In chapter 8 of Mitchell's and Steel's FSIT, they prove a central fine structural result, which basically states that if $\mathcal{M}$ is 1-small, $k$-sound, $k$-iterable premouse then the $k+1$-strandard parameter is $k+1$-solid/universal over $\mathcal M$.


The proof is not too complicated but on page 76 they state that if $\rho\_{k+1}^{\mathcal M}> lh(E)$ where $E$ is some extender on the $\mathcal M$-sequence then $\mathcal H= \mathcal H\_{k+1}^{\mathcal M}(\alpha\_s)$ (I omitted the rest of the parameter and $u$ ($u$ is the tuple having the solidity witnesses and the appropriate projectums whenever defined)from the hull so that it is easier to read, this is well written in details page 74) then $\mathcal H$ is an initial segment of $\mathcal M$. I can't see why this is true. Could anyone help me figure out the reason behind this? Thx.



A question about Mitchell/Steel Fine Structure and Iteration Trees
The assumption is that $\rho=\rho\_{k+1}^M>\text{lh}(E)$ for \emph{all} $E$, not just some $E$. But looking at your message of March 6/7, it looks like you are indeed reading it as ``all''. Regarding the question in that message, $H$ agrees with $M$ below $\rho$ because $\rho\leq\alpha\_s$, and all ordinals below $\alpha\_s$ were put into the uncollapsed hull when forming $H$. So if $\pi:H\to M$ is the uncollapse, then $\text{crit}(\pi)\geq\rho$. (Note you could also just argue that $M=J\_\gamma(M|\rho)$ for some ordinal $\gamma$, and $M|\rho$ is also passive. Since $\text{crit}(\pi)\geq\rho$, this gives that $H=J\_\beta(M|\rho)$ for some $\beta\leq\gamma$.)


