# Confusion on Proof of Marker Lemma 2.1.14, $T \vDash \phi \implies \exists \Delta \subseteq T, |\Delta| \in \mathbb{N} \land \Delta \vDash T$

Here is lemma 2.1.14 from Marker's Model Theory restated exactly,

$$\textit{If \ T \vDash \phi, then \Delta \vDash T for some finite \Delta \subseteq T.}$$

To make sure I understand it properly I'll restate the lemma in my own words:

Let $$T$$ be an $$\mathcal L$$-theory, and $$\phi$$ and $$\mathcal L$$-sentence where $$T \vDash \phi$$. Then there exists a finite subset $$\Delta$$ of $$T$$ where $$T$$ is a logical consequence of $$\Delta$$, that is, $$\Delta \vDash T$$.

The provided proof of this lemma is very short, so I'll reproduce that exactly too:

Suppose not. Let $$\Delta \subseteq T$$ be ﬁnite. $$\color{red}{\underline{\text{Because \Delta \not\vDash \phi}}}$$, $$\Delta \cup \{\neg \phi\}$$ is satisﬁable. Thus, $$T \cup \{\neg\phi\}$$ is ﬁnitely satisﬁable and, by the Compactness Theorem, $$T \not \vDash \phi$$.

I've emphasized in red and underlined the portion that I cannot understand. By "Suppose not", we're fixing a $$\phi$$ and assuming that $$T \vDash \phi$$, but that for all finite $$\Delta$$, we have $$\Delta \not \vDash T$$. That is, for each $$\Delta$$, there is some $$\varphi_{\Delta} \in T$$, where $$\Delta \not \vDash \varphi_{\Delta}$$. I have no idea how we can claim that $$\Delta \not \vDash \phi$$, since it's fixed at the time of the "Suppose not". The way the proof is structured, when assuming the negative for contradiction, seems to imply that for all finite $$\Delta \in T$$ we have $$\Delta \not \vDash \phi$$, which I don't understand.

The other steps of the proof are fine as far as I can see. But I'm not sure if the error is in my misunderstanding or if the error is in the text, as this text has a number of errors.

If someone can clear up this specific step in the proof that would be much appreciated.

• I suspect the $\Delta\vDash T$ is a typo which should read $\Delta\vDash \phi$? Otherwise the lemma appears to be saying that every theory is finitely axiomatizable,.. Apr 8, 2020 at 5:29
• That makes much more sense. It’s a shame there are so many errors in this book. That also of course clears up the question. Feel free to post that as an answer. So I guess the consequence of the compactness theorem here is that, every sentence is finitely axiomatizable? Apr 8, 2020 at 5:30
• Yes, it is a typo: it must be "then $\Delta \vDash \phi$". And yes: due Completeness for FOL, we have $T \vDash \phi$ iff $T \vdash \phi$. But a derivation is a finite sequence; thus, it can use only a finite number of premises (axioms). Apr 8, 2020 at 5:50

As indicated in the comments, this is a typo and the conclusion should be $$\Delta\models \phi.$$ It says if a sentence is a logical consequence of a theory, then there is a finite subtheory that it is a logical consequence of.
This is easier to understand from the deductive side. If a sentence is provable from $$T,$$ then since proofs are finite, it only uses a finite number of assumptions from $$T,$$ and thus it's provable from a finite subtheory. And the completeness theorem says that $$T\models \phi$$ if and only if $$\phi$$ is provable from $$T.$$