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This is a claim in Shelah's Proper and Improper Forcing, more specifically Claim 2.3(1) of Chapter X (p. 484). The proof of the claim is "Easy" but I cannot quite figure it out. There must be some trick that I am missing.

A forcing notion $P$ is $RUCar^{V}$-semiproper iff there is a cardinal $\kappa$, such that for all cardinal $\lambda \geq \kappa$, for all $N$ countable elementary submodel of $(H(\lambda), \in)$ with $P \in N$, if $p \in P \cap N$ then there is $q$ ($\in P$) extending $p$ satisfying the following:

for every regular uncountable cardinal $\theta \in N$ and $P$-name $\pi \in N$ of an element of $\theta$, $q \Vdash_{P}$ "there is an ordinal $\alpha \in N$ with $\pi < \alpha < \theta$".

Shelah claims every $RUCar^{V}$-semiproper forcing notion is proper. Is it really easily observable?

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Given such an $N$ and $q$, you want to see that $q$ is $(N, P)$-generic. This is equivalent to $q$ forcing $N[G]\cap On = N\cap On$. (Corollary 2.13 page 106 of Proper and Improper Forcing)

Suppose $\dot\alpha$ is a $P$-name for an ordinal with $\dot\alpha\in N$. Then $\{\xi\in On: \text{ some $r\in P$ forces $\dot\alpha$ to equal $\xi$}\}$ is in $N$, and its supremum is in $N$ as well. Thus, there is an ordinal $\xi\in N$ such that

$\Vdash_P \dot\alpha<\xi$.

Now let $q$ be as in your question, and let $G$ be a generic subset of $P$ containing $q$. Assume by way of contradiction that $N\cap On $ is a proper subset of $N[G]\cap On$.

Let $\alpha=\min(N[G]\cap On\setminus N)$, and let $\beta$ be $\min(N\cap On\setminus\alpha)$. Notice that $\beta$ exists by our discussion above.

We know that $N\cap\alpha = N[G]\cap\alpha$ by choice of $\alpha$ and using this it is not hard to see that $\beta$ must be a regular cardinal.

Now our assumption on $q$ will give us an ordinal in $N\cap\beta$ above $\alpha$, contradicting the choice of $\beta$.

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