I am trying to show the relationship between logical consequence of Gamma and being true in every model of Gamma

Question

A problem in my textbook is the following.

We say $\Gamma \models A $ if the following holds: If $ I $ be any interpretation of $ L $ and $ \phi $ is any assignment that satisifies $ \Gamma ~~( \phi(B) = T \leftrightarrow B \in \Gamma),$ then $ \phi $ satisfies A.

1) if $ \Gamma \models A $, then $ A $ is true in every model of $ \Gamma $.

2) if every formula in $ \Gamma $ is a sentence, and if $ A $ is true in every model of $ \Gamma $, then $ \Gamma \models A $.

3).the formula $ \forall x_1 Rx_1 $ is true in every model of $ \{ Rx_1 \} $, yet $ \forall x_1 Rx_1 $ is not a logical consequence of $ Rx_1 $

$ ~~ $

For the first part what I said was,

Assume $ \Gamma \models A $, yet there is model, M, of $ \Gamma $ such that $ A $ is not true. (1)

By definition of logical consequence, we have that for any $ \phi $ that satisfies $ \Gamma $ it follows that $ \phi(A) = T $. (2)

As $ M $ is a model, we have that $ \phi(B) = T $ for all $ B \in \Gamma ~~$. (3)

$ \Rightarrow ~ \phi(A) = T ~ in ~M~~~$ (4), from line 2 and 3

$ \bot ~~ \Rightarrow $ A is true in every model M.

$~$

Now, looking at the other question, my answer has completeny ignored the logical quantifiers, and whether A is a sentence of not didn't matter. Is my proof for part 1) wrong? what am I missing? How should I proceed for the other parts? What does it really mean to be true in every model?

Thank you in advance.

Answer 1

In this type of problems we have to use all the definition: what does it mean for a formula $A$ to be true in a model $M$ ? That for every $\phi$ we have that $M, \phi \vDash A$ ($\phi$ satisfies $A$ in $M$).

For 1), if $M$ is a model of $\Gamma$, this means that: for every $B \in \Gamma$ and every $\phi$, we have: $M,\phi \vDash B$.

But $\Gamma \vDash A$, i.e. $M,\phi \vDash A$, for every $\phi$. And this holds for every $M$ that is a model of $\Gamma$.

Thus:

for every model of $\Gamma$ and every $\phi$ we have $M,\phi \vDash A$.

2) What about sentences ?

Now the key property is that if $B$ is a sentence and $M$ is a model of $B$, then $M,\phi \vDash B$, fo revery $\phi$.

Let $M$ a model of $\Gamma$: this means that $M,\phi \vDash \Gamma$, for every $\phi$ (because all formulas in $\Gamma$ are sentences).

But $A$ is true in very model of $\Gamma$, i.e. $M,\phi \vDash A$, for very $\phi$ and every $M$ that is a model of $\Gamma$.

Thus:

$\Gamma \vDash A$.

3) is a counter-example showing that the proviso about $\Gamma$ (all formulas in $\Gamma$ are sentences) is necessary.

By 1) we have that $\forall x Rx$ is true in every model of $Rx$, because if $M$ is a model of $Rx$ this means that $M, \phi \vDash Rx$, for every $\phi$.

But thus also every $x$-variant of $\phi$ will satisfy $Rx$, and thus $M,\phi \vDash \forall xRx$.

Consider now a simple intepretation using $\mathbb N$ as domain and intepret $Rx$ as $(x=0)$.

Let $\phi$ such that $\phi(x)=0$; clearly $\mathbb N, \phi \vDash (x=0)$.

But $\mathbb N$ is not a model of $(x=0)$, because not every $\phi$ satisfies it.

And obviously $\forall x (x=0)$ is not true in $\mathbb N$.

Thus:

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