Thanks, that makes sense! Could you say a little about why the weak union axiom holds? I've been struggling to prove that from your definitions. I was hoping that $h^F(X|z,w)\subseteq h^F\left(X|z\right)$ would hold, but I don't think that $h^F\left(X|z\right)$ satisfies the second condition in the definition of conditional history for $h^F(X|z,w)$.

I'm confused by the definition of conditional history, because it doesn't seem to be a generalisation of history. I would expect $h^F\left(X|\varnothing \right)=h^F\left(X\right)$, but both of the conditions in the definition of $h^F\left(X|\varnothing \right)$ are vacuously true if $E=\varnothing$. This is independent of what $H$ is, so $h^F\left(X|\varnothing \right)=\varnothing$. Am I missing something?