Given two $n$-element structures, $\mathcal{A}$ and $\mathcal{B}$, which can be distinguished by a sentence of $k$-variable first-order logic ($\mathcal{L}^k$), what is the minimum $f(n)$ such that there is guaranteed to be a sentence $\phi \in \mathcal{L}^k$ with at most $f(n)$ quantifiers, such that $\mathcal{A} \models \phi$ but $\mathcal{B} \not \models \phi$? We present various results related to this question obtained by using the recently introduced QVT games. In particular, we show that when we limit the number of variables, there can be an exponential gap between the quantifier depth and the quantifier number needed to separate two structures. Through the lens of this question, we will highlight some difficulties that arise in analysing the QVT game and some techniques which can help to overcome them. As a consequence, we show that $\mathcal{L}^{k+1}$ is exponentially more succinct than $\mathcal{L}^{k}$. We also show, in the setting of the existential-positive fragment, how to lift quantifier depth lower bounds to quantifier number lower bounds. This leads to almost tight bounds.

翻译：给定两个$n$元素结构$\mathcal{A}$和$\mathcal{B}$，它们能被$k$变量一阶逻辑($\mathcal{L}^k$)的一个语句区分，那么保证存在一个具有至多$f(n)$个量词的语句$\phi \in \mathcal{L}^k$，使得$\mathcal{A} \models \phi$但$\mathcal{B} \not \models \phi$的最小$f(n)$是多少？我们通过最近引入的QVT博弈给出了与这一问题相关的多种结果。特别地，我们证明，当限制变量数量时，区分两个结构所需的量词深度与量词数量之间可能存在指数级差距。通过这一问题的视角，我们将突出分析QVT博弈时出现的一些困难，以及有助于克服这些困难的若干技术。作为推论，我们证明$\mathcal{L}^{k+1}$比$\mathcal{L}^{k}$指数级更简洁。我们还证明，在存在正片段的设定下，如何将量词深度下界提升为量词数量下界，从而得到几乎紧致的界。