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Congratulations for the wonderful community you have in here. I've been trying to understand Henkin's proof of the semantic completeness of PL using Hunter's book and other internet sources. Although I can understand the details (or I think so), I still can't get the whole picture. Can someone please give me an intuitive summary of the main insight behind Henkin's proof? Just in order to see how the little details are interconnected in the whole picture. This is my attempt to do a brief summary of the kind I am asking you:

First, we assume that $\Gamma\vDash A$

From this, it follows (although it is not clear for me why, so please explain) that there does not exist a model for $\Gamma \cup \{\thicksim A\}$

Now, there is a theorem (which Henkin proves) which says that: "If $\Gamma$ is a p-consistent set of formulae in PL, then there is a model for $\Gamma$". Then, by the contrapositive of this theorem and modus ponens, it follows that $\Gamma \cup \{\thicksim A\}$ is p-inconsistent (intuitively, this set is inconsistent because it has no model). To summarize what we did until now: we have shown that the set $\Gamma \cup \{\thicksim A\}$ is p-inconsistent. So far, so good.

Why does this helps us? Because, as Henkin proves:

A set $\Gamma \cup\{\thicksim A\}$ is p-inconsistent iff $\Gamma\vdash A$

From that theorem, the fact that $\Gamma \cup\{\thicksim A\}$ is p-inconsistent, and modus ponens, we can conclude that $\Gamma\vdash A$.

QED

Is this right? I am missing something? Thanks!

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    We need the source clearly spelled out, and preferably typed in full into your question, with a specific and clear explanation of exactly what your concern is. I get the gist from this, but not any sense of how to really help... – 2017-01-18
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    Hello, I added more details to my question. Cheers! – 2017-01-18

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Just to explain that very first step:

$\Gamma \vDash A$ means that whenever all sentences in $\Gamma$ are true, $A$ will be true as well, which is the same as saying that for any model $M$: If $M \vDash \Gamma$ (i.e. If $M$ sets all sentences of $\Gamma$ to true), then $M \vDash A$ (i.e. $M$ sets $A$ to true). But this means that there cannot be a model for $\Gamma$ that sets $A$ to false, and hence not a model for $\Gamma$ that sets $\neg A$ to true. Hence, there is no model that sets all of the sentences in $\Gamma$ to true and that sets $\neg A$ to true as well. So, there is no model for $\Gamma \cup \{ \neg A \}$.