[proofplan] The proof is a direct unwinding of the definitions. Each factorization-level formula is assumed to agree with the same observational-law functional $\Psi$ whenever its induced law lies in $\mathcal P$. Since the two factorizations induce the same observational law $P$, both formulas evaluate to $\Psi(P)$, and identifiability identifies this value with $\Theta(P)$. [/proofplan]

[step:Apply the defining agreement of $\Phi_1$ with the observational-law functional]Since $F_1\in\mathfrak D_1$ and $P_{F_1}=P\in\mathcal P$, the defining property of the factorization-level identifying formula \begin{align*} \Phi_1:\mathfrak D_1\to\mathcal T \end{align*} applies to $F_1$. Hence \begin{align*} \Phi_1(F_1)=\Psi(P_{F_1}). \end{align*} Using the assumed equality $P_{F_1}=P$, this becomes \begin{align*} \Phi_1(F_1)=\Psi(P). \end{align*}[/step]

[guided]We first isolate exactly which hypothesis is being used. The object $F_1$ is an element of the domain $\mathfrak D_1$ of the map \begin{align*} \Phi_1:\mathfrak D_1\to\mathcal T. \end{align*} The factorization $F_1$ induces the observational probability measure $P_{F_1}$ on $(\mathcal O,\mathcal A)$, and the theorem assumes \begin{align*} P_{F_1}=P \end{align*} for a measure $P\in\mathcal P$. Therefore $P_{F_1}\in\mathcal P$. The defining property of a factorization-level identifying formula in this theorem says that whenever $F\in\mathfrak D_1$ and $P_F\in\mathcal P$, the formula $\Phi_1(F)$ agrees with the observational-law functional evaluated at the induced law: \begin{align*} \Phi_1(F)=\Psi(P_F). \end{align*} Applying this statement with $F=F_1$ gives \begin{align*} \Phi_1(F_1)=\Psi(P_{F_1}). \end{align*} Finally, because $P_{F_1}=P$ as probability measures on $(\mathcal O,\mathcal A)$, the right-hand side is exactly $\Psi(P)$. Thus \begin{align*} \Phi_1(F_1)=\Psi(P). \end{align*}[/guided]

[step:Identify $\Psi(P)$ with the target $\Theta(P)$] Because $P\in\mathcal P$ and $\Theta$ is identifiable from the observational law through \begin{align*} \Psi:\mathcal P\to\mathcal T, \end{align*} the defining identifiability equality gives \begin{align*} \Theta(P)=\Psi(P). \end{align*} Combining this with the previous step yields \begin{align*} \Phi_1(F_1)=\Theta(P)=\Psi(P). \end{align*} [/step]

[step:Apply the same defining agreement to $\Phi_2$] Since $F_2\in\mathfrak D_2$ and $P_{F_2}=P\in\mathcal P$, the defining property of \begin{align*} \Phi_2:\mathfrak D_2\to\mathcal T \end{align*} gives \begin{align*} \Phi_2(F_2)=\Psi(P_{F_2}). \end{align*} Using the assumed equality $P_{F_2}=P$, we obtain \begin{align*} \Phi_2(F_2)=\Psi(P). \end{align*} [/step]

[step:Conclude that both factorization formulas have the same value] The first two steps gave \begin{align*} \Phi_1(F_1)=\Theta(P)=\Psi(P), \end{align*} and the preceding step gave \begin{align*} \Phi_2(F_2)=\Psi(P). \end{align*} By transitivity of equality in the target space $\mathcal T$, \begin{align*} \Phi_1(F_1)=\Theta(P)=\Psi(P)=\Phi_2(F_2). \end{align*} This is precisely the asserted invariance under replacing the factorization by another factorization with the same induced observational law. [/step]