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

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

(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:

$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

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

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)

(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)

Edit:

**(It was not my intention to scare a lot of nice mice)**

**(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)**

PLEASEdo not deface your own questions. People have put work in answering it, and removing the text makes their effort go to waste. – Mariano Suárez-Alvarez♦ Nov 19 '12 at 4:54