ccc forcing that adds no real which dominates all ground model reals is $\mathfrak{b}$-nondominating

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I'm reading Canjar's "Mathias Forcing which does not add dominating reals", where he defines a $\lambda$-cc forcing to be $\lambda$-nondominating if whenever $D$ is a family of reals in $V[G]$ with $|D|< \lambda$ we can find a real in $V$ which is not dominated by any real in $D$. He then argues that any ccc partial order which does not add a real which dominates all ground model reals is $\mathfrak{b}$-nondominating. The argument should be easy:

Let $D$ be a family of reals in $V[G]$ with $|D|<\mathfrak{b}$. $(*)$ Thus, by defintion of $\mathfrak{b}$, take a real $f$ in $V[G]$ which bounds $D$. Now, by assumption $f$ does not dominate all ground model reals, so there is a real $g$ in $V$ which is not dominated by $f$. By our choice of $g$ no real in $D$ can dominate $f$.

Now, my problem is in $(*)$ we assume that the forcing does not decrease the bounding number, because if $\mathfrak{b}^{V[G]} < \mathfrak{b}$ the argument would fail, so my question boils down to why a ccc forcing which does not add a real which dominates all ground model reals cannot decrease the bounding number.

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The forcing adding $\omega_1$ many Cohen reals adds no dominating reals but forces $\mathfrak{b} = \omega_1$. Thus you are right that your argument does not work.

The actual argument is different:

Suppose that in $V^{\mathbb{P}}$, $D \subseteq \omega^\omega$ and $\vert D \vert < \mathfrak{b}^V$. Let's say $D = \{f_\alpha : \alpha < \lambda \}$ where $\lambda < \mathfrak{b}^V$. Then since each $f_\alpha$ is not dominating over $V$ there is $g_\alpha \in V$ so that $g_\alpha \not <^* f_\alpha$. Now in order to decide what $g_\alpha$ is we would have to pass to a condition. But since $\mathbb{P}$ is ccc, there is a countable set $G_\alpha \in V$ so that the trivial condition forces $$\exists g \in G_\alpha (g \not <^*\dot f_\alpha).$$

Now in $V$ we can form the set $G := \bigcup_{\alpha < \lambda} G_\alpha$ which has size $< \mathfrak{b}^V$. In particular there is $h \in V$ so that $$\forall g \in G (g <^* h).$$ Putting things together, we find that $$\Vdash_{\mathbb{P}} \forall \alpha < \lambda (h \not <^* \dot f_\alpha).$$