_{\mathfrak{B} \tilde{\otimes} 2} \perp \tilde{\otimes} \mathbf{Y} \neq \perp \tilde{\otimes} \perp. \quad (7.21) $$

As a conclusion, the state $\Sigma$ is Bell non-local and the pairs $(\phi_1, \phi_2)$ et $(\rho_1, \rho_2)$ are pairs of incompatible real measurements. $\square$

## 8. Conclusion

My ongoing research program [11][12][14] seeks to devise a method for overcoming the previously mentioned no-go theorems [38][26][39], which concern the impossibility of constructing a tensor product suitable for describing composite systems starting from the logical formulation of single systems. The approach pursued relies on a formalism closely related to that originally developed by G. Birkhoff and J. von Neumann, but give a central role to the existence of a tensor product to describe experiments on compound systems.

In Section 2, after presenting the formalism of States/Effects Chu spaces together with their morphisms, we introduced the notion of real structure, distinguishing between real states and hidden states. We then specified how determinism and indeterminism are defined in this setting. Sections 3 and 5 refined the definition of real structure by showing that generalized state spaces including hidden states arise (i) through a natural completion process applied to real indeterministic state spaces, and (ii) in connection with contextual empirical models derived from the operational description of such real spaces. Section 4 established the conditions tensor products of state spaces must satisfy to allow the construction of compound systems from their components, and highlighted that the simplest solution also accounts for composite deterministic (classical) systems. Section 6 extended the discussion with a proposal for representing compound indeterministic systems, making use of the ontic completions developed in Section 3. This proposal remained general, and Section 7 offered a more precise study of the iterated tensor product of basic indeterministic systems. The resulting structure proved to approximate as closely as possible that of irreducible Hilbert geometries although this type of structure had not been imposed ab initio in our construction. Finally, Section 8 demonstrated that our framework explicitly reproduces key quantum-like phenomena, such as the no-broadcasting theorem and the existence of Bell non-local states in our framework. Our primary objective has therefore been fully achieved, which is a result of paramount importance in quantum logic.
# 9. Appendix

Let us first consider $\sigma_1, \sigma_2, \sigma_3, \sigma_4$ all distinct elements in $(\check{\mathfrak{S}} \setminus \overline{\mathfrak{S}})$.

Using properties (6.44) and $\sigma_1 \asymp \sigma_2, \sigma_2 \asymp \sigma_3, \sigma_3 \asymp \sigma_4, \sigma_1 \asymp \sigma_4$ and (6.50), we note that we can always choose $\pi_i, \rho_i \in \overline{\mathfrak{S}}$ for $i = 1, 2, 3, 4$ such that

$$ \forall i = 1, 2, 3, 4, \qquad \pi_i, \rho_i \in \Theta^{\overline{\mathfrak{G}}}(\sigma_i), \quad \sigma_i = \pi_i \llcorner_{\overline{\mathfrak{G}}} \rho_i \qquad (9.1) $$

$$ \pi_1 = \pi_2, \quad \pi_3 = \pi_4, \quad \rho_2 = \rho_3, \quad \rho_1 = \rho_4 \qquad (9.2) $$

Let us now assume that there exists $\lambda \in \mathfrak{G}^{\check{\mathfrak{S}}}$ such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$.
In other words,

$$ \lambda \overline{\mathfrak{G}} \pi_2, \quad \lambda \overline{\mathfrak{G}} \pi_3. \qquad (9.3) $$

The consistency relation $\sigma_2 \asymp \sigma_4$ means that there exists $\varepsilon_{24} \in \Theta^{\overline{\mathfrak{G}}}(\sigma_2) \cap \Theta^{\overline{\mathfrak{G}}}(\sigma_4)$ and $\varepsilon_{24} = \sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_4$ with $\sigma_2 := \pi_2 \sqcup_{\overline{\mathfrak{G}}} \rho_2$ and $\sigma_4 := \pi_3 \sqcup_{\overline{\mathfrak{G}}} \rho_1$. We will consider that $\varepsilon_{24} \neq \pi_2, \varepsilon_{24} \neq \rho_2, \varepsilon_{24} \neq \pi_3$ and $\varepsilon_{24} \neq \rho_1$. From $\varepsilon_{24} = \sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_4$ and $\sigma_2 \overline{\sqcap}_{\overline{\mathfrak{G}}} \pi_2$ and $\sigma_4 \overline{\sqcap}_{\overline{\mathfrak{G}}} \pi_3$, we deduce

$$ \varepsilon_{24} \overline{\sqcap}_{\overline{\mathfrak{G}}} (\pi_2 \sqcap_{\overline{\mathfrak{G}}} \pi_3). \qquad (9.4) $$

Moreover, from (6.3) (if $\lambda \in (\check{\mathfrak{S}} \setminus \overline{\mathfrak{S}})$) or from (6.34) (if $\lambda \in \overline{\mathfrak{S}}^{\text{pure}}$), we deduce

$$ (\pi_2 \sqcap_{\overline{\mathfrak{G}}} \pi_3) \overline{\sqcap}_{\overline{\mathfrak{G}}} \pi_2, \pi_3 \qquad (9.5) $$

and then, using $\varepsilon_{24} \neq \pi_2$ and $\varepsilon_{24} \neq \pi_3$

$$ (\varepsilon_{24} \sqcap_{\overline{\mathfrak{G}}} \pi_2) = (\varepsilon_{24} \sqcap_{\overline{\mathfrak{G}}} \pi_3) = (\pi_2 \sqcap_{\overline{\mathfrak{G}}} \pi_3) \qquad (9.6) $$

From $\pi_2 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_{24} = \pi_3 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_{24}$, we deduce, using (6.3), that $\varepsilon_{24} = (\pi_2 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_{24}) \sqcup_{\overline{\mathfrak{G}}} (\rho_2 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_{24}) = (\pi_3 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_{24}) \sqcup_{\overline{\mathfrak{G}}} (\rho_2 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_{24})$. As a consequence, $\varepsilon_{24}$ is an element of $\Theta^{\overline{\mathfrak{G}}}(\sigma_2) \cap \Theta^{\overline{\mathfrak{G}}}(\sigma_3)$. However, we have already $\rho_2 \in \Theta^{\overline{\mathfrak{G}}}(\sigma_2) \cap \Theta^{\overline{\mathfrak{G}}}(\sigma_3)$ and $\varepsilon_{24} \neq \rho_2$. Using (6.44) we then deduce that $\sigma_2 = \sigma_3$. This case has been excluded by assumption.

As a conclusion, this configuration is excluded.

Let us now consider that $\sigma_1, \sigma_2, \sigma_3 \in (\check{\mathfrak{S}} \setminus \overline{\mathfrak{S}})$ and $\sigma_4 \in \overline{\mathfrak{S}}^{\text{pure}}$. Using properties (6.44) and $\sigma_1 \asymp \sigma_2, \sigma_2 \asymp \sigma_3, \sigma_3 \asymp \sigma_4, \sigma_1 \asymp \sigma_4$ and (6.50)(6.3), we note that we can always choose $\pi_i, \rho_i \in \overline{\mathfrak{S}}$ for $i = 1, 2, 3$ such that

$$ \forall i = 1, 2, 3 \qquad \pi_i, \rho_i \in \Theta^{\overline{\mathfrak{G}}}(\sigma_i), \quad \sigma_i = \pi_i \llcorner_{\overline{\mathfrak{G}}} \rho_i \qquad (9.7) $$

$$ \pi_1 = \pi_2, \quad \rho_2 = \rho_3, \quad \sigma_4 \overline{\sqcap}_{\overline{\mathfrak{G}}} \pi_3, \quad \sigma_4 \overline{\sqcap}_{\overline{\mathfrak{G}}} \rho_1. \qquad (9.8) $$

Let us now assume that there exists $\lambda \in \mathfrak{G}^{\check{\mathfrak{S}}}$ such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$.
We have then

$$ \lambda \overline{\sqcap}_{\overline{\mathfrak{G}}} \pi_2, \quad \lambda \overline{\sqcap}_{\overline{\mathfrak{G}}} \pi_3. \qquad (9.9) $$

The consistency relation $\sigma_2 \asymp \sigma_4$ means that there exists $\varepsilon_2 \in \Theta^{\overline{\mathfrak{G}}}(\sigma_2)$ such that $\varepsilon_2 \overline{\sqcap}_{\overline{\mathfrak{G}}} \sigma_4$. Analogously, the consistency relation $\sigma_1 \asymp \sigma_3$ means that there exists $\varepsilon_1 \in \Theta^{\overline{\mathfrak{G}}}(\sigma_1) \cap \Theta^{\overline{\mathfrak{G}}}(\sigma_3)$ such that $\varepsilon_1 = \sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_3$.
Let us now choose $\chi := \sigma_1$ and $\chi := \sigma_2$ and $\chi := \sigma_3$ with $\kappa := \sigma_4$ in (3.26). We obtain

$$ \rho_1 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \pi_2), (\sigma_4 \sqcap_{\mathbb{G}} \varepsilon_1) \qquad (9.10) $$

$$ \varepsilon_2 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \pi_2), (\sigma_4 \sqcap_{\mathbb{G}} \rho_2) \qquad (9.11) $$

$$ \pi_3 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \rho_2), (\sigma_4 \sqcap_{\mathbb{G}} \varepsilon_1). \qquad (9.12) $$

From $\rho_1 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \pi_2)$ and $\sigma_4 \underset{\mathbb{G}}{\overline{\sqcap}} \rho_1$ we deduce $(\rho_1 \sqcap_{\mathbb{G}} \pi_2) = (\sigma_4 \sqcap_{\mathbb{G}} \pi_2)$. From $\varepsilon_2 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \pi_2)$ and $\varepsilon_2 \underset{\mathbb{G}}{\overline{\sqcap}} \sigma_4$, we deduce $(\varepsilon_2 \sqcap_{\mathbb{G}} \pi_2) = (\sigma_4 \sqcap_{\mathbb{G}} \pi_2)$.

From $\rho_1 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \varepsilon_1)$ and $\sigma_4 \underset{\mathbb{G}}{\overline{\sqcap}} \rho_1$ we deduce $(\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\sigma_4 \sqcap_{\mathbb{G}} \varepsilon_1)$. From $\pi_3 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \varepsilon_1)$ and $\pi_3 \underset{\mathbb{G}}{\overline{\sqcap}} \sigma_4$, we deduce $(\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1) = (\sigma_4 \sqcap_{\mathbb{G}} \varepsilon_1)$.

From $\varepsilon_2 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \rho_2)$ and $\sigma_4 \underset{\mathbb{G}}{\overline{\sqcap}} \varepsilon_2$ we deduce $(\rho_2 \sqcap_{\mathbb{G}} \varepsilon_2) = (\sigma_4 \sqcap_{\mathbb{G}} \rho_2)$. From $\pi_3 \underset{\mathbb{G}}{\overline{\sqcap}} (\sigma_4 \sqcap_{\mathbb{G}} \rho_2)$ and $\pi_3 \underset{\mathbb{G}}{\overline{\sqcap}} \sigma_4$, we deduce $(\pi_3 \sqcap_{\mathbb{G}} \rho_2) = (\sigma_4 \sqcap_{\mathbb{G}} \rho_2)$.

Let us summarize

$$ (\rho_1 \sqcap_{\mathbb{G}} \pi_2) = (\varepsilon_2 \sqcap_{\mathbb{G}} \pi_2) \qquad (9.13) $$

$$ (\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1) \qquad (9.14) $$

$$ (\rho_2 \sqcap_{\mathbb{G}} \varepsilon_2) = (\pi_3 \sqcap_{\mathbb{G}} \rho_2). \qquad (9.15) $$

Using $(\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) \underset{\mathbb{G}}{\overline{\sqcap}} \rho_1$ inherited from (6.3), and $(\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1)$, we deduce

$$ (\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1) = (\pi_3 \sqcap_{\mathbb{G}} \rho_1) \qquad (9.16) $$

From $\varepsilon_1 = \sigma_1 \sqcap_{\mathbb{G}} \sigma_3$ we know that

$$ \varepsilon_1 \underset{\mathbb{G}}{\overline{\sqcap}} (\pi_2 \sqcap_{\mathbb{G}} \pi_3) \qquad (9.17) $$

On another part, from $\lambda = \pi_2 \sqcap_{\mathbb{G}} \pi_3$ and using (6.3) (if $\lambda \in (\tilde{\mathfrak{G}} \setminus \tilde{\mathfrak{G}})$) or using (6.34) (if $\lambda \in \tilde{\mathfrak{G}}^{\text{pure}}$), we deduce

$$ (\pi_2 \sqcap_{\mathbb{G}} \pi_3) \underset{\mathbb{G}}{\overline{\sqcap}} \pi_2, \pi_3. \qquad (9.18) $$

and then $(\varepsilon_1 \sqcap_{\mathbb{G}} \pi_3) = (\pi_2 \sqcap_{\mathbb{G}} \pi_3)$. Let us summarize

$$ (\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\pi_3 \sqcap_{\mathbb{G}} \rho_1) = (\varepsilon_1 \sqcap_{\mathbb{G}} \pi_3) = (\pi_2 \sqcap_{\mathbb{G}} \pi_3) \underset{\mathbb{G}}{\overline{\sqcap}} \pi_2, \pi_3. \qquad (9.19) $$

Now, from $(\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) \underset{\mathbb{G}}{\overline{\sqcap}} \pi_2$ we have immediately

$$ (\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\rho_1 \sqcap_{\mathbb{G}} \pi_2). \qquad (9.20) $$

Now, we can summarize our results. From $(\rho_1 \sqcap_{\mathbb{G}} \pi_2) = (\varepsilon_2 \sqcap_{\mathbb{G}} \pi_2)$, $(\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1)$ and $(\rho_1 \sqcap_{\mathbb{G}} \varepsilon_1) = (\rho_1 \sqcap_{\mathbb{G}} \pi_2)$, we obtain $(\varepsilon_2 \sqcap_{\mathbb{G}} \pi_2) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1)$. Hence, we have obtained

$$ (\varepsilon_2 \sqcap_{\mathbb{G}} \pi_2) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1), \quad (\varepsilon_2 \sqcap_{\mathbb{G}} \rho_2) = (\pi_3 \sqcap_{\mathbb{G}} \rho_2) \qquad (9.21) $$

and then, in particular,

$$ (\varepsilon_2 \sqcap_{\mathbb{G}} \pi_2) \sqcup_{\mathbb{G}} (\varepsilon_2 \sqcap_{\mathbb{G}} \rho_2) = (\pi_3 \sqcap_{\mathbb{G}} \varepsilon_1) \sqcup_{\mathbb{G}} (\pi_3 \sqcap_{\mathbb{G}} \rho_2). \qquad (9.22) $$
On another part, we recall that, using properties (6.50) (6.3) and $\varepsilon_2 \in \Theta^{\overline{\mathfrak{G}}}(\pi_2 \sqcap_{\overline{\mathfrak{G}}}) \rho_2) = \Theta^{\overline{\mathfrak{G}}}(\sigma_2)$ and $\pi_3 \in \Theta^{\overline{\mathfrak{G}}}(\varepsilon_1 \sqcap_{\overline{\mathfrak{G}}} \rho_2) = \Theta^{\overline{\mathfrak{G}}}(\sigma_3)$, we have

$$ \varepsilon_2 = (\varepsilon_2 \sqcap_{\overline{\mathfrak{G}}} \pi_2) \sqcup_{\overline{\mathfrak{G}}} (\varepsilon_2 \sqcap_{\overline{\mathfrak{G}}} \rho_2), \quad \pi_3 = (\pi_3 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_1) \sqcup_{\overline{\mathfrak{G}}} (\pi_3 \sqcap_{\overline{\mathfrak{G}}} \rho_2) \quad (9.23) $$

We then conclude that

$$ (\sigma_2 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) = \varepsilon_2 = \pi_3 = (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \sigma_4) \quad (9.24) $$

but this degenerate case has been excluded by assumption. As a conclusion, this configuration is excluded.

Let us now consider that $\sigma_1, \sigma_2 \in (\check{\mathfrak{G}} \setminus \overline{\mathfrak{G}})$ and $\sigma_3, \sigma_4 \in \overline{\mathfrak{G}}^{\text{pure}}$. Using properties (6.44) and $\sigma_1 \asymp \sigma_2, \sigma_2 \asymp \sigma_3, \sigma_3 \asymp \sigma_4, \sigma_1 \asymp \sigma_4$ and (6.50)(6.3), we note that we can always choose $\pi_i, \rho_i \in \overline{\mathfrak{G}}$ for $i = 1, 2$ such that

$$ \forall i = 1, 2 \qquad \pi_i, \rho_i \in \Theta^{\overline{\mathfrak{G}}}(\sigma_i), \quad \sigma_i = \pi_i \sqcup_{\overline{\mathfrak{G}}} \rho_i \quad (9.25) $$

$$ \pi_1 = \pi_2, \quad \sigma_3 \sqsupseteq_{\overline{\mathfrak{G}}} \rho_2, \quad \sigma_4 \sqsupseteq_{\overline{\mathfrak{G}}} \rho_1. \quad (9.26) $$

Let us now assume that there exists $\lambda \in \mathfrak{G}^{\check{\mathfrak{G}}}$ such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$. We have then

$$ \lambda \sqsupseteq_{\overline{\mathfrak{G}}} \pi_2, \quad \lambda \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \sigma_4). \quad (9.27) $$

The consistency relation $\sigma_2 \asymp \sigma_4$ means that there exists $\varepsilon_2 \in \Theta^{\overline{\mathfrak{G}}}(\sigma_2)$ such that $\varepsilon_2 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_4$. Analogously, the consistency relation $\sigma_1 \asymp \sigma_3$ means that there exists $\varepsilon_1 \in \Theta^{\overline{\mathfrak{G}}}(\sigma_1)$ such that $\varepsilon_1 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_3$.

Once again, we insist on the fact that $\varepsilon_1 \neq \pi_1$ (resp. $\varepsilon_1 \neq \rho_1$) because this would mean $\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_3 = \varepsilon_1 = \pi_1 = \sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_2$ (resp. $\sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_3 = \varepsilon_1 = \rho_1 = \sigma_1 \sqcap_{\overline{\mathfrak{G}}} \sigma_4$). These degenerate cases have been excluded by assumption. We will also assume obviously $\varepsilon_2 \neq \pi_2$ and $\varepsilon_2 \neq \rho_2$.

Let us now choose $\chi := \sigma_2$ and $\chi := \sigma_1$ with $\kappa := \sigma_3$ in (3.26), and simultaneously $\chi := \sigma_1$ and $\chi := \sigma_2$ with $\kappa := \sigma_4$ in (3.26). We obtain

$$ \rho_2 \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2), (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_2) \quad (9.28) $$

$$ \varepsilon_1 \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2), (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \rho_1) \quad (9.29) $$

$$ \rho_1 \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2), (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \varepsilon_1) \quad (9.30) $$

$$ \varepsilon_2 \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2), (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \rho_2) \quad (9.31) $$

From $\varepsilon_1 \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2)$ and $\varepsilon_1 \sqsubseteq_{\overline{\mathfrak{G}}} \sigma_3$, we deduce $(\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2) = (\varepsilon_1 \sqcap_{\overline{\mathfrak{G}}} \pi_2)$.

From $\rho_1 \sqsupseteq_{\overline{\mathfrak{G}}} (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2)$ and $\sigma_4 \sqsupseteq_{\overline{\mathfrak{G}}} \rho_1$, we deduce $(\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2) = (\rho_1 \sqcap_{\overline{\mathfrak{G}}} \pi_2)$.

On another part, from (6.3) (if $\lambda \in (\check{\mathfrak{G}} \setminus \overline{\mathfrak{G}})$) or from (6.34) (if $\lambda \in (\overline{\mathfrak{G}}^{\text{pure}})$, we deduce

$$ (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2) \sqcap_{\overline{\mathfrak{G}}} (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2) \sqsupseteq_{\overline{\mathfrak{G}}} \pi_2. \quad (9.32) $$

As long as $\sigma_2 \neq \sigma_3$, we have $\sigma_3 \sqsupseteq_{\overline{\mathfrak{G}}} \pi_2$. Analogously, we have $\sigma_4 \sqsupseteq_{\overline{\mathfrak{G}}} \pi_2$. Hence,

$$ (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2) = (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2) = (\sigma_3 \sqcap_{\overline{\mathfrak{G}}} \pi_2) \sqcap_{\overline{\mathfrak{G}}} (\sigma_4 \sqcap_{\overline{\mathfrak{G}}} \pi_2) \sqsupseteq_{\overline{\mathfrak{G}}} \pi_2. \quad (9.33) $$
As a result, we obtain

$$ (\varepsilon_1 \sqcap_{\bar{\mathfrak{G}}} \pi_2) = (\rho_1 \sqcap_{\bar{\mathfrak{G}}} \pi_2). \qquad (9.34) $$

Hence, we have obtained a contradiction with the property (6.4).
As a conclusion, this configuration is excluded.

Let us now consider that $\sigma_1, \sigma_4 \in (\mathfrak{G} \setminus \bar{\mathfrak{G}})$ and $\sigma_2, \sigma_3 \in \bar{\mathfrak{G}}^{\text{pure}}$. Using properties (6.44) and $\sigma_1 \asymp \sigma_2, \sigma_2 \asymp \sigma_3, \sigma_3 \asymp \sigma_4, \sigma_1 \asymp \sigma_4$ and (6.50)(6.3), we note that we can always choose $\pi_i, \rho_i \in \bar{\mathfrak{G}}$ for $i = 1, 2$ such that

$$ \forall i = 1, 4 \qquad \pi_i, \rho_i \in \Theta^{\bar{\mathfrak{G}}}(\sigma_i), \quad \sigma_i = \pi_i \sqcup_{\bar{\mathfrak{G}}} \rho_i \qquad (9.35) $$

$$ \rho_1 = \rho_4, \quad \sigma_2 \bar{=} \pi_1, \quad \sigma_3 \bar{=} \pi_4. \qquad (9.36) $$

Let us now assume that there exists $\lambda \in \mathfrak{G}^{\mathfrak{G}}$ such that $\mathfrak{r}^U(\lambda, \sigma_1, \sigma_2)$ and $\mathfrak{r}^U(\lambda, \sigma_3, \sigma_4)$.
We have then

$$ \lambda \bar{=} \pi_1 \qquad \lambda \bar{=} \pi_4. \qquad (9.37) $$

The consistency relation $\sigma_2 \asymp \sigma_4$ means that there exists $\varepsilon_4 \in \Theta^{\bar{\mathfrak{G}}}(\sigma_4)$ such that $\varepsilon_4 \bar{=} \sigma_2$. Analogously, the consistency relation $\sigma_1 \asymp \sigma_3$ means that there exists $\varepsilon_1 \in \Theta^{\bar{\mathfrak{G}}}(\sigma_1)$ such that $\varepsilon_1 \bar{=} \sigma_3$.

Once again, we insist on the fact that $\varepsilon_1 \neq \pi_1$ (resp. $\varepsilon_1 \neq \rho_1$) because this would mean $\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_3 = \varepsilon_1 = \pi_1 = \sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_2$ (resp. $\sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_3 = \varepsilon_1 = \rho_1 = \sigma_1 \sqcap_{\bar{\mathfrak{G}}} \sigma_4$). These degenerate cases have been excluded by assumption. We will also assume obviously $\varepsilon_4 \neq \pi_4$ and $\varepsilon_4 \neq \rho_1$.

Let us now choose $\chi := \sigma_1$ and $\chi := \sigma_4$ with $\kappa := \sigma_2$ in (3.26), and simultaneously $\chi := \sigma_1$ and $\chi := \sigma_4$ with $\kappa := \sigma_3$ in (3.26). We obtain

$$ \pi_1 \bar{=} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \rho_1), (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \varepsilon_1) \qquad (9.38) $$

$$ \varepsilon_4 \bar{=} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \pi_4), (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \qquad (9.39) $$

$$ \varepsilon_1 \bar{=} (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \pi_1), (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \qquad (9.40) $$

$$ \pi_4 \bar{=} (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \rho_1), (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \varepsilon_4). \qquad (9.41) $$

From $\pi_1 \bar{=} (\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \rho_1)$ and $\pi_1 \bar{=} \sigma_2$, we deduce $(\sigma_2 \sqcap_{\bar{\mathfrak{G}}} \rho_1) = (\pi_1 \sqcap_{\bar{\mathfrak{G}}} \rho_1)$.

From $\pi_4 \bar{=} (\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \rho_1)$ and $\pi_4 \bar{=} \sigma_3$, we deduce $(\sigma_3 \sqcap_{\bar{\mathfrak{G}}} \rho_1) = (\pi_4 \sqcap_{\bar{\mathfrak{G}}} \rho_1)$.

We have then obtained

$$ \varepsilon_4 \bar{=} (\pi_1 \sqcap_{\bar{\mathfrak{G}}} \rho_1), \quad \varepsilon_1 \bar{=} (\pi_4 \sqcap_{\bar{\mathfrak{G}}} \rho_1). \qquad (9.42) $$

From

$$ (\pi_1 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \bar{=} \pi_1, \rho_1, \quad (\pi_4 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \bar{=} \pi_4, \rho_1 \qquad (9.43) $$

we deduce

$$ (\rho_1 \sqcap_{\bar{\mathfrak{G}}} \varepsilon_4) = (\pi_1 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \bar{=} \rho_1, \quad (\rho_1 \sqcap_{\bar{\mathfrak{G}}} \varepsilon_1) = (\pi_4 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \bar{=} \rho_1. \qquad (9.44) $$

On another part, from $\lambda \bar{=} \pi_1, \pi_4$, we know that $\lambda \bar{=} \rho_1$ because, using (6.50), we would deduce that $\lambda = (\rho_1 \sqcup_{\bar{\mathfrak{G}}} \pi_1) = \sigma_1$ and $\lambda = (\rho_1 \sqcup_{\bar{\mathfrak{G}}} \pi_4) = \sigma_4$ which has been excluded. Then, we deduce, using $(\pi_1 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \bar{=} \rho_1$ and $(\pi_4 \sqcap_{\bar{\mathfrak{G}}} \rho_1) \bar{=} \rho_1$

$$ (\pi_1 \sqcap_{\bar{\mathfrak{G}}} \rho_1) = (\pi_4 \sqcap_{\bar{\mathfrak{G}}} \rho_1) = (\lambda \sqcap_{\bar{\mathfrak{G}}} \rho_1) \qquad (9.45) $$
As a result, we obtain

$$ (\rho_1 \sqcap_{\mathfrak{G}} \varepsilon_4) = (\pi_4 \sqcap_{\mathfrak{G}} \rho_1), \quad (\rho_1 \sqcap_{\mathfrak{G}} \varepsilon_1) = (\pi_1 \sqcap_{\mathfrak{G}} \rho_1). \quad (9.46) $$

Hence, we have obtained a contradiction with the property (6.4).

As a conclusion, this configuration is excluded.

Let us now consider that $\sigma_1, \sigma_3 \in (\mathfrak{G} \setminus \mathfrak{G})$ and $\sigma_2, \sigma_4 \in \mathfrak{G}^{pure}$. Using properties (6.44) and $\sigma_1 \asymp \sigma_2, \sigma_2 \asymp \sigma_3, \sigma_3 \asymp \sigma_4, \sigma_1 \asymp \sigma_4$ and (6.50), we note that we can always choose $\pi_i, \rho_i \in \mathfrak{G}$ for $i = 1, 2$ such that

$$ \forall i = 1, 3 \quad \pi_i, \rho_i \in \Theta_{\mathfrak{G}}(\sigma_i), \quad \sigma_i = \pi_i \sqcap_{\mathfrak{G}} \rho_i \quad (9.47) $$

$$ \sigma_4 \sqcap_{\mathfrak{G}} \rho_1, \quad \sigma_4 \sqcap_{\mathfrak{G}} \pi_3 \quad (9.48) $$

$$ \sigma_2 \sqcap_{\mathfrak{G}} \rho_3, \quad \sigma_2 \sqcap_{\mathfrak{G}} \pi_1. \quad (9.49) $$

Let us now assume that there exists $\lambda \in \mathfrak{G}^{\mathfrak{G}}$ such that $\tau^U(\lambda, \sigma_1, \sigma_2)$ and $\tau^U(\lambda, \sigma_3, \sigma_4)$. We will exclude the trivial case $\pi_1 = \sigma_1 \sqcap_{\mathfrak{G}} \sigma_2 = \sigma_3 \sqcap_{\mathfrak{G}} \sigma_4 = \pi_3$, i.e. we will assume $\pi_1 \neq \pi_3$. We will also exclude the trivial cases $\pi_1 = \sigma_1 \sqcap_{\mathfrak{G}} \sigma_2 = \sigma_2 \sqcap_{\mathfrak{G}} \sigma_3 = \rho_3$ or $\pi_3 = \sigma_3 \sqcap_{\mathfrak{G}} \sigma_4 = \sigma_1 \sqcap_{\mathfrak{G}} \sigma_4 = \rho_1$, i.e. we will then assume $\pi_1 \neq \rho_3$ and $\pi_3 \neq \rho_1$. We have then

$$ \lambda \sqcap_{\mathfrak{G}} \pi_1, \pi_3. \quad (9.50) $$

The consistency relation $\sigma_2 \asymp \sigma_4$ contains nothing. On another part, the consistency relation $\sigma_1 \asymp \sigma_3$ means that there exists $\varepsilon \in \Theta_{\mathfrak{G}}(\sigma_1) \cap \Theta_{\mathfrak{G}}(\sigma_3)$.

Let us now choose $\chi := \sigma_1$ and $\chi := \sigma_3$ with $\kappa := \sigma_2$ in (3.26), and simultaneously $\chi := \sigma_3$ and $\chi := \sigma_1$ with $\kappa := \sigma_4$ in (3.26).

$$ (\sigma_2 \sqcap_{\mathfrak{G}} \rho_1), (\sigma_2 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \pi_1 \quad (9.51) $$

$$ (\sigma_2 \sqcap_{\mathfrak{G}} \pi_3), (\sigma_2 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \rho_3 \quad (9.52) $$

$$ (\sigma_4 \sqcap_{\mathfrak{G}} \rho_3), (\sigma_4 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \pi_3 \quad (9.53) $$

$$ (\sigma_4 \sqcap_{\mathfrak{G}} \pi_1), (\sigma_4 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \rho_1 \quad (9.54) $$

From $(\sigma_2 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \pi_1, (\sigma_2 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \rho_3$ and $\sigma_2 \sqcap_{\mathfrak{G}} \rho_3$ and $\sigma_2 \sqcap_{\mathfrak{G}} \pi_1$ (and also $(\sigma_4 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \pi_3, (\sigma_4 \sqcap_{\mathfrak{G}} \varepsilon) \sqsubseteq_{\mathfrak{G}} \rho_1$ and $\sigma_4 \sqcap_{\mathfrak{G}} \rho_1$ and $\sigma_4 \sqcap_{\mathfrak{G}} \pi_3)$ we deduce

$$ (\sigma_2 \sqcap_{\mathfrak{G}} \varepsilon) = (\pi_1 \sqcap_{\mathfrak{G}} \varepsilon) = (\rho_3 \sqcap_{\mathfrak{G}} \varepsilon) \quad (9.55) $$

$$ (\sigma_4 \sqcap_{\mathfrak{G}} \varepsilon) = (\pi_3 \sqcap_{\mathfrak{G}} \varepsilon) = (\rho_1 \sqcap_{\mathfrak{G}} \varepsilon). \quad (9.56) $$

On another part, from $\lambda \sqcap_{\mathfrak{G}} \pi_1, \pi_3$, we know that $\lambda \sqsupseteq_{\mathfrak{G}} \varepsilon$ because, using (6.50), we would deduce that $\sigma_1 = (\varepsilon \sqcup_{\mathfrak{G}} \pi_1) = (\varepsilon \sqcup_{\mathfrak{G}} \pi_3) = \sigma_3$ which has been excluded. Then, we deduce, using $(\pi_1 \sqcap_{\mathfrak{G}} \varepsilon) \sqsupseteq_{\mathfrak{G}} \varepsilon$ and $(\pi_3 \sqcap_{\mathfrak{G}} \varepsilon) \sqsupseteq_{\mathfrak{G}} \varepsilon$

$$ (\pi_1 \sqcap_{\mathfrak{G}} \varepsilon) = (\pi_3 \sqcap_{\mathfrak{G}} \varepsilon) = (\lambda \sqcap_{\mathfrak{G}} \varepsilon) \quad (9.57) $$

We obtain finally

$$ (\pi_1 \sqcap_{\mathfrak{G}} \varepsilon) = (\rho_1 \sqcap_{\mathfrak{G}} \varepsilon), \quad (\pi_3 \sqcap_{\mathfrak{G}} \varepsilon) = (\rho_3 \sqcap_{\mathfrak{G}} \varepsilon) \quad (9.58) $$

which contradicts (6.4).

As a conclusion, this configuration is excluded.