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Let $P, Q \in V$ be such that $P$ is strongly proper and $Q$ is ccc. Does $Q$ continue to be ccc after forcing with $P$?

Since strongly proper forcings do not add new branches to $\omega_1$-trees, they do preserve the ccc-ness of some partial orders: Suslin trees, the poset to specialise an Aronszajn tree.

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Yes. The proof of Claim 3.8 in Neeman's paper on forcing with side conditions shows that:

(*) if $P$ is strongly proper and $Q$ is proper, $M$ is a countable elementary submodel, $p$ is an $(M,P)$ strong master condition, and $q$ is an $(M,Q)$ master condition, then $p$ forces that $\check{q}$ is an $(M[\dot{G}_P], Q)$-master condition.

(this is an abstraction of a lemma of Sy Friedman)

Now consider also that the following 3 properties of a poset $Q$ are equivalent (this is an old result, I don't recall where, but it's easy to prove):

  1. $Q$ is c.c.c.
  2. $1_Q$ is a master condition for stationarily many countable models
  3. $1_Q$ is a master condition for club-many countable models

So back to your question: assume $P$ is strongly proper and $Q$ is c.c.c. Then there is a club $C$ of countable models $M$ such that $P$ is strongly proper w.r.t. $M$, and $1_Q$ is an $(M,Q)$ master condition (the latter uses equivalence of 1 with 3). Let $G_P$ be $(V,P)$-generic. Using a genericity argument, in $V[G_P]$ the set of $M \in C$ for which $G_P$ includes a strong master condition is stationary; let $S \in V[G_P]$ denote this stationary set. Also $S':= \{ M[G_P] \ : \ M \in S \}$ is stationary. Since $S \subseteq C$, then for every $M \in S$, $1_Q$ was an $(M,Q)$-master condition in the ground model. Then by (*), $V[G_P]$ believes that $1_Q$ is a master condition for every model in the stationary set $S'$. Hence by the equivalence of 1 and 2 above, $V[G_P]$ believes that $Q$ is c.c.c.

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  • $\begingroup$ Thanks. The result about ccc forcing can be found in Mekler, 'CCC forcing without combinatorics', though he mentions that this is implicit in Shelah's work. $\endgroup$ – tci Apr 27 '17 at 14:22

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