In "The Kunen-Miller Chart (Lebesgue Measure, The Baire Property, Laver Reals and Preservation Theorems for Forcing)" by Haim Judah and Saharon Shelah
JSL Vol. 55, No. 3 (Sep., 1990), pp. 909-927 ([JdSh308] in Shelah's numbering) the authors remark on the final page that adding a pair of Hechler reals makes the union of the meager sets coded by the ground model meager. My questions are:


1. How does one prove this? (and/or is there a citation with the proof, in the paper none is given that I can find)
2. Does this not happen when one adds only one Hechler real?


I suppose more generally my question is as follows: clearly in the Hechler model $\mathrm{add}(\mathcal M) = \mathfrak{c}$ by Miller-Truss theorem that $\mathrm{add}(\mathcal M) = \min\{\mathfrak{b}, \mathrm{cov}(\mathcal M)\}$ since both dominating reals and Cohen reals are added but I don't quite see how this proliferates down to the finite steps. Is there a more direct way to see how Hechler forcing effects $\mathrm{add}(\mathcal M)$?



Effect of adding one Hechler real versus adding two on the meager ideal
If $c$ is Cohen over $V$, and $d$ is dominating over $V[c]$ (not necessarily Hechler-generic), then in $V[c][d]$ there is a meager set covering all meager sets from $V$. Hence 2 successive Hechler reals make the union of all old meager sets meager. 


(I suspect that this is not true if you just add one Hechler real.)


Proof: (This is implicit in Bartoszynski-Judah 2.2, and implicit or perhaps even explicit in other papers, such as Miller or Truss. I seem to remember that Andreas Blass invented a notion of "composition" of Galois-Tukey relations, which gives a general framework for arguments such as the one I give below.)


For each meager set $M\subseteq 2^\omega$ coded in $V$ there are functions $f:\omega\to \omega$ and $x\in 2^\omega$ (again in $V$) such that $M$ is contained in $ M\_{f,x}:= \{ y\in 2^\omega\mid \forall^ \infty n: x\restriction I\_n\not= y\restriction I\_n\}$, where $I\_n:=[f(n), f(n+1))$. 


In $V[c]$ there are infinitely many $n$ such that $c\restriction I\_n = x\restriction I\_n$. (As $c$ is Cohen). So there is an increasing sequence $(n\_k)$ such that
$x\restriction I\_{n\_k} = c\restriction I\_{n\_k}$ for all $k$. (The sequence $(n\_k:k\in \omega)$ depends on $x$, of course. But all we will use about it in the next paragraph is that it is an element of $V[c]$.) 


In $V[c][d]$ we may wlog assume that $d$ strongly dominates $V[c]$, e.g. by first assuming $d(k)>k$ for all $k$, and then replacing $d$ with $k\mapsto d^{(k)}(0)$ ($k$-th iterate). Let $J\_\ell$ be the interval $[ d(\ell), d(\ell+1))$. As these intervals are very long (compared with anything from $V[c]$), almost every $J\_\ell$ contains at least one intervall of the form $I\_{n\_k}$. 


I claim that $M\_{f,x} \subseteq M\_{d,c}$. So let $y\in M\_{f,x}$ (in $V[c][d]$). For notational simplicity assume that $\forall n: y\restriction I\_n \not= x\restriction I\_n$. Since $x$ and $c$ agree on all the $I\_{n\_k}$, we also have $y\restriction I\_{n\_k}\not= c\restriction I\_{n\_k}$. For any (sufficiently large) $\ell $ we can find $k$ such that $J\_\ell$ contains $I\_{n\_k}$; so we also have $y\restriction J\_\ell\not= c\restriction J\_\ell$. Hence $y\in M\_{d,c}$. 


