According to the soundness and completeness theorems, the answers to both questions should be the same. It is easy to see that $\Gamma$ is satisfiable. Consider $\mathfrak{A}=(\{0,1\};P)$ where $P^{\mathfrak{A}}=\{<1>\}$ and $s:V\rightarrow|\mathfrak{A}|$ such that $s(v_{i})=1$ . Then, certainly, $\vDash_{\mathfrak{A}}Pv_{i}[s]$ , $\vDash_{\mathfrak{A}}Pv_{1}[s(v_{1}|1)]$ and $\not\vDash_{\mathfrak{A}}Pv_{1}[s(v_{1}|0)]$ , hence, $\not\vDash_{\mathfrak{A}}\forall v_{1}Pv_{1}[s]$ .

Note. At first, it may seem to some that $\Gamma$ is neither consistent nor satisfiable. Indeed, the first formula says that $P$ does not hold for every element in the domain, and then it lists all the possible variables, saying that $P$ holds for all of them. The problem with this approach is that what the first formula says is that, indeed, there is an element in the domain for which $P$ does not hold. However, after that, $\Gamma$ does not list all the elements of the domain, rather, it contains an infinite list of independent variables saying that whatever element is assigned to each of them, $P$ holds for the element. In other words, the rest of $\Gamma$ just says that there is an element for which $P$ holds. In other words, $\Gamma$ is somewhat (not really logically) equivalent to $\Gamma'=\{\exists x\neg Px,\exists yPy\}$ , in the sense that for every domain for which $\Gamma$ can be satisfied with some $s$ , $\Gamma'$ (consisting of sentences only) is satisfied, and vice versa.