My book lists this rule as one of the rules of making formulae:

Let $\phi$ be a formula and $x$ be a variable, then $\exists x \phi$ is a formula

Now let $P(x)$ be a formula with $x$ a free variable, it seems to me that using this rule twice, one can conclude that $\exists x\exists x P(x)$ is a formula. I am willing to accept this as just a string of symbols, but I'd like to know it's "meaning" if there is one.

Thank you

  • $\begingroup$ Surely your book also gives you semantic rules that allow you to determine the meanings of formulae? $\endgroup$ – Chris Eagle Jun 26 '13 at 23:04

This is just to add to Alex Kocurek's perfectly correct answer. It might well help to explicitly recall the semantics for wffs of the form $\exists v\varphi$. The basic idea is

$\exists v\varphi$ is true [relative to some fixing of the domain and some given interpretation of the non-logical vocabulary, of course] iff there is some object $o$ in the domain such that $\varphi$ comes out true when the free occurrences of $v$ in $\varphi$ are treated as parameters denoting $o$.

If there are no free occurrences of $v$ in $\varphi$ this is trivially equivalent to: $\exists v\varphi$ is true iff $\varphi$ is true. So in the given example, $\exists x\exists xPx$ is indeed true just when $\exists xPx$ is true.


You're right that it's an odd string of symbols, but the reason it's odd is because it's redundant. If you "rebind" a variable after it's already been bound, the the innermost quantifier takes precedence. In this case, $\exists x \exists x P(x)$ is redundant, and is equivalent to $\exists x P(x)$. If, for instance, you had $\forall x \exists x P(x)$, this would also be equivalent to $\exists x P(x)$; again, here, the universal quantifier is redundant.

  • $\begingroup$ OK. Sounds reasonable +1 $\endgroup$ – Amr Jun 26 '13 at 23:08
  • $\begingroup$ I think $\exists x\exists x P(x)$ has the same meaning as $\exists y\exists x P(x)$ (The latter formula has a meaning for me) $\endgroup$ – Amr Jun 26 '13 at 23:10
  • $\begingroup$ Yes, @Amr The two formulas are equivalent. Perhaps even better is to say that $\exists x\exists x P(x)$ is equivalent to $\exists x\exists y P(y)$. Of course, they are equivalent to $\exists xP(x)$, as Alex indicates. $\endgroup$ – Andrés E. Caicedo Jun 26 '13 at 23:50
  • $\begingroup$ @Amr The two do have the same meaning, but be sure not to fall into the common misconception that $\exists x\exists y \phi(x)$ implies that $x$ and $y$ refer to different individuals. The formula $\exists x\exists y (x = y)$ is true, for instance. $\endgroup$ – Joshua Taylor Jun 27 '13 at 15:15

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