Tableau Confutation contains Finite Tableau Confutation

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

Let $T$ be a tableau confutation of $\mathbf H$.

Then there exists a finite rooted subtree of $T'$ that is also a tableau confutation of $\mathbf H'$.

Proof
For each node $v \in T$, let $p (v)$ be the path from $v$ to $r_T$, the root of $T$.

This path is unique by Path in Tree is Unique.

Let $\mathcal V$ be the subtree of $T$ consisting those nodes $v$ of $T$ such that $p (v)$ is not contradictory.

Suppose that $\mathcal V$ were infinite.

Then by König's Tree Lemma, $\mathcal V$ has an infinite branch $\Gamma$.

Since $\mathcal V \subseteq T$, it follows that $\Gamma$ is also a branch of $T$.

However, by construction, it is impossible that $\Gamma$ is contradictory.

This contradicts that $T$ is a tableau confutation.

Hence $\mathcal V$ is finite.

Next, define a finite propositional tableau $T'$ by:


 * $v \in T' \iff \pi (v) \in \mathcal V$

that is, the rooted tree formed by $\mathcal V$ and all its children.

Then by construction, for each leaf node $v$ of $T'$, we have that $v \notin \mathcal V$.

That is, that $p (v)$ is a contradictory branch of $T'$.

By Leaf of Rooted Tree is on One Branch, every branch of $T'$ is contradictory.

Hence $T'$ is a tableau confutation of $\mathbf H'$, as desired.

Comment
Some sources define a tableau confutation to be a finite propositional tableau.

This result establishes that this distinction is not of material importance.

Also see

 * Definition:Tableau Confutation