# Robinson's theorem- a proof.

If $\Delta, \Delta'$ are satisfable sets of sentences while $\Delta \cup \Delta'$ is not satisfable then exists the such sentence $\phi$ that $\Delta \models \phi$ and $\Delta' \models \neg \phi$

Let's assume that there is no such $\phi$. Let take a finite subset: $\Delta_0 = \Delta \cup \Delta'$. Let's show that $\Delta_0$ is satisfable. Let's assume that is it unsatisfable. Then, no model of $\Delta'$ satisfies $\Delta_0$. It means that there exists a such $\phi$ that $\Delta \models \phi$ and $\Delta' \models \neg \phi$. What is a $\phi$? $\phi$ is just a conjuction of sentences from $\Delta$ ( $\Delta$ is finite). So $\Delta_0$ was satisfable. By compactness theorem $\Delta \cup \Delta'$ is satisfable. Contraddiction.

How does that follow?

$\Delta \models \phi$. It is obvious because of construction of $\phi$.

$\Delta' \models \neg \phi$

No model of $\Delta'$ satisfies $\Delta_0$ where $\Delta_0 = (\Delta \cap \Delta_0) \cup (\Delta_0 \cap \Delta')$. It means that no model of $\Delta' \models \Delta_0 \cap \Delta \iff \text{ no model of } \Delta' \models \phi$. Therefore every model $M$ of $\Delta'$ $$M \models \neg \phi$$ and so $\Delta'\models \neg \phi$

But why does it follow from the fact that that $\Delta_0$ is satisfiable? Again, see my above counterexample for this:

Please note that at the beginnig I assume that there is no such $\phi$ as it is expected in the problem. Then, I want to prove that every finite subset is satisfable. So, I take $\Delta_0$ and assume that it is unsatisfable. Then I show that it follows that a such $\phi$ exists. So, contradiction. $\Delta_0$ must be satisfable.

... but \Delta_0 is still not satisfiable.

Yes, but note that there exists $\phi = P$

Right?

You overlooked an essential part of the conclusion of Robinson's theorem: All of the non-logical symbols (relation symbols, function symbols and constant symbols) that occur in $\phi$ must occur in $\Delta$ and also in $\Delta'$. That is, $\phi$ uses only then intersection of the vocabularies of $\Delta$ and $\Delta'$.

Wrong.

Let $\Delta = \{ P \}$ and $\Delta' = \{ \neg P \}$

Then $\Delta_0 = \{ P,\neg P \}$ would be a finite subset of $\Delta \cup \Delta'$, but obviously $\Delta_0$ is not satisfiable.

OK, so we know something went wrong in your proof. ... but where?

You say:

It means that there exists a such $ϕ$ that $\Delta \vDash \phi$ and $\Delta' \vDash \neg \phi$. What is a $\phi$? $\phi$ is just a conjuction of sentences from $\Delta$ ( $\Delta$ is finite). So $\Delta_0$ was satisfable.

Now, your elaboration of your argument for why $\Delta \vDash \phi$ and $\Delta' \vDash \neg \phi$ certainly looks correct. ... But why does it follow from the fact that that $\Delta_0$ is satisfiable? Again, see my above counterexample for this:

With $\Delta = \{ P \}$ and $\Delta' = \{ \neg P \}$ and $\Delta_0 = \{ P,\neg P \}$, we have $\phi = P$, and we have $\{ P \} \vDash P$ (so indeed $\Delta \vDash \phi$) and $\{ \neg P \} \vDash \neg P$ (so indeed $\Delta \vDash \neg \phi$), but $\Delta_0$ is still not satisfiable.

• I've edited. :) – user376326 Jan 27 '17 at 20:25
• @Logic Thanks, I'll take a look at your elaboration ... but there must still be something wrong, since with my example $\Delta_0 = \{ P, \neg P \}$ is still not satisfiable. – Bram28 Jan 27 '17 at 20:30
• Thanks :). I've edited again. – user376326 Jan 27 '17 at 20:47