This is Shelah's simpler notion of 'completeness' for not adding reals via forcing. Here $P$ and $Q$ are forcing notions.

Let $S_{\aleph_0}(\kappa)$ be the set of all countable subsets of a cardinal $\kappa$. A family of subsets of $S_{\aleph_0}(\kappa)$, $\varepsilon$, is said to be nontrivial iff there is a cardinal $\mu$, for all $\lambda > \mu$, there is a countable $N \preccurlyeq H(\lambda)$ such that $\varepsilon \in N$ and $N \cap \kappa \in A$ for every $A \in \varepsilon \cap N$. We call such an $N$ suitable for $\varepsilon$. By a theorem, we can always assume $S_{\aleph_0}(\kappa) \in \varepsilon$.

$p := \{ p_n : n < \omega \}$ is a generic sequence for $(N, P)$ iff for each $n$, $p_n \in P \cap N$, $p_{n+1} \geq$ (extends) $p_n$, and $p$ intersects all dense open sets $\chi \in \mathcal{P}(P) \cap N$. We say the pair $(N, P)$ is complete iff every generic sequence for $(N, P)$ has an upper bound in $P$.

We can now define $\varepsilon$-completeness. $P$ is $\varepsilon$-complete for some nontrivial subset $\varepsilon$ of $S_{\aleph_0}(\kappa)$, iff there is a cardinal $\mu$, for all $\lambda > \mu$, for every $N \preccurlyeq H(\lambda)$, $N$ suitable for $\varepsilon$ such that $P \in N$, the pair $(N, P)$ is complete.

Shelah mentions in passing, in his book, that if $Q$ is not $\varepsilon$-complete whereas $P$ is, then forcing with $P$ will not make $Q$ $\varepsilon$-complete in the generic extension. Why is this so? Proving the same statement with the word 'proper' in place of '$\varepsilon$-complete' is easy, but even with some modifications, I am stuck.

My reference for this is Shelah's 'Proper and Improper Forcing', Chapter 5, Section 1. http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pl/1235419814

Much thanks in advance!