I am reading Rod Girle's Modal Logics and Philosophy.

And I have a problem with one of the answers of the exercises.

In the exercise question, the reader has to provide a counterexample showing the invalidity of the sentence; (((∃x)(a ≠ x) & Fa) ⊃ (∃y)Fy)

The counterexample I provided is,

a∈D(n), b∈D(n), a∈/Q(n), b∈Q(n) Fa (n) = 1, Fb (n) = 0

I am pretty sure that my counterexample works. My problem is that the simpler counterexample Rod Girle provides doesn't seem to work.

His counterexample is; "World n has nothing in Q(n) but a ∈ D(n) and Fa (n) = 1 ( ∃x)(a ≠ x) (n) = 1; Fa (n) = 1; (∃y)Fy (n) = 0 ((1 & 1) ⊃ 0) (n) = 0 (n)"

I can't understand how this works. To be a counterexample, a model should make (∃x)(a ≠ x) (n) true, but how can (∃x)(a ≠ x) be true in n if there's nothing in Q(n)? Am I missing something about free logic? Please help.


I think you are right.

See page 133:

In free logic, the quantifiers have existential import, but constants do not. The formula $(Fa ⊃ (∃x)Fx)$ is not valid.

Why so ? Because we have to consider a "world" where $Fa$ is true but $(∃x)Fx$ is false, because the constant $a$ has not existential import.

We can check it verifying that the proof tree deos not close:

1) $\lnot (Fa ⊃ (∃x)Fx)$

2) $Fa$

3) $\lnot (∃x)Fx$

4) $(x) \lnot Fx$.

But we cannot use (FUI) to instantiate 4) with $\lnot Fa$ because we do not have $(∃x)(x = a)$.

Consider now:

$((∃x)(a ≠ x) \& Fa) ⊃ (∃y)Fy$.

Counterexample : World $n$ is empty, i.e. $Q(n)=\emptyset$ but $a ∈ D(n)$.

It seems to me that it does not work; in an empty domain, every existentially quantified formula $(∃x)Fx$ is false (there are no elements in $Q$ to satisfy it) and every universally quantified formula $(x)Fx$ is true (there are no elements in $Q$ to falsify it).

Thus, $(x)(x=a)$ is true because there are no objects in the domain of quantification $D$, and so $\lnot (x)(a=x)$, i.e. $(∃x) \lnot (a = x)$ is false.

Due to $a ∈ D(n)$, we have: $Fa [n] = 1$ but: $(∃x)(a ≠ x) [n] = 0$. $Q(n)$ is empty, and thus: $(∃y)Fy [n] = 0$.

Conclusion: $((0 \& 1) ⊃ 0) [n] = 1$.

We may "fix" it considering a one-element domain $Q$ with $b \in Q$ and not $Fb$.

In this case, we still have: $Fa [n] = 1$, due to $a ∈ D(n)$, but: $(∃x)(a ≠ x) [n] = 1$. Finally, not $Fb$, and thus: $(∃y)Fy [n] = 0$.

Again, we can check that the proof tree does not close:

1) $\lnot [ \ ((∃x)(a ≠ x) \& Fa) ⊃ (∃y)Fy \ ]$

2) $(∃x)(a ≠ x) \& Fa$

3) $\lnot (∃y)Fy$

4) $(∃x)(a ≠ x)$ --- from 2)

5) $Fa$ --- from 2)

6) $(a \ne b)$ --- from 4 by (FEI)

7) $(∃z)(z = b)$ --- from 4 by (FEI)

8) $(y) \lnot Fy$ --- from 3)

and we are stuck; we can apply (FUI) to 8) because we have : $(∃z) (z = b)$ to get $\lnot Fb$, but this does not close with 5).


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