# Can the conjunction of any two $\exists\forall$-sentences be expressed equivalently as a $\exists \forall$-sentence?

We say that a sentence is a $$\exists\forall$$-sentence iff it is of the form $$\exists w_1\cdots\exists w_m\forall v_1\cdots\forall v_n \theta(\bar{w},\bar{v})$$ where $$\theta$$ is quantifier-free. My question is, given any two $$\psi,\varphi$$ $$\exists\forall$$-sentences, does there exist another $$\exists\forall$$-sentence $$\gamma$$ such that $$\psi\land\varphi\equiv\gamma$$?

I've only managed to almost prove the case when both $$\psi$$ and $$\varphi$$ have the same amount of existential and universal quiantifiers, but I don't see how to proceed in the general case.

The main problem I have is that $$(\exists x\alpha)\land(\exists x\beta)\not\equiv\exists x(\alpha\land\beta)$$, however it is true that $$\exists x(\alpha\land\beta)$$ logically implies $$(\exists x\alpha)\land(\exists x\beta)$$.

Thanks for any help.

• Am I missing something simple? If $\exists x . \forall y . \theta_1$ and $\exists u . \forall v . \theta_2$, where $x,y,u,v$ all are disjoint collections of variables, then why doesn't $\exists x, u . \forall y, v . \theta_1 \land \theta_2$ work? Commented Feb 8, 2021 at 20:12

The formula $$\exists \vec{w} \forall \vec{v} \left[P(\vec{w},\vec{v})\right] \land \exists \vec{x} \forall \vec{y} \left[Q(\vec{x},\vec{y})\right]$$ is equivalent to $$\exists \vec{w},\vec{x} \forall \vec{v},\vec{y} \left[P(\vec{w},\vec{v}) \land Q(\vec{x},\vec{y})\right].$$ This assumes that all the $$\vec{w},\vec{v},\vec{x},\vec{y}$$ are distinct, but you can always achieve that by renaming.