One of the logical axioms in a Hilbert system for first order logic is the quantifier axiom (Q7 in the wikipedia article):

$\phi\to\forall x\phi$

whenever $\phi$ is a formula and $x$ is a variable not occurring free in $\phi$.

What would be an example in some simple model where $x$ is free in $\phi$ and the resulting axiom would not be valid?


2 Answers 2


In the supposed case, consider $\phi$ states '$x$ is an even number'. Then, the proposition purported to be axiom would state:

If $x$ is an even number, all $x$'s (e.g., all natural numbers) are even.

Hence, there would be illegitimate generalisation.

We take heed of the detail that these are propositions in the object language, not to be confused with the rule of universal generalisation.


Let formula $ϕ(x)$ be satisfied by object $p,$ but not object $q,$ in the domain of discourse (e.g., $\{p,q\}$); then $ϕ(p)$ is true but $∀xϕ(x)$ is false, that is, $$ϕ(p)\to ∀xϕ(x)$$ is false.

So, in this model, $$ϕ(x)\to ∀xϕ(x)$$ is not definitely true.

Thus, the open formula $$ϕ(x)\to ∀xϕ(x)$$ is not valid.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .