% !TEX root =  ../main.tex
\section{Proofs}\label{app:proofs}

\interventionlemma*
\begin{equation}
    \begin{tikzcd}[row sep=1.5cm, column sep=1.5cm]
        \scm{}(I) \arrow[r, "\mymixing"] \arrow[d, "\catidentity_{\scm{}(I)}"'] & \scm{}(I^\prime) \arrow[d, "\intervention"] \\
        \scm{}_{\intervention}(I) \arrow[r, "\mymixing_\intervention"'] & \scm{}_\mathcal{H}({I^\prime)}
    \end{tikzcd} \Longrightarrow  \mymixing_\intervention = \intervention \circ \mymixing\,. 
    \tag{(7)}
\end{equation}
\begin{proof}
    Let us start from the exogenous component.
    Since the intervention \intervention, that could be either soft or hard, acts only on the endogenous variables, the probability space corresponding to the exogenous part remains unchanged. 
    That is, for the object $\scm{}(I)$ we simply have
    \begin{equation}
        \eta^\intervention_I \coloneqq \catidentity_{\scm{}(I)}\,.
    \end{equation}
    This means that for any measure $\mu$ on the exogenous, the pushforward along $\eta^\intervention_I$ is exactly $\mu$.

    Now, let us consider the endogenous component and recall the collections $\mathcal{F}_\hard$ and $\mathcal{F}_\soft$ in \Cref{sec:cat_scm_ck}. 
    If the intervention is hard, than we represent it as the collection $\mathcal{F}_\intervention \coloneqq \mathcal{F}_\hard$.
    If soft, with $\mathcal{F}_\intervention \coloneqq \mathcal{F}_\soft$.
    The collection $\mathcal{F}_\intervention$ entails a new collection of mixing functions; specifically \emph{(i)} $\mymixing_\hard$ for the hard intervention case, \emph{(ii)} $\mymixing_\soft$ for the soft one (cf. \Cref{sec:cat_scm_ck}).
    At this point, by invertibility of \mymixing, \Cref{eq:nat_transf_inter_SCMcat} trivially follows by setting the second component $\eta^\intervention_{I^\prime}$ equal to $\intervention=\mymixing_\intervention \circ \mymixing^{-1}$.
\end{proof}

\convexspaceprob*
\begin{proof}
    According to Def. 3.1 in \cite{fritz2009convex}, we have to demonstrate that $cc_{\lambda}(\chi_1,\chi_2)$ satisfies: \emph{(i)} $cc_0(\chi_1,\chi_2)=\chi_2$, \emph{(ii)} $cc_0(\chi_1,\chi_1)=\chi_1$; \emph{(iii)} $cc_{\lambda}(\chi_1,\chi_2)=cc_{\bar{\lambda}}(\chi_2,\chi_1)$; and $cc_{\lambda}(cc_{\mu}(\chi_1,\chi_2),\chi_3)=cc_{\widetilde{\lambda}}(\chi_1, cc_{\widetilde{\mu}}(\chi_2,\chi_3))$ where:
    \begin{equation}
        \widetilde{\lambda}=\lambda \mu, \quad \quad \quad \widetilde{\mu} = \begin{cases}
            \frac{\lambda \bar{\mu}}{\bar{\lambda \mu}}, \quad \text{if } \lambda\mu \neq 1\,,\\
            \text{arbitrary}, \quad \text{if } \lambda=\mu=1\,.
        \end{cases}
    \end{equation}
    Since the convex set $\Delta_{(\myendogenousvals,\Omega)}$ is a subset of a vector space, the previous properties follow from the axioms of vector space.
\end{proof}

\csprobcat*
\begin{proof}    
    From Definition 3.2 in \cite{fritz2009convex} we have that convex spaces as in \Cref{lem:convexspace_prob} together with morphisms commuting with \(cc_\lambda\) form a category. Hence we have to demonstrate that affine measurable maps commute with $cc_\lambda$.
    Recall that a map $f: \Delta_{(\myendogenousvals,\, \Omega)} \to \Delta_{(\myendogenousvals^\prime,\, \Omega^\prime)}$
    is affine if for every $\chi_1,\,\chi_2 \in \Delta_{(\myendogenousvals,\, \Omega)}$ and every $\lambda \in [0,1]$ it satisfies
    $f\left(cc_\lambda(\chi_1,\chi_2)\right) = cc_\lambda\left(f(\chi_1),f(\chi_2)\right).$
    Hence, by definition, any affine measurable map necessarily commutes with $cc_\lambda$. 
\end{proof}

\convexcomb*
\begin{proof}
    Consider $\scm{}=\langle (\myexogenousvals,\, \Upsilon, \zeta), \, (\myendogenousvals,\, \Omega, \chi)\, , \mymixing \rangle$, $\lambda \in [0,1]$, and w.l.o.g. the probability measures $\chi_{\mathcal{S}_1}$ and $\chi_{\mathcal{S}_2}$ entailed by two (measurable) soft interventions $\eta^{{S}_1}$ and $\eta^{\mathcal{S}_2}$ run on \scm{}.
    Hence, the convex combination
    \begin{equation}
        \chi_{\mathcal{S}_3} = cc_{\lambda}(\chi_{\mathcal{S}_1}, \chi_{\mathcal{S}_2}) = \lambda \chi_{\mathcal{S}_1} + \bar{\lambda} \chi_{\mathcal{S}_2}
    \end{equation}
    is a proper soft-interventional probability measure entailed by $\mymixing_{\soft_3}=\{m^{\soft_3}_1, \ldots, m^{\soft_3}_n\}$ according to \Cref{eq:scm_sint_push}, where $m^{\soft_3}_i=\lambda m^{\soft_1}_i + \bar{\lambda} m^{\soft_2}_i$.
    Indeed we have
    \begin{equation}
        \begin{aligned}
            \lambda \chi_{\mathcal{S}_1} + \bar{\lambda} \chi_{\mathcal{S}_2} &=\bigtimes_{X_i\in \myendogenous} \Big( \lambda m^{\soft_1}_i\left(\mu_i \left( \myexogenousvals_i \times \myexogenousvals^{\ancestors_i} \right) \right) +\\
            & \quad \quad \quad \quad + \bar{\lambda} m^{\soft_2}_i\left(\mu_i \left( \myexogenousvals_i \times \myexogenousvals^{\ancestors_i} \right) \right) \Big)\\
            &=\bigtimes_{X_i\in \myendogenous} \left(\lambda m^{\soft_1}_i + \bar{\lambda} m^{\soft_2}_i\right)\left(\mu_i \left( \myexogenousvals_i \times \myexogenousvals^{\ancestors_i} \right) \right)\\
            &= \bigtimes_{X_i\in \myendogenous} m^{\soft_3}_i\left(\mu_i \left( \myexogenousvals_i \times \myexogenousvals^{\ancestors_i} \right) \right) = \chi_{\mathcal{S}_3}\,.\\
        \end{aligned}
    \end{equation}
    Also, recall that a convex combination of measurable functions is still measurable.
    Starting from \Cref{eq:scm_obs_push,eq:scm_hint_push,eq:nat_transf_inter_SCMcat}, it is straightforward to see that the same holds also when we consider observational and hard-interventional measures within the convex combination.
\end{proof}

\caaffinefunctions*
\begin{proof}
    Consider two SCMs \scm{} and $\scm{\prime}$ related by an IC \alphaabs as in \cref{def:IC_alpha_abstraction_scmcat}, and interventions $\mathcal{I}_1^\prime$ and $\mathcal{I}_2\prime$ on $\scm{\prime}$, either hard or soft, corresponding to interventions $\mathcal{I}_1$ and $\mathcal{I}_2$ on $\scm{}$.
    Then, given $\intervention_3^\prime$ convex combination of $\mathcal{I}_1^\prime$ and $\mathcal{I}_2\prime$, we have
    \begin{equation}\label{eq:proof_alpha_affine}
        \begin{aligned}
            \chi'_{\intervention_3^\prime} &= \alphamap{\myendogenous}\left(\chi_{\intervention_3}\right)\\
            &\stackrel{(a)}{=} \alphamap{\myendogenous}\left(\lambda \chi_{\intervention_1}+ \bar{\lambda} \chi_{\intervention_2}\right) \\
            &\stackrel{(b)}{=} \lambda \chi'_{\intervention_1^\prime}+ \bar{\lambda} \chi'_{\intervention_2^\prime} = \chi'_{\intervention_3^\prime}\,.
        \end{aligned}
    \end{equation}
    where in \emph{(a)} we apply \Cref{th:convex_comb_prob_meas}, and in \emph{(b)} the linearity of the pushforward.
\end{proof}

\encodingfunctor*
\begin{proof}
    The mapping on objects is well-defined since \Cref{th:convex_comb_prob_meas} guarantees that convex combinations of probability measures associated with observational and interventional states of \scm{} are valid soft-interventional probability measures.
    On morphisms, \Cref{th:ca_affine_functions} implies that \alphamap{\myendogenous} is a proper morphism in \CSprob.
    Interventional consistency is inherited because in \NI every morphism is an IC \alphaabs.
\end{proof}