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Say, for example, I have this tableau:

\turnstile ∃x∀ySxy

How would one go about explaining how the infinite countermodel,

infinite countermodel

or the finite countermodel,

finite countermodel

actually work? In other words, how when looking through the tableau should you apply the truth values in order to prove your counter-interpretation?

Thanks for you help!

n.b. tableau and countermodels from Priest(2001).

1 Answers1

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Similar to your previous post…

If the tableau does not close, we can use the atomic formulas in it to define an interpretation satisfying the initial formula.

If we stop the tableau at $\lnot Sab$ we have the information needed to define an interpretation $\mathcal I$ with domain $I = \{ a,b \}$ (the ground terms used by the tableau up to that node).

In order to satisfy formula $\lnot Sab$, we need that $(a,b) \notin S^{\mathcal I}$.

This means that we may define : $S^{\mathcal I} = \{ (a,a), (b,b) \}$ to complete the interpretation $\mathcal I$.

Consider now the initial formula; its negation is $\exists x \forall y Sxy$. This formula is falsified by the above interpretation because there is no element $x$ in the domain $I$ such that $Sxy$ holds for all $y$ (i.e. for $a$ and $b$).

This means that the above interpretation satisfies the initial formula and if a formula is satisfiable, its negation is not valid.

Conclusion: having found a tableau that does not close, we have used it to manufacture a counter-model for the formula $\exists x \forall y Sxy$, and this means that the formula is not valid.

  • Thanks again for your quick and lucid answer Mauro! I'm having some trouble understanding what you mean by this statement: "This formula is falsified by the above interpretation because there is no element in the domain such that holds for all ". How does x not fit in the domain ? Does it not just represent some term? –  Oct 16 '19 at 11:24
  • @ev350 - it means that we cannot find an element in the domain such that every element in the domain is $S$-related to it. I need some way to name it... – Mauro ALLEGRANZA Oct 16 '19 at 11:58
  • I understand now, I went over it a few times. I'm wondering now though, how do you then deal with a situation that is not a tautology, but has some premise? Do you consider both formulae and then work out using the material conditional? –  Oct 16 '19 at 12:01
  • @ev350 - the Tableau method works also for logical consequence: $\Gamma \vDash \alpha$. The Tableau must start with all formulas in $\Gamma$ plus $\lnot \alpha$. Again, if the tree closes, this means that the initial set is unsatisfibale which is equiv to say that every interpretation satisying all of $\Gamma$ will satisfy also $\alpha$. See also the following post. – Mauro ALLEGRANZA Oct 16 '19 at 12:15