# equivalent statement of compactness theorem

My book on logic says that there are two equivalent compactness theorems:

1. Let $$\Gamma$$ be a set of propositional formulas, and let $$\phi$$ be a formula. If $$\Gamma\models \phi$$, then there is a finite subset $$\Gamma_0\subset\Gamma$$ such that $$\Gamma_0\models\phi$$.

2. If for each finite $$\Gamma_0\subset\Gamma$$ we have a model, then for $$\Gamma$$ we have a model.

Now for $$2\implies 1$$, my book writes the following: Suppose 2. is true. Let $$\Gamma$$ be a set of propositional formulas, and let $$\phi$$ be a formula such that $$\Gamma\models\phi$$. Then there is no model for $$\Gamma\cup\{\neg\phi\}$$, and because of our assumption, there exists a finite $$\Gamma_0\subset\Gamma$$ such that there is no model for $$\Gamma_0\cup\{\neg\phi\}$$. But then we have $$\Gamma_0\models\phi$$.

I don't understand how our assumption leads to the existence of $$\Gamma_0\subset\Gamma$$ such that $$\Gamma_0\cup\{\neg\phi\}\models\perp$$. I'm confused, because $$\Gamma_0\cup\{\neg\phi\}$$ has no model, so it seems like we can't apply the assumption, except for $$\Gamma$$ separately, but I don't see how.

Since your $\Gamma_0 \cup \{\neg \phi\}$ has no model we have $\Gamma_0 \cup \{\neg \phi\} \models \bot$ (remind that $\Theta\models\varphi$ if every model of $\Theta$ is a model of $\varphi$ and that $\bot$ has no model by definition).

Since $\Gamma_0 \cup \{\neg \phi\}$ has no model it follows that every model of $\Gamma_0$ is not a model of $\neg \phi$, it makes $\phi$ false. So every such model must make $\phi$ true, hence $\Gamma_0 \models \phi$.

Suppose 2. is true.

Let $$\Gamma$$ be a set of propositional formulas, and let $$\phi$$ be a formula such that $$\Gamma\models\phi$$.

Then there is no model for $$\Gamma\cup\{\neg\phi\}$$, and because of our assumption, there exists a finite $$\Gamma_0\subset\Gamma$$ such that there is no model for $$\Gamma_0\cup\{\neg\phi\}$$.

Otherwise, consider the finite subsets of $$\Gamma \cup \{\phi\}$$.

If it is $$\Gamma_0 \subset \Gamma$$, if it does not have a model, then $$\Gamma_0 \vDash \phi$$ vacuously.

If it is in the form of $$\Gamma_0 \cup \{\neg \phi\}$$, then by "otherwise", it has a model.

So by 2, $$\Gamma \cup \{\neg \phi\}$$ would have a model, contradiction.

But then we have $$\Gamma_0\models\phi$$.