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.