Tableau Extension Lemma

Lemma
Every finite propositional tableau with a finite root $$\mathbf{H}$$ can be extended to a finite finished tableau with the same hypothesis set.

Proof
Let $$\mathbf{A}$$ be a propositional WFF at a node $$t$$ such that:


 * $$\mathbf{A}$$ is not a basic WFF;


 * There is a non-contradictory branch through $$t$$ on which $$\mathbf{A}$$ is not used.

Such a WFF $$\mathbf{A}$$ will be called unused.

Note that a tableau is finished iff there are no unused WFFs in the tableau.

Now, let $$\mathbf{H}$$ be a finite hypothesis set, which is to remain unchanged throught the course of this proof.

Suppose we have a particular finite tableau $$T$$ whose root is $$\mathbf{H}$$.

Let $$u \left({T}\right)$$ be the length of the longest unused WFF in $$T$$.

If $$T$$ is finished, we set $$u \left({T}\right) = 0$$.

Since there are only finitely many WFFs in $$T$$, the number $$u \left({T}\right)$$ must exist.

Let $$R \left({n}\right)$$ be the proposition that every finite propositional tableau with root $$\mathbf{H}$$ and with $$u \left({T}\right) < n$$ can be extended to a finite finished tableau.

That is, $$R \left({n}\right)$$ is an assertion that this lemma is true whenever $$u \left({T}\right) < n$$.

The statement $$R \left({1}\right)$$ is true, because a tableau $$T$$ such that $$u \left({T}\right) < 1$$ is already finished.

So, let us assume the truth of $$R \left({k}\right)$$.

Let us choose a finite tableau $$T$$ with root $$\mathbf{H}$$ and $$u \left({T}\right) < k+1$$.

We extend $$T$$ to a new tableau $$T'$$ by using every unused WFF $$\mathbf{A}$$ in $$T$$ once on every non-contradictory branch through $$\mathbf{A}$$.

Each of the unused WFFs in $$T$$ is used in the new tableau $$T'$$.

Also, each new WFF which was added when forming $$T'$$ has a length less than $$u \left({T}\right)$$, because the added WFFs are always shorter than the used WFFs.

So $$u \left({T'}\right) < u \left({T}\right) < k + 1$$, so $$u \left({T'}\right) < k$$.

So, by the induction hypothesis $$R \left({k}\right)$$, there is a finite finished extension $$T''$$ of $$T'$$.

$$T''$$ is also a finished extension of $$T$$.

Hence $$R \left({k+1}\right)$$, and the result follows by strong induction.