Assume V=L. Is there a semi-proper notion of forcing that is not proper?

Namba forcing isn't semi-proper in L, and Prikry forcing isn't even available there.

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Assume V=L. Is there a semi-proper notion of forcing that is not proper?

Namba forcing isn't semi-proper in L, and Prikry forcing isn't even available there.

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The following is proved as Claim 2.1 in Shelah's paper ``Forcing axiom failure for any $\lambda>\aleph_1$''

Theorem.There is a forcing notion of size $2^{\aleph_2}$ which is not proper but is $\{\aleph_1\}$-semi proper.

In fact Shelah proves more: for every regular $κ>ℵ_1$ there is a forcing notion $P$ of cardinality $2^κ$ such that there is a stationary set $S⊆[2^κ]^{ℵ_0}$ not preserved by $P$ and if $χ>2^κ$, $p∈P∈N≺(H(χ),∈), N$ countable, then there is $q∈P$ stronger than $p$ and $q$ forces that $N∩κ$ is an initial segment of $N[G]∩κ.$

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The poset described in the proof of Lemma 37.11 of Jech's book seems like it might be relevant (this is the poset Shelah used in getting equivalence of MM with SPFA). Namely, consider some $X \subset P_{\omega_1}(H_{\omega_2})$. Define $X^\perp$ to be the set of $M \in P_{\omega_1}(H_{\omega_2})$ which not only fail to be in $X$, but cannot even be $\omega_1$-end-extended to an element of $X$ (i.e. there is no $N \in X$ such that $M \subseteq N$ and $M \cap \omega_1 = N \cap \omega_1$). So $X^\perp$ is a subset of the complement of $X$. The natural poset to shoot a club through $X \cup X^\perp$ is semiproper. So if one can (in ZFC, or under V=L) find an $X$ so that $X \cup X^\perp$ is costationary, it would fail to be proper.

Edit: I think the following works. Fix a stationary, costationary $S \subset \omega_2 \cap \text{cof}(\omega)$ (here I mean stationary and costationary within $\text{cof}(\omega)$). Let $X$ be the set of countable $N \prec H_{\omega_2}$ such that $\text{sup}(X \cap \omega_2) \in S$. I claim that the complement of $X \cup X^\perp$ is stationary. Let $\mathfrak{A} = (H_{\omega_2}, \in, \dots)$ be an algebra on $H_{\omega_2}$. We need to find some countable $M \prec \mathfrak{A}$ such that the sup of $M$ with $\omega_2$ is not in $S$, but $M$ can be $\omega_1$-end extended (elementarily) to something whose sup is in $S$. Fix $W \prec W' \prec \mathfrak{A}$ where $\omega_1 \subset W$, $W$ and $W'$ are transitive of size $\omega_1$, $W \cap \omega_2$ and $W' \cap \omega_2$ are both $\omega$-cofinal, $W \cap \omega_2 \notin S$, but $W' \cap \omega_2 \in S$ (this can be done since $S$ is stationary and costationary within $\text{cof}(\omega)$). Also make sure $W \in W'$. Pick some countable $M' \prec W'$ with $W \in M'$ such that $M' \cap ORD$ is cofinal in $W' \cap \omega_2$; in particular the sup of $M'$ with $\omega_2$ is in $S$, so $M' \in X$. Let $M:= M' \cap W$. Since $W \in M'$ then $M'$ sees a cofinal function from $\omega$ to $W \cap \omega_2$, so $M \cap ORD$ will be cofinal in $W$. In particular $M \notin X$, and $M'$ is an $\omega_1$-end-extension of $M$.