I am currently working with "The Foundations of Mathematics" by Kunen to understand the proof for Gödel's completeness theorem due to Henkin.
When adding the witnessing terms, there is one thing I can't seem to understand. First we show that given a set of sentences $\Sigma$ in the language $\mathscr{L}$, we may add $\delta$-many new constant symbols to $\mathscr{L}$. We call this new languag $\mathscr{L}_{\delta} := \mathscr{L}\cup \{c_{\alpha} | \alpha <\delta\}$. $\delta$ is an arbitrary ordinal. Now with the aid of these constants, we can add witnesses to $\Sigma$ ($\Sigma_{\delta}:=\Sigma \cup \{\exists x \varphi_{\alpha}(x) \rightarrow \varphi_{\alpha}(c_{\alpha}) | \alpha < \delta\}$) for the existential sentences $\exists x \varphi_{\alpha}(x)$ of $\mathscr{L}_{\alpha} := \mathscr{L}\cup \{c_{\xi} | \xi <\alpha\}$, s.t. $$ Con_{\vdash,\mathscr{L}}(\Sigma) \Rightarrow Con_{\vdash,\mathscr{L}_{\delta}}(\Sigma_{\delta}) $$
This is Lemma II.12.19 in the book. Now $\Sigma_{\delta} $ has witnesses for the $\exists x \varphi_{\alpha}(x), \alpha < \delta$ but it may not have witnesses for all existential sentences of the new language $\mathscr{L}_{\delta}$, so we need to remedy this.
This is done with Lemma II.12.20 which states that for a consistent set of sentences $\Sigma$ in $\mathscr{L}$, you can actually find $\Sigma'$ and $\mathscr{L}'$, s.t. $\Sigma'$ has witnesses in all of $\mathscr{L}'$ and $Con_{\vdash,\mathscr{L}'}(\Sigma')$. Here we add $\kappa \cdot \omega$ new constants to $\mathscr{L}$, where $\omega$ is the first limit ordinal and $\kappa = max\{|\mathscr{L}|,\aleph_0\}$. The part that I don't understand is why do we know that there are exactly $\kappa$ existential sentences of $\mathscr{L}_{\kappa \cdot n} := \mathscr{L}\cup \{c_{\alpha} | \alpha <\kappa \cdot n\}$? Also I don't quite understand what that statement actually means. Does it say there are exactly $\kappa$ existential sentences that cannot be stated in $\mathscr{L}_{\kappa \cdot (n-1)}$ and don't need symbols of $\mathscr{L}_{\kappa \cdot (n+1)} \setminus \mathscr{L}_{\kappa \cdot n}$?
How can we determine how many existential sentences can be stated in a language? Especially if we do not know the language. For example if we consider the language of ordered rings $\mathscr{L} = \{1,0,<,+,-,\cdot\}$, we may state a sentence of the form $\exists x (x < 1+\dots +1)$ with $n$ ones being added up for every $n < \omega$. However, if we look at $\mathscr{L}=\{<\}$, we could not do that as we neither have the addition $+$ nor the constant $1$ at hand.
Any help would be very much appreciated. Thank you :)
We know there are at most $\kappa$ existential sentences, because each existential sentence is a finite sequence of symbols from an alphabet consisting of the basic logical symbols, a countable infinity of variable letters, the symbols of $\mathscr L$, and the $\kappa\times_c n=\kappa$ new constant letters. This alphabet has size $\kappa$, so the number of existential sentences is no larger than $$ \sum_{n\in\omega} \kappa^n = \sum_{n\in\omega} \kappa = \omega\times_c\kappa = \max(\omega,\kappa) = \kappa $$ (This is basic cardinal arithmetic).
On the other hand, there are also at least $\kappa$ existential sentences, even before we begin using the new constant letters. For example, there are $\aleph_0$ formulas of the form $\exists x(x=x\land x=x\land \cdots \land x=x)$. This is already enough, unless $\mathscr L$ is uncountable such that $\kappa>\aleph_0$. But in that case we can also take $\exists x\,p(x,\ldots, x)$ for each predicate letter $p\in\mathscr L$, $\exists x(x=f(x,\ldots,x))$ for each function letter $f\in\mathscr L$, and $\exists x(x=c)$ for each constant letter $c\in\mathscr L$.
Cantor-Schröder-Bernstein now gives that there are exactly $\kappa$ existential sentences.
In fact, however, the proof would still work well enough if there were fewer than $\kappa$ existential sentences; just choose some of the $\varphi_\alpha$s to be equal! We need a witness for each existential sentence, but it doesn't harm anything if, for lack of enough properties to witness, some of the existential sentences get more than one witness constant.
That's not what it actually says (and not what is needed; see above remarks about reusing $\varphi_\alpha$s), but it may be most intuitive to imagine that.
In that case you would need to show that there are at least $\kappa$ sentences that can be made with $\mathscr{L}_{\kappa\cdot n}$ but not with $\mathscr L_{\kappa\cdot(n-1)}$. But the sentences $\exists x(x=c_\alpha)$ for every $\alpha$ between $\kappa\cdot(n-1)$ and $\kappa\cdot n$ will do fine for that.
As long as the language allows you to form $\exists x(x=c_\alpha)$ for each different $c_\alpha$, there will be enough sentences. The size of $\mathcal L$ doesn't matter for this -- except if there are more than countably many symbols in $\mathscr L$, in which case they're going to push $\kappa$ itself upwards.
If $=$ is not a primitive logical symbol for you, it needs to be tacitly assumed that $\mathscr L$ does contain at least one predicate letter of nonzero arity, such that every term can be put into a wff somehow. But if it doesn't, then the completeness theorem holds vacuously, because everything collapses into propositional logic at best.