For each litter \( L \), there is an injection

by a cardinality argument. Define the relation \( R \) on atoms by the constructor

This is one-to-one and has disjoint image and coimage from \( \xi ^{\mathcal A}\).

We now show that if \( (a_1, a_2) \in R \) and \( (N_1, N_2) \in \xi ^{\mathcal N}\), then \( a_1 \in N_1 \) if and only if \( a_2 \in N_2 \). Let \( (N_1, N_2), (N_1', N_2') \in \xi ^{\mathcal N}\), and let \( a \in N_1 \setminus \operatorname {\mathsf{LS}}(N_1^\circ ) \setminus \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\). Suppose that \( a \in N_1' \); we must show \( i_{N_2^\circ }(a) \in N_2' \). If \( N_1^\circ \neq {N_1’}^\circ \), then \( \operatorname {\mathsf{interf}}(N_1, N_1') = N_1 \cap N_1' \), so \( a \in \operatorname {\mathsf{interf}}(N_1, N_1') \subseteq \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\), a contradiction. So \( N_1^\circ = {N_1’}^\circ \), giving \( N_2^\circ = {N_2’}^\circ \), so \( i_{N_2^\circ }(a) = i_{{N_2’}^\circ }(a) \in N_2' \) by definition.

Conversely, suppose that \( i_{N_2^\circ }(a) \in N_2' \); we must show \( a \in N_1' \). Note that by definition, \( i_{N_2^\circ }(a) \in N_2 \). So if \( N_2^\circ \neq {N_2’}^\circ \), we would have \( i_{N_2^\circ }(a) \in N_2 \cap N_2' = \operatorname {\mathsf{interf}}(N_2, N_2') \subseteq \operatorname {\mathsf{im}}\xi ^{\mathcal A}\), a contradiction. Hence \( N_2^\circ = {N_2’}^\circ \) and \( N_1^\circ = {N_1’}^\circ \). Thus, if \( a \notin N_1' \), we would have \( a \in N_1 \mathrel {\triangle }N_1' = \operatorname {\mathsf{interf}}(N_1, N_1') \subseteq \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\), again a contradiction.

Hence \( \zeta = (\xi ^{\mathcal A}\sqcup R, \xi ^{\mathcal N}) \) is a base action and satisfies the conclusion.

Without loss of generality (as extensions are transitive), let \( \xi \) satisfy the conclusion of proposition 4.28.

Let \( L \) be an arbitrary litter that whose litter set does not contain an atom in \( \operatorname {\mathsf{im}}\xi ^{\mathcal A}\) or \( \bigcup \operatorname {\mathsf{im}}\xi ^{\mathcal N}\). Define an injection

by a cardinality argument. Note that \( i \) has domain disjoint from \( \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\) and image disjoint from \( \operatorname {\mathsf{im}}\xi ^{\mathcal A}\).

We show that \( (\xi ^{\mathcal A}\sqcup \operatorname {\mathsf{graph}}i, \xi ^{\mathcal N}) \) is a base action. It suffices to check that if \( (a_1, a_2) \in \operatorname {\mathsf{graph}}i \) and \( (N_1, N_2) \in \xi ^{\mathcal N}\), then \( a_1 \in N_1 \) if and only if \( a_2 \in N_2 \). As \( a_2 \in \operatorname {\mathsf{LS}}(L) \), we have \( a_2 \notin N_2 \). Suppose that \( a_1 \in N_1 \). We know that there is a near-litter \( N \in \operatorname {\mathsf{coim}}\xi ^{\mathcal N}\) such that \( a_1 \in \operatorname {\mathsf{LS}}(N^\circ ) \setminus N \setminus \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\). If \( N^\circ = N_1^\circ \), then \( a_1 \in N \mathrel {\triangle }N_1 = \operatorname {\mathsf{interf}}(N, N_1) \subseteq \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\), a contradiction, hence \( a^\circ = N^\circ \neq N_1^\circ \). But then as \( \xi \) satisfies the conclusion of proposition 4.28, we have \( N_1 \setminus \operatorname {\mathsf{LS}}(N_1^\circ ) \subseteq \operatorname {\mathsf{coim}}\xi ^{\mathcal A}\), which again is a contradiction.

Apply proposition 4.28 to \( \xi \) to obtain \( \xi _1 \); apply proposition 4.28 again to \( \xi _1^{-1} \) to obtain \( \xi _2 \); apply proposition 4.29 to \( \xi _2 \) to obtain \( \xi _3 \), and finally apply proposition 4.29 again to \( \xi _3^{-1} \) to obtain \( \xi _4 \), our target.