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.