This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident: 

For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements 

$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names) 

and

$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha < \beta \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \delta \in \check{\lambda}\backslash\beta \ (\sup(ran(f)) \le \delta) $$

(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))

Now, here comes the question: Does the following conjunction:

 $\exists \lambda = cf(\lambda) > \omega\ \exists\ \mathbb{P}$ such that

 - $\vert \mathbb{P} \vert = \lambda^{+}$
 - $\forall \mu \  (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$ 
 - $\neg Cof(\lambda, \mathbb{P})$

imply that $\lambda$ is measurable? (Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)

**But more importantly can formal assertions about what can be forced to hold in a generic-extension have true large cardinal strength?**