Understanding the statement of Robinson's Consistency Theorem The following is from https://ncatlab.org/nlab/show/Robinson+consistency+theorem.
Note: As Alex pointed out, the "if and only if" in the theorem statement is wrong. However, I am only interested in the $T_1 \cap T_2$ complete $\implies$ $T_1 \cup T_2$ consistent direction anyway.

I have some questions regarding the statment of the theorem. (Forgive me if they sound naive, for I am still new to logic.)

#1: When it says $T_1 \cup T_2$ is "consistent," what does this even mean? Is "consistent" the same as "satisfiable" in this context? And are we considering $T_1 \cup T_2$ as a set of $L_1 \cup L_2$ sentences, or are we only concerned with $\{\sigma \in T_1 \cup T_2 : \sigma$ is an $(L_1 \cap L_2)$-sentence$\}$?


#2: Does "complete" in this context just mean for every $(L_1 \cap L_2)$-sentence $\sigma$ we have $\sigma \in T_1 \cap T_2 \vee \neg \sigma \in T_1 \cap T_2$?


#3: The one sentence proof "Apply the Craig interpolation theorem. QED." makes me wonder if I'm overthinking this. I know Craig's theorem says that if $\sigma_1 \models \sigma_2$ ($\sigma_i$ is an $L_i$ sentence) then there is a $\sigma$ ($L_1 \cap L_2$ sentence) such that $\sigma_1 \models \sigma$ and $\sigma \models \sigma_2$. I'm guessing it might help connecting $T_1$ and $T_2$ with $T_1 \cap T_2$, but I don't see how it helps with $T_1 \cup T_2$.

 A: First, the statement as written is incorrect: it's not an "if and only if". Completeness of $T_1\cap T_2$ implies consistency of $T_1\cup T_2$, but not the other way around.
As an aside, the nLab is a great resource for category theory, but it's probably not the best place to learn about logic (other than categorical logic).
#1:  A theory is a set of sentences in some language $L$. We're looking at the theory $T_1\cup T_2$, which is a set of sentences in the language $(L_1\cup L_2)$. A theory is consistent if it does not  prove $\bot$ (a contradiction). A theory is satisfiable if it has a model. By Gödel's Completeness Theorem, a first-order theory is consistent if and only if it is satisfiable, so these two terms are often used interchangeably.
#2: I prefer to say an $L$-theory $T$ is complete if $T\models \varphi$ or $T\models \lnot \varphi$ for every $L$-sentence $\varphi$ (i.e., $T$ entails $\varphi$ or its negation, not necessarily contains $\varphi$ or its negation). This is related to decision of whether or not to assume theories are always closed under entailment by convention.
#3: Here's how to derive Robinson Joint Consistency as a consequence of Craig Interpolation. Suppose $T_1\cap T_2$ is complete, and assume for contradiction that $T_1\cup T_2$ is inconsistent. By the compactness theorem, a finite subset of $T_1\cup T_2$ is inconsistent, say $\varphi_1,\dots,\varphi_n,\psi_1,\dots,\psi_m$, with all $\varphi_i\in T_1$ and all $\psi_j\in T_2$. Let $\varphi = \bigwedge_{i=1}^n \varphi_i$ and $\psi = \bigwedge_{j = 1}^m \psi_j$. Then $\varphi\land \psi$ is inconsistent, so  $\varphi\models  \lnot \psi$. By Craig Interpolation, there is some $(L_1\cap L_2)$-sentence $\theta$ such that $\varphi\models \theta$ and $\theta\models \lnot \psi$. Note that $\lnot\theta\models \lnot \varphi$.
Now since $(T_1\cap T_2)$ is complete, either $T_1\cap T_2\models \theta$ or $T_1\cap T_2\models \lnot\theta$. In the first case, $T_2\models\theta$, so $T_2\models \lnot\psi$, but $T_2\models \psi$, contradicting consistency of $T_2$. In the second case, $T_1\models \lnot\theta$, so $T_1\models \lnot \varphi$, but $T_1\models \varphi$, contradicting consistency of $T_1$.
