Supervaluational semantic equivalence

Question

I'm working on Ted Siders Logic for Philosophers and I'm stuck on exercise 3.15.

I need to show the global and local definitions of supervaluational semantic consequence are equivalent. How do I go about getting started on this?

Answer 1

HINT

Local implies global.

According to the "local" definition of semantic consequence [Exercise 3.14, page 103] :

$\phi$ is a supervaluational semantic consequence of $\Gamma$ iff for every trivalent interpretation $\mathscr I$, and every precisification $\mathscr C$ of $\mathscr I$, if $VC_{\mathscr C} (\gamma) = 1$, for each $\gamma \in \Gamma$, then $VC_{\mathscr C} (\phi) = 1$.

We can "semi-formalize" the above definition as follows :

$\forall \mathscr I \, \, \forall \mathscr C \, , \, $ if $VC_{\mathscr C} (\gamma) = 1 \, , \, $ for each $\gamma \in \Gamma$, then $VC_{\mathscr C} (\phi) = 1$.

Using the following property of quantifiers :

$\forall x(\alpha \to \beta) \to (\forall x \alpha \to \forall x \beta)$

we can derive from the above definition :

$\forall \mathscr I \, , \, $ if $\forall \mathscr C \, \, VC_{\mathscr C} (\gamma) = 1 \, , \, $ for each $\gamma \in \Gamma$, then $\forall \mathscr C \, \, VC_{\mathscr C} (\phi) = 1$.

But, applying the definition of page 101 :

if $VC_{\mathscr C} (\phi) = 1$ for every precisification $\mathscr C$ of $\mathscr I$, we say that $\phi$ is supertrue in $\mathscr I$

we can rewrite the above condition as :

$\forall \mathscr I \, , \, $ if $\gamma$ is supertrue in $\mathscr I$, for each $\gamma \in \Gamma$, then $\phi$ is supertrue in $\mathscr I$

and this one is the "global" definition of supervaluational semantic consequence : $\Gamma \vDash_S \phi$ of page 101.