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

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?

• Out of interest, what textbook is this? – 雨が好きな人 Apr 7 at 22:58

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 $$M,\phi \vDash A$$.

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:

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