In first order logic we have the following inference rule :

 Γ ⇒ F(v)
 Γ ⇒ ∀vF(v)

where v is not a free variable of any formula in Γ. I understand that there is a condition F(v) and variable v satisfies this condition. However how come all the v's can satisfy this condition ? What is the logic behind this inference ?

  • $\begingroup$ If the "name" $v$ is "generic" (i.e. nothing is known or suppoesd about it : this is the reason for the proviso : $v$ not free in $\Gamma$) then if $F$ holds of it, then $F$ holds for all. $\endgroup$ – Mauro ALLEGRANZA Dec 28 '16 at 13:23

Maybe it will be helpful if you think about it from a semantic point of view. Suppse that if $\Gamma\models\varphi$, and a model ($\mathcal{M},v)\models\Gamma$. Then we have $(\mathcal{M},v)\models\varphi$. Our goal is to show $(\mathcal{M},v)\models\forall_x\varphi$. We also know that $\varphi$ is a axiom, or an elemnent of $\Gamma$, or got from $MP$ from a syntactic point of view.

Suppose is $\varphi$ is an axiom, of course, $(\mathcal{M},v)\models\forall_x\varphi$.

Suppose is $\varphi$ an elemnent of $\Gamma$, or got from MP, then we also have $\mathcal({M},v)\models\forall_x\varphi$. Because we can prove that: if x is not a free variable of $\varphi$, $(\mathcal{M},v)\models\varphi\leftrightarrow(\mathcal{M},v)\models\forall_x\varphi$ by induction on the compexity of $\varphi$.


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