Is the term $$\forall x p(x) \rightarrow \forall x q(x)$$ equal to $$\forall x (p(x) \rightarrow \forall x q(x))$$ or $$(\forall x p(x)) \rightarrow (\forall x q(x))$$

In other words: What is the operator precedence of $\forall$ and $\exists$? Does it make a difference?

  • 1
    $\begingroup$ You have to use parentheses; $∀x(p(x) \to ∀xq(x))$ is equivalent to : $∃xp(x) \to ∀xq(x)$. Thus, in $∀xp(x) \to ∀xq(x)$ we assume that the scope of the leftmost quantifiers is only $p(x)$. In other words, the quantifiers has the same priority that $\lnot$, i.e. higher than the binary connectives. $\endgroup$ Feb 16, 2015 at 16:54
  • $\begingroup$ @MauroALLEGRANZA Do you want to say that $\forall x p(x) \rightarrow \forall x q(x)$ is not a valid formula? $\endgroup$ Feb 16, 2015 at 17:04
  • $\begingroup$ It is, Moose. The second choice is equivalent to your posted statement, but not the first choice you give.$$\forall x p(x) \rightarrow \forall x q(x) \equiv (\forall x p(x)) \rightarrow (\forall x q(x))$$ $\endgroup$
    – amWhy
    Feb 16, 2015 at 17:19
  • $\begingroup$ Of course it is admissible; but if you ask if it is "well-formed" you have to check with your specifications. If you ask if is a "correct" abbreviation, my answer is : YES, but I read it as : $((∀xp(x)) → (∀xq(x)))$. $\endgroup$ Feb 16, 2015 at 17:30

1 Answer 1


There is no "precedence" between quantifiers: in a formula like $\forall x \ \exists y \ \varphi$, the "inner" $\exists$ is in the scope of the "outer" $\forall$.

The recursive definition of formula for FOL is (having defined term) more or less this :

(i) $t_1=t_2$ and $P^n(t_1,\ldots,t_n)$ are atomic formulas, where $t_1,\ldots,t_n$ are terms and $P^n$ is a $n$-ary predicate symbol;

(ii) if $\varphi, \psi$ are formulas, then $\lnot \varphi, \varphi \land \psi, \varphi \lor \psi, \varphi \to \psi$ are formulas;

(iv) if $\varphi$ is a formula, then $((∀x)\varphi), ((∃x)\varphi)$ are formulas.

Then we can introduce abbreviations for readibility; see :

For parentheses we will omit mention of just as many as we possibly can. Toward that end we adopt the following conventions:

  1. Outermost parentheses may be dropped. For example, $∀x α → β$ is $(∀x α → β)$.

  2. $\lnot, ∀$, and $∃$ apply to as little as possible. For example,

$¬α ∧ β$ is $((¬α) ∧ β)$, and not $¬(α ∧ β)$;

$∀x α → β$ is $(∀x α → β)$, and not $∀x(α → β)$;

$∃x α ∧ β$ is $(∃x α ∧ β)$, and not $∃x(α ∧ β)$.

In such cases we might even add gratuitous parentheses, as in $(∃x α)∧β$.

  1. ∧ and ∨ apply to as little as possible, subject to item 2. For example, $¬α ∧ β → γ$ is $((¬α) ∧ β) → γ$.

  2. When one connective is used repeatedly, the expression is grouped to the right. For example, $α → β → γ$ is $α → (β → γ)$.

Thus, we have a "formal" specification for the syntax: it must be unambiguous, i.e. "processable" by a machine.

  • 4
    $\begingroup$ Thank you for the detailed answer. ((2) was what I needed to know) $\endgroup$ Feb 16, 2015 at 17:31

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