0
$\begingroup$

One of the rules of formation for the language of set theory is

If $x$ is a variable and $P$ is a ${\square}$, then ${\forall}x(P)$ is a WFF

The reason I wrote ${\square}$ is that I have heard two versions of this rule: that $P$ is any WFF and that $P$ is an open formula.

So which version is the correct one?

$\endgroup$
1
$\begingroup$

The standard definition of well-formed formula (short: wff) for first-order language is:

  • every atomic formula is a wff

  • if $\varphi, \psi$ are wff, then $(\lnot \varphi), (\varphi \lor \psi), (\varphi \to \psi), (\varphi \land \psi)$ are wff

  • if $\varphi$ is a wff and $x_i$ a variable, then $(\forall x_i \varphi)$ and $(\exists x_i \varphi)$ are wff.

Usual conventions for omitting parentheses are straightforward.

Note In order to "apply" a quantifier to a formula $\varphi$, it is not mandatory that the quantified variable is free in $\varphi$. (Of course, quantifying a variable that it is not in the scope of the quantifier has no "practical" effects, i.e. it does not affect the "meaning" of the formula.)


See :

| cite | improve this answer | |
$\endgroup$

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.