1
$\begingroup$

Suppose $\Gamma$ is an infinite set of formulas in FOL. It is given that for a structure, $M$: If $\Gamma$ is satisfiable in $M$ then $\varphi$ is valid in $M$.

Why is it true that there's some finite $\Gamma ' \subset \Gamma$, such that $\Gamma ' \vDash \varphi$?

It reminds the compactness theorem in a way, but it's not quite the same.

  • 0
    $\Gamma$ and non-$\phi$ has no model, i.e. (by completeness theorem) you can derive a contradiction from a finite subset of $\Gamma$ and from non-$\phi$ – 2017-02-03

2 Answers 2

2

It is indeed a consequence of compactness. HINT: If $\Gamma\models\varphi$, then $\Gamma\cup\{\neg\varphi\}$ is unsatisfiable. What does the contrapositive of the Compactness Theorem tell you about a theory which is unsatisfiable?

  • 0
    I'm sorry, didn't see your answer. – 2017-02-03
  • 1
    @Leo163 No reason to apologize, simultaneous answers happen! (I upvoted yours btw.) – 2017-02-03
  • 0
    Oh nice. By the compactness theorem there's some finite $\Gamma ' \cup \{\lnot \varphi\}$ which isn't satisfiable which implies $\Gamma '\vDash \varphi$ – 2017-02-03
2

From the Compactness Theorem, if there is no finite $\Gamma'\subset\Gamma$ with $\Gamma'\models\varphi$, then $\Gamma\not\models\varphi$. From your hypothesis, it follows that $\Gamma$ cannot be satisfiable. But then, again by Compactness, there is a finite $\Gamma'\subset\Gamma$ with $\Gamma'\vdash\perp$, which means $\Gamma'\models\varphi$.

If otherwise there is a finite $\Gamma'$ as you ask, the thesis follows immediately.

  • 0
    What do you mean by $\Gamma'\vdash\perp$? – 2017-02-03