Tuesday, 6 August 2013

Does the Truth of a Ramsey Sentence Reduce to its Observational Consequences?

From Ladyman, "Structural Realism" (SEP 2007; revision 2009):
As Demopoulos and Friedman point out, if $\Pi$ is consistent, and if all its purely observational consequences are true, then the truth of the corresponding Ramsey sentence follows as a theorem of second-order logic or set theory (provided the initial domain has the right cardinality—and if it does not then consistency implies that there exists one that does).
The wording is nearly the same as that of Demoupolos & Friedman 1985, "Bertrand Russell's Analysis of Matter: Its Historical Context and Contemporary Interest":
... if our theory is consistent, and if all its purely observational consequences are true, then the truth of the Ramsey-sentence follows as a theorem of set theory or second-order logic, provided our initial domain has the right cardinality--if it doesn't, then the consistency of our theory again implies the existence of a domain that does.
But in any case this is not quite right (though it can sort of be modified and fixed: see below). Let $\Theta$ be a finitely axiomatized theory in a language with an O/T distinction. Suppose that $\Re(\Theta)$ is its Ramsey sentence. Suppose that all observational consequences of $\Theta$ are true. It does not follow that "the truth of [$\Re(\Theta)$] follows as a theorem of second-order logic".

A counterexample to this is given in Ketland 2004, "Empirical Adequacy and Ramsification" (BJPS). That counter-example involved the Friedman-Sheard truth theory $FS$. But a simpler counter-example, which I learnt from reading a 1978 paper by van Benthem ("Ramsey Eliminability") around the time when my paper was published, is this. Suppose we think of numbers in $\mathbb{N}$, along with their properties definable in terms of $0$ and $S$, as observational. Let $I = (\mathbb{N}, 0, S)$ be the intended interpretation. So, $(L, I)$ is the interpreted observational language. So, $\phi$ is true if and only if $I \models \phi$. Let our observational theory be $Th_L((\mathbb{N}, 0, S))$.

Let us introduce a new theoretical predicate $F$. Let our additional axioms be
$F(0)$
$\forall x(F(x) \to F(s(x)))$
$\exists x \neg F(x)$ 
This theory is not finitely axiomatizable, but it has a finitely axiomatized conservative extension by adding a new satisfaction predicate. Call the extended language $L^{+}$. Let $T$ be the resulting theory in $L^{+}$. Note two things:
  • $T$ is consistent (using a compactness argument).
  • $T$ is $\omega$-inconsistent. 
So, $T$ has no model whose observational part is isomorphic to $(\mathbb{N}, 0, S)$. All of $T$'s models are non-standard. Let $\Re(T)$ be its Ramsey sentence. Then
(i) All observational consequences of $T$ are true.
(ii) But the truth of $\Re(T)$ does not follow. In fact, $\Re(T)$ is not true.
Given that the intended interpretation of $L$ is $I = (\mathbb{N}, 0, S)$, it follows that
$\Re(T)$ is true if and only if $I$ can be expanded to a model of $T$. 
But $T$ is $\omega$-inconsistent, and therefore no expansion of $(\mathbb{N}, 0, S)$ is a model of $T$. So, $\Re(T)$ is not true. QED.

I noticed this around 2000 or 2001, and worried about how to fix it for quite a while, so as to get something that approximates the claim in Demopoulos & Friedman 1985.

There is a fix, which is the one I used in 2004, "Empirical Adequacy and Ramsification". It involved changing the structure of the language used to formulate scientific theories to a 2-sorted language, including a separate range of variables for unobservable objects. In this setting, even given a fixed observational interpretation $I$, one can extend the domain (of a model) representing the unobservable objects and define various relations etc., as need be. But even so, one still requires that the original theory $T$ must have an empirically correct model $\mathcal{A}$: i.e., a model whose purely observational part is isomorphic to the intended one $I$. It is not sufficient that merely all theorems of $T$ in the observational language be true in $I$.

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