2

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.

James
  • 21
  • Might it be that $x$ is the name of some particular member of the model, and it is one of which $Fx$ is true? ${}\qquad{}$ – Michael Hardy Mar 02 '15 at 20:03
  • I updated the original post. I don't think this solution works since Fx is true in a model iff Fx is true for each variable assignment for the model. So if I'm thinking correctly, Fx is true iff each member of the domain is in the extension of F, but then this is just to say AxFx. So it seems we cant create a model in which Fx is true and AxFx is false. – James Mar 02 '15 at 20:49
  • If the exercise asks "to show that $Fx$ does not semantically entail $\forall x \ Fx$" this amount to find a model M and a variable assignment $g$ for $M$ such that $V_{M,g}(Fx)=1$ and $V_{M,g}(\forall x Fx)=0$. – Mauro ALLEGRANZA Mar 03 '15 at 07:12

1 Answers1

0

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

In :

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 :

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

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 $\mathbb N$.

Michael Hardy
  • 1
Mauro ALLEGRANZA
  • 94,169
  • So you conclude that PHI is true in M iff AxPHI is true. But then we have the same problem right? Plug in Fx for PHI, so Fx is true in M iff AxFx is true in that M, so it's impossible to provide a model in which Fx is true and AxFx is false. But exercise 4.3d asks one to do just that. – James Mar 02 '15 at 21:11
  • @James - sorry, but I think that we have different editions... On mine, Exercise 4.3 (d) is different. – Mauro ALLEGRANZA Mar 02 '15 at 21:35
  • @James - but in your question you are saying : "I'm asked to show that $Fx$ does not semantically entail $\forall xFx$", and this is proved in my answer. This is not the same as "to provide a model in which $Fx$ is true and $\forall xFx$ is false". – Mauro ALLEGRANZA Mar 02 '15 at 21:38