Tableau Confutation means No Model

Lemma

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

If $\mathbf H$ has a tableau confutation, then $\mathbf H$ has no model.

Proof

Let:

• $\mathbf H$ be a hypothesis set;

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

Suppose $\mathbf H$ has a model $\mathcal M$.

Then by Model of a Branch of a Propositional Tableau, there is a branch $\Gamma$ of $T$ where each of the WFFs in $\mathbf H$ are modelled by $\mathcal M$.

Since, by hypothesis, every branch of $T$ is contradictory, there is a WFF $\mathbf A$ such that both $\mathbf A$ and $\neg \mathbf A$ are on $\Gamma$.

But this is a contradiction, as No Model Satisfies a WFF and its Negation.

Hence there can be no such model $\mathcal M$.

$\blacksquare$

Sources

• H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability (1996): $\S 1.8$: Lemma $1.8.2$.