semantic consequence in first order logic

Question

I'm asked to show that Fx does not semantically entail AxFx. However in the preceding paragraph the author tells me Fx is true in a model iff AxFx is true in that model. So how am I supposed to provide a model in which Fx is true and AxFx is false?

Here the definition of semantic consequence is from Ted Sider's Logic for Philosophy: PHI is a semantic consequence of a set of well formed formulas GAMMA if and only if for every model M and every variable assignment g for M, if each member of GAMMA is true, then PHI is true.

Answer 1

It depends on the details of the definitions of "true in a model" and "semantical entailment".

In :

• Herbert Enderton, A Mathematical Introduction to Logic (2nd ed - 2001), page 88, we have that :

DEFINITION. Let $\Gamma$ be a set of wffs, $\varphi$ a wff. Then $\Gamma$ logically implies $\varphi$, written $\Gamma \vDash \varphi$, iff for every structure $\mathfrak A$ for the language and every function $s : Var → |\mathfrak A|$ such that $\mathfrak A$ satisfies every member of $\Gamma$ with $s$, $\mathfrak A$ also satisfies $\varphi$ with $s$.

According to this definition, it is not true in general that $\varphi(x) \vDash \forall x\, \varphi(x)$.

Consider as $\varphi$ the formula of first-order arithmetic : $(x=0)$. It is clear that with an $s$ such that $s(x)=0$ :

$\mathbb N \vDash (x=0)[s]$;

but clearly $\forall x\,(x=0)$ is not satisfied by $s$ in $\mathbb N$, and thus :

$(x=0) \nvDash \forall x\,(x=0)$.

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As you can see in :

• Theodore Sider, Logic for Philosophy (2009), page 115

this is the same as :

Definition of semantic consequence : $\phi$ is a PC-semantic consequence of set $\Gamma$ of wffs (“$\Gamma \vDash_{PC} \phi$”) iff for every PC-model $\mathscr M$ and every variable assignment $g$ for $\mathscr M$, if $V_{\mathscr M,g} (\gamma) = 1$ for each $\gamma \in \Gamma$, then $V_{\mathscr M,g} (\phi) = 1$.

See page 114 :

Definition of truth in a model: $\phi$ is true in PC-model $\mathscr M$ iff $V_{\mathscr M,g} (\phi) = 1$, for each variable assignment $g$ for $\mathscr M$.

Thus, we have that $\phi$ is true in $\mathscr M$ iff $\forall x \, \phi$ is.

If we apply this definition to my example above, we have that $(x=0)$ is not true in $\mathbb N$ simply because it is not true that $V_{\mathbb N,g} (x=0) = 1$, for each variable assignment $g$ : it is enough to consider a $g$ such that $g(x)=1$.

And thus, we have that both $(x=0)$ and $\forall x\,(x=0)$ are false in $\mathbb N$.