# satisfiability under special model

Let $\sigma$ a dictionary and let $\mathcal{M}$ a model such that for all $a\in D^\mathcal{M}$ there exists some ground term $s$ such that $s^\mathcal{M}=a$.

For a formula of the form $\forall x\alpha$ (not necessarily a statement), Prove or Disprove:

$\mathcal{M}\vDash \forall x\alpha\iff \forall \text{ ground term } s:\mathcal{M}\vDash\alpha[s/x]$

I tried to prove the above.

Let $v$ be some assignment in $\mathcal{M}$. We get that

$\mathcal{M},v\vDash \forall x\alpha\iff$

for every $d\in D^\mathcal{M}:\mathcal{M},v[d/x]\vDash\alpha\iff$

$\forall \text{ ground term } s:\mathcal{M},v[s^\mathcal{M}/x]\vDash\alpha\iff$

$\forall \text{ ground term } s:\mathcal{M},v[\bar{v}(s)/x]\vDash\alpha\iff$

$\forall \text{ ground term } s:\mathcal{M},v\vDash\alpha[s/x]$

Is it true? If not what is false? must we have that $\forall x\alpha$ is a statement and not just any genral formula?

• Correct; the key point is that every element $a$ of the domain of the model has a "name". – Mauro ALLEGRANZA Jan 1 '18 at 16:42
• @MauroALLEGRANZA Is the statment also true for $\exists x\alpha$, i.e. $\mathcal{M}\vDash \exists x\alpha\iff \exists \text{ ground term } s:\mathcal{M}\vDash\alpha[s/x]$. I 'proved' it exactly as I did with $\forall x\alpha$ – Don Fanucci Jan 1 '18 at 16:44