Finished Set Lemma

Lemma
Let $\Delta$ be a finished set of propositional WFFs.

Then:
 * $\Delta$ has a model
 * Any model of the set of basic WFFs in $\Delta$ is a model of all the WFFs in $\Delta$.

Proof
Note that the set of basic WFFs in $\Delta$ has at least one model.

Let $\mathcal N$ be the model defined as follows:
 * $p_{\mathcal N} = \begin{cases}

T & : p \in \Delta \\ F & : p \notin \Delta \end{cases}$

Because $\Delta$ is finished, it is not contradictory, and hence $p_{\mathcal N} = F$ if $\neg p \in \Delta$.

Hence, any model $\mathcal M$ where: is a model of the basic WFFs in $\Delta$.
 * each $p$ occurring in $\Delta$ is true;
 * each $p$ such that $\neg p$ occurs in $\Delta$ is false

So, given one model of the set of basic WFFs in $\Delta$, we can get another one by changing the truth values of any propositional symbol $q$ such that neither $q$ nor $\neg q$ occur on $\Delta$.

So, let $\mathcal M$ be a model of the set of basic WFFs in $\Delta$.

We need to show that $\mathcal M \models \Delta$.

That is, that $\mathcal M \models \mathbf C$ for each $\mathbf C \in \Delta$.

Now, let $R \left({n}\right)$ be a propositional function on the set of natural numbers $\N$ such that:
 * $R \left({n}\right) = T$ iff: for every WFF $\mathbf C$, if $\mathbf C \in \Delta$ and $\mathbf C$ has length at most $n$, then $\mathcal M \models \mathbf C$.

It is clear that $R \left({0}\right), R \left({1}\right), R \left({2}\right)$ are true because every WFF length 2 or less is basic, and $\mathcal M$ models every basic WFF in $\Delta$.

So, assume $R \left({k}\right)$ is true for some $k \in \N$.

Suppose $\mathbf C$ has length at most $k+1$ and belongs to $\Delta$.

By examining each of the cases in the definition of finished set of propositional WFFs, we see that since $\mathcal M$ models every WFF in $\Delta$ of length at most $k$, then $\mathcal M$ models $\mathbf C$.

Thus $R \left({k+1}\right)$ is true.

Thus by strong induction, $R \left({n}\right)$ is true for all $n \in \N$.

Hence the result.