Countable Hypothesis Set has Finished Tableau

Lemma
Let $$\mathbf H$$ be a countable set of propositional WFFs.

Then there exists a finished tableau whose root node is $$\mathbf H$$.

Proof
The Tableau Extension Lemma shows that each finite hypothesis set $$\mathbf H$$ is the root of some finished tableau.

It remains, then, to show that the result still applies when $$\mathbf H = \left\{{\mathbf A_1, \ldots, \mathbf A_n, \ldots}\right\}$$ is countably infinite.

Let $$\mathbf H_n = \left\{{\mathbf A_1, \ldots, \mathbf A_n}\right\} \subset \mathbf H$$.

We will say that a finite tableau $$T_n$$ with root $$\mathbf H$$ is finished for $$\mathbf H_n$$ if the tableau $$T'_n$$ is finished where:
 * $$T'_n$$ is the same as $$T_n$$ except:


 * $$T'_n$$ has the root $$\mathbf H_n$$ instead of $$\mathbf H$$.

We can use the Tableau Extension Lemma countably many times, and get a sequence of finite tableaus $$T_0, T_1, \ldots, T_n, \ldots$$ such that:
 * $$T_0$$ has only a root node;


 * For each $$n > 0$$, $$T_n$$ is an extension of $$T_{n-1}$$ such that $$T_n$$ is finished for $$\mathbf H_n$$;


 * For each $$n > 0$$, $$T_n$$ has the property that no contradictory branch $$\Gamma$$ of $$T_{n-1}$$ gets extended when forming $$T_n$$.

Now, let $$T$$ be the union $$T = \bigcup_{k=0}^\infty T_k$$.

Let $$\Gamma$$ be a branch of $$T$$.

Suppose $$\Gamma$$ is contradictory, with complementary pair $$\mathbf A, \neg \mathbf A$$.

Then $$\exists n \in \N$$ such that both $$\mathbf A$$ and $$\neg \mathbf A$$ are in $$T_n$$.

Then $$\Gamma \cap T_n$$ is already a contradictory branch of $$T_n$$.

So, by our method of construction of $$T$$, this branch $$\Gamma \cap T_n$$ is never extended past stage $$n$$, so $$\Gamma = \Gamma \cap T_n$$ and $$\Gamma$$ is finite.

On the other hand, suppose $$\Gamma$$ is not contradictory.

Then the construction ensures that $$\Gamma$$ is a finished branch.

So $$T$$ is a finished tableau whose root node is $$\mathbf H$$.