Structure theorems of Bravyi-Vyalyi and zero conditional mutual information

A fundamental result in quantum information is that of Bravyi and Vyalyi (Lemma 8, https://arxiv.org/abs/quant-ph/0308021).

Result 1: It states that for commuting hermitian operators $H_{ab}\otimes \text{I}_c, \text{I}_a\otimes H_{bc}$ acting on particles $a,b,c$ (where $\text{I}$ is the identity operator) there exists a decomposition of the Hilbert space of particle $b$ into direct sum of Hilbert spaces $\mathcal{H}_{b_1}, \mathcal{H}_{b_2},\ldots$ such that

1. each Hilbert space $\mathcal{H}_{b_i}$ has a tensor product form $\mathcal{H}^a_{b_i}\otimes \mathcal{H}^c_{b_i}$ and
2. the operator $H_{ab}\otimes \text{I}_c$, when restricted to the Hilbert space $\mathcal{H}_{b_i}$, acts trivially on $\mathcal{H}^c_{b_i}$
3. similarly the operator $\text{I}_a\otimes H_{bc}$ , when restricted to the Hilbert space $\mathcal{H}_{b_i}$, acts trivially on $\mathcal{H}^a_{b_i}$.

In a vague form: particle $b$ can be split into three particles $b_1,b_2,b_0$ such that $H_{ab} = \sum_i |i\rangle\langle i|_{b_0}\otimes H^i_{ab_1}\otimes \text{I}_{b_2}$ and $H_{bc} = \sum_i |i\rangle\langle i|_{b_0}\otimes \text{I}_{b_1}\otimes H^i_{b_2c}$.

Result 2: There is yet another fundamental result in quantum information due to Hayden et.al. (Theorem 6, https://arxiv.org/abs/quant-ph/0304007). Given spins (or registers or particles) $A,B,C$, it concerns quantum state $\rho_{ABC}$ that satisfies $\text{I}(A:C|B) = S(AB)+S(BC)-S(B)-S(ABC)=0$, where $S(.)$ is the von-Neumann entropy. It is very easy to see that this result has striking similarity with the previous result of Bravyi and Vyalyi.

My question: Is there a connection between the two? Such a connection maybe interesting, as non-negativity of conditional mutual information has recently been improved by Fawzi and Renner (https://arxiv.org/abs/1410.0664).

Partial connection: I am able to see one direction (Result 2 implies Result 1) in the special case when the operators $H_{ab}, H_{bc}$ are projectors. In this case, define $P_{ab} = \text{I}- H_{ab}$ and $P_{bc} = \text{I}- H_{bc}$ as `ground space projectors' of these operators. Finally, consider the joint ground space (which takes a very simple form due to commutativity): $P_{abc} = P_{ab}P_{bc}$. Take $\rho_{abc} = \frac{P_{abc}}{\text{Tr}(P_{abc})}$ as maximally mixed state inside $\rho_{abc}$. Now, suppose I loose the particle $c$. Then there is a very easy way to recover the state $\rho_{abc}$: bring back particle $c$ in maximally mixed state and measure with projector $P_{bc}$. Repeat upon failure.

After applying this operation till you succeed, one gets back $\rho_{abc}$ because measuring with projector $P_{bc}$ does not affect the $P_{ab}$ part (due to commutativity). This means that there is a quantum operation that takes $\rho_{ab}$ to $\rho_{abc}$ and directly implies that $\text{I}(A:C|B)=0$ in the state $\rho_{abc}$. Then we can apply Result 2 and find that the state $\rho_{abc}$ has the structure as given by the result. This means that such structure must also be true for $P_{ab},P_{bc}$ (I don't have a proof, but sounds plausible), through which we recover Result 1. But I don't know what happens when the operators $H_{ab},H_{bc}$ are not projectors. Maybe we consider their thermal state?

Edit: In fact, it can be shown (Equation 2.1, https://arxiv.org/abs/1403.6102) that the condition $\text{I}(A:C|B)=0$ is equivalent to the statement that $\rho_{ABC} = e^{log(\rho_{AB}) + log(\rho_{BC}) - log(\rho_{B})}$. One could exploit this to obtain Result 2 from Result 1, by choosing operators $H_{AB},H_{BC}$ such that $H_{AB} + H_{BC} = log(\rho_{AB}) + log(\rho_{BC}) - log(\rho_{B})$ and the operators commute. But it is not clear a priori how commutativity can be ensured. In other direction, to prove Result 1 from Result 2, one could set $\rho_{abc} = e^{-H_{ab}-H_{bc}}$ and try to prove that one can recover particle $c$ by acting only on particle $b$ (using some form of Monte-carlo algorithm maybe).