Could wf $\exists x \mathscr B(x)$ be logically valid? Definitely. For instance, when wf $\forall x\mathscr B(x)$ is logically valid. But, could wf $\exists x \mathscr B(x)$ be logically valid when wf $\forall x\mathscr B(x)$ is not logically valid? Then where is exist an interpretation with denumerable sequence not satisfying wf $\mathscr B(x)$. Then we can look into another interpretation with domain consisting only with elements not satisfying wf $\mathscr B(x)$. But then in that second interpretation wf $\exists x \mathscr B(x)$ is not logically valid. My thoughts look valid. But, in the book of Elliott Mendelson “Introduction to Mathematical Logic” (Fifth Edition) on page 97 in Proposition 2.28 we see “Assume that $\vdash \exists_1 u \mathscr B(u, y_1, \cdots, y_n)$” and I seriously doubt that he means that $\mathscr B(u, y_1, \cdots, y_n)$ must be logically valid. How to solve that contradiction?
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Here's an example where $\exists_1 u P(u,y)$ is "logically valid" but $\forall u P(u,y)$ is not. Let $P(u,y)$ be the (well-formed) formula $u=y$.
Of course (under first-order logic with equality) there exists unique $u$ such that $u=y$ without it necessarily being true that every $u$ is equal to $y$.