I asked this question first on math SE and was told that it would better fit here. So: 


The following concept is due to Shelah and I have some issues with a claim using this notion:
Suppose that $\nu$ is a limit ordinal and that $P\_\nu$ is an iteration of forcing notions. We say that a $P\_\nu$ name $\dot{\alpha}$ of an ordinal is $prompt$ iff the following two things hold:


1. $\Vdash\_\nu \dot{\alpha} \le \nu$
2. If $p \Vdash\_\nu "\dot{\alpha} = \xi"$ then even $p \upharpoonright \xi ^\smallfrown 1\_\nu \upharpoonright [\xi, \nu) \Vdash\_\nu \dot{\alpha} = \xi$ ( $1\_\nu$ should be the largest element of the iteration, and $\xi$ is the hacek name of an ordinal though I refused to write the hacek)


Then the following two things should hold:


1. If $\dot{\alpha}$ is prompt $\eta \le \nu$ and if $p \Vdash\_\nu \eta \le \dot{\alpha}$ then $p \upharpoonright \eta ^\smallfrown 1\_\nu \upharpoonright [\eta, \nu) \Vdash\_\nu \eta \le \dot{\alpha}$
2. If $\dot{\alpha\_i}$ are prompt then so is the supremum $sup$ and the minimum $min$


I have problems proving those two assertions so any help would be highly appreciated. Thank you!



Question about prompt names of ordinals
Here's proof for conclusion 1; I think 2 will then follow fairly easily. For brevity, I'll write things like $p\upharpoonright\eta$ when I really mean its extension by 1 to domain $\nu$. Suppose $\eta$ were a counterexample to 1. Since $p\upharpoonright\eta$ fails to force $\eta\leq\dot\alpha$, it must have an extension $q$ forcing $\dot\alpha$ to have some specific value $\xi<\eta$. Apply promptness to infer that $q\upharpoonright\xi$ already forces this value $\xi$ for $\dot\alpha$. But, since $\xi<\eta$ and $q$ extends $p\upharpoonright\eta$, the conditions $q\upharpoonright\xi$ and $p$ are compatible. Since the former forces $\dot\alpha$ to have value $\xi$ while the latter forces $\eta\leq\dot\alpha$, this is a contradiction.


EDIT: I think I was too optimistic in expecting conclusion 2 to "follow fairly easily", so I'm adding information about that. Notice first that what I wrote above remains correct if we replace the inequalities $\eta\leq\dot\alpha$ and $\xi<\eta$ by $\eta<\dot\alpha$ and $\xi\leq\eta$, respectively. (Of course, $\eta<\dot\alpha$ is equivalent to $\eta+1\leq\dot\alpha$, so I could apply the preceding paragraph directly, but then I'd get a conclusion about $p\upharpoonright(\eta+1)$, whereas I really want $p\upharpoonright\eta$.) 


Next, let me establish a "dual" (i.e., order-reversed) version of conclusion 1, namely that if $\dot\alpha$ is prompt and $p$ forces $\dot\alpha\leq\eta$, then $p\upharpoonright\eta$ already forces the same. To prove this, suppose it fails, and let $q$ be an extension of $p\upharpoonright\eta$ forcing $\eta<\dot\alpha$. By what I proved above, $q\upharpoonright\eta$ suffices to force $\eta<\dot\alpha$. But $q\upharpoonright\eta$ is an extension of $p\upharpoonright\eta$ and is therefore compatible with $p$. That's absurd since compatible conditions can't force contradictory things about the ordering of $\eta$ and $\dot\alpha$.


Of course, this dual version of conclusion 1 also has an analog for conditions $p$ that force $\dot\alpha<\eta$.


At last, I'm ready to prove the part of conclusion 2 that deals with the supremum, say $\dot\beta$, of some prompt names $\dot\alpha\_i$. So suppose, toward a contradiction, that $p$ forces $\dot\beta=\xi$ but $p\upharpoonright\xi$ doesn't force this. There are two possibilities: Either some extension of $p\upharpoonright\xi$ forces some $\dot\alpha\_i$ to be strictly above $\xi$, or some extension of $p\upharpoonright\xi$ forces some ordinal $\eta<\xi$ to be an upper bound for all the $\dot\alpha\_i$'s. In either case, let $q$ be such an extension of $p\upharpoonright\xi$.


Consider the first case: $q$ forces $\xi<\dot\alpha\_i$ for a certain index $i$. Then, by one of the versions of conclusion 1, $q\upharpoonright\xi$ already forces the same inequality. But $q\upharpoonright\xi$ is an extension of $p\upharpoonright\xi$ and is therefore compatible with $p$, which forces the opposite. So this case cannot occur.


There remains the case that, for a certain $\eta<\xi$, the condition $q$ forces $\dot\alpha\_i\leq\eta$ for every index $i$. By a dual version of conclusion 1, as proved above, $q\upharpoonright\eta$ forces all these inequalities and therefore forces $\dot\beta\leq\eta$. But $q\upharpoonright\eta$ is compatible with $p$, which forces the opposite. So this case is also impossible, and the proof of conclusion 2 for the supremum is complete.


Finally, since I was (I hope) careful to use, in this proof for the supremum, only versions of conclusion 1 for which the dual is also available, we can dualize this whole proof to get the result for the infimum, which is of course the minimum since we're dealing with a well-rodering.


