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?

  • 2
    $\begingroup$ What does ‘wf’ mean? I only know it as meaning ‘well-formed’, but that doesn't seem to make sense here. $\endgroup$ – Zhen Lin Aug 3 '11 at 13:08
  • $\begingroup$ Yes. It does mean well-formed. Why it does not make sense here? $\endgroup$ – Victor Victorov Aug 3 '11 at 13:19
  • 1
    $\begingroup$ 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. $\endgroup$ – Zhen Lin Aug 3 '11 at 13:22
  • $\begingroup$ 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. $\endgroup$ – Victor Victorov Aug 3 '11 at 13:47
  • $\begingroup$ 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. $\endgroup$ – Zhen Lin Aug 3 '11 at 14:05

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 $y$.

  • 1
    $\begingroup$ I appreciated it. You are right. I am ashamed that didn’t get myself. $\endgroup$ – Victor Victorov Aug 3 '11 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.