Doing an modal logic exercise. The tableau for both the argument and its negation are open. Am I doing something wrong?

Question

I am doing an exercise from Graham Priest's textbook "An Introduction to Nonclassical Logic", and I think I'm doing something wrong. The exercise is to determine the validity of α=β, ◇Pα, ⊢ ◇Pβ. I am using constant domain modal logic. The lowercase Greek letters represent non-rigid descriptors, whereas lowercase Latin letters represent rigid designators. Capital letters represent predicates. The accessibility relation is the same as system K (so it is not reflexive, symmetrical, transitive, or Euclidean). My tableau is as follows:

1. α=β, 0
2. ◇Pα, 0
3. ¬◇Pβ, 0
4. α=a, 0 (Assigning a rigid constant to descriptor)
5. β=b, 0 (Assigning a rigid constant to descriptor)
6. a=β, 0 (Identity, 1, 4)
7. a=b, 0 (Identity, 5, 6)
8. □¬Pβ, 0 (¬◇, 3).
9. 0R1 10.Pα, 1 (2, ◇)
10. α=c, 1 (Assigning a rigid constant to descriptor)
11. ¬Pβ, 1 (8, □)
12. β=d, 1 (Assigning a rigid constant to descriptor)
13. Pc, 1 (Identity, 10, 11)
14. ¬Pd, 1 (Identity, 12, 13)
15. a=b, 1 (Identity Invariance, 7)

I think that this tableau is correct and therefore this argument is invalid. However, I am a little confused because when I attempt to do a tableau of the same argument, except the conclusion is not negated, I am still left with an open tableau. Because the logic I am using has a classical metatheory, and thus bivalence and the law of excluded middle hold, shouldn't I end up with a closed tableau? Here is my tableau for the negation of the original problem in the textbook:

1. α=β, 0
2. ◇Pα, 0
3. ◇Pβ, 0
4. α=a, 0 (Assigning a rigid constant to descriptor)
5. β=b, 0 (Assigning a rigid constant to descriptor)
6. a=β, 0 (Identity, 1, 4)
7. a=b, 0 (Identity, 5, 6)
8. 0R1
9. Pα, 1 (◇, 2)
10. α=c, 1 (Assigning a rigid constant to descriptor)
11. Pc, 1 (Identity, 9, 10)
12. 0R2
13. Pβ, 2 (◇, 3)
14. β=d, 2 (Assigning a rigid constant to descriptor)
15. Pd, 2 (Identity, 13, 14)
16. a=b, 1 (Identity invariance)
17. a=b, 2 (Identity invariance)

I must have made a mistake somewhere and one of these tableaus should close. Perhaps I misunderstand some of the rules of this logic. If anybody can show me where I have made a mistake I would greatly appreciate it. Thank you.

P.S. This is not homework, I am just refreshing myself on modal logic in preparation for further study in philosophy.

Answer 1

The deduction is not valid.

Since $\alpha$ and $\beta$ are not rigid descriptors, it could be that $\alpha=\beta$ in a world, but not in the accessible world where $P\alpha$ holds. So there is no reason to expect that $P\beta$ is possible. One can make a countermodel with two worlds, where $\alpha=\beta$ in the first world, but not in the second world, where $P\alpha$ holds and $P_\beta$ fails in all worlds.

Answer 2

As it has already been mentioned, it is not the case that $\alpha=\beta$ and $\lozenge P(\alpha)$ entail $\lozenge P(\beta)$. However, there is one more important point to address.

You ask

I am a little confused because when I attempt to do a tableau of the same argument, except the conclusion is not negated, I am still left with an open tableau. Because the logic I am using has a classical metatheory, and thus bivalence and the law of excluded middle hold, shouldn't I end up with a closed tableau?

The answer to this is negative. It is perfectly possible in classical logic that a set of formulas $\Gamma$ does not entail a formula $\phi$, nor its negation $\neg\phi$. In fact, if there is no formula $\phi$ s.t. $\Gamma\not\models\phi$ and $\Gamma\not\models\neg\phi$, then $\Gamma$ is called a complete theory.

Of course, not every theory is complete: the simplest example is to take $\Gamma=\{\chi:\chi\text{ is valid}\}$ (or, equivalently, make $\Gamma$ empty) — in this case, $\Gamma\not\models P(\alpha)$ nor $\Gamma\not\models\neg P(\alpha)$ for any variable $\alpha$. There are other examples, e.g., $S(a)$ entails neither $Q(c)$ nor $\neg Q(c)$, etc. Similarly, $\{\alpha=\beta,\lozenge P(\alpha)\}$ is not complete.

A theory being incomplete does not contradict the law of excluded middle. This is because the entailment concerns all models (in all models where assumptions hold, the conclusion must also be true). On the other hand, the law of excluded middle says that in every model, either $\psi$ or $\neg\psi$ is true but, of course, does not say that any of them is true in all models.