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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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What does ‘wf’ mean? I only know it as meaning ‘well-formed’, but that doesn't seem to make sense here. – Zhen Lin Aug 3 '11 at 13:08
Yes. It does mean well-formed. Why it does not make sense here? – Victor Victorov Aug 3 '11 at 13:19
What does ‘wf ...’ mean? Is it a sentence asserting that the formula is well-formed? Because your question seems to make much more sense when all the ‘wf’ are taken out. – Zhen Lin Aug 3 '11 at 13:22
English is not my native language. So I try to follow Mendelson. He is often writes “wf” for instance page 65: “If wf $\mathscr B$ and its negation $\neg \mathscr B$ …”. If you share with me when it appropriate to use I appreciate it. – Victor Victorov Aug 3 '11 at 13:47
I would write ‘If a wff $\scr B$ and its negation $\lnot B$ ...’. The abbreviation ‘wff’ stands for ‘well-formed formula’ and is a noun, whereas ‘wf’ is an adjective. – Zhen Lin Aug 3 '11 at 14:05
up vote 2 down vote accepted

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 $y$.

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I appreciated it. You are right. I am ashamed that didn’t get myself. – Victor Victorov Aug 3 '11 at 13:33

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