How does Popper overcome this critique?

Suppose P and Q are falsifiable theories (in the Popperian sense). Then it seems to me that 'P and Q' is a falsifiable theory (we can refute it by refuting A, or by refuting B), and so too is 'P or Q' (we can refute it by refuting both A and B). However, it seems to me that, even if P and Q are falsifiable theories, the sentence 'if P, then Q' needn't be.

That's kind of weird, because for example the statement "if P and Q, then Q" is a logical tautology. Thus, its clearly true. But a naive reading of Popper seems to suggest that, for most choices of P and Q, this claim is unfalsifiable, and therefore unscientific.

How does one overcome this critique from a Popperian standpoint?

• Is P unprovable? If P can be demonstrated, then 'if P then Q' is not unfalsifiable. It can be falsified by demonstrating a case in which P is demonstrated but Q is refuted. Commented Aug 19, 2013 at 14:26
• What do you mean by 'if P then Q' ? Commented Aug 31, 2013 at 8:10
• Do you want to say it is weird that a logical tautology is not falsifiable? Commented Aug 31, 2013 at 8:13
• @KenB: Are there provable empirical theories? Commented Aug 31, 2013 at 8:33

1 Answer

If theories P and Q are falsifiable, then:

(1) there exists a finite set of observation sentences Γ such that ¬ P is a logical consequence of Γ,
(2) there exists a finite set of observation sentences Σ such that ¬ Q is a logical consequence of Σ.

Fact 1. If P is falsifiable, then (P ∧ Q) is also falsifiable, for any theory Q.

Proof. Suppose P is falsifiable. Then, by (1), there exists some Γ that implies ¬ P. But since Γ implies ¬ P, it also implies (¬ P ∨ ¬ Q), which is logically equivalent to ¬ (P ∧ Q ).                                           ■

Problem 2. If P is falsifiable, is then (P ∨ Q) also falsifiable, for any falsifiable theory Q?

Remark. I think the falsifiability of (P ∨ Q) doesn't follow from the falsifiability of P and Q, but so far my attempts to prove it have failed (see Updates, Sept. 3). Another way of formulating the problem is this: from the existence of falsifiers for P and Q can we infer the existence of a falsifier for (P ∨ Q)? To prove this, it will be sufficient to show that: the union of falsifying models for P and Q is a falsifying model for (P ∨ Q). The difficulty here, as pointed out by miracle173, is that we don't know whether the resulting set is consistent, so we can't infer that such a combined model exists.

Fact 3. Even if P and Q are falsifiable, (P → Q) needn't be.

Proof. Consider a model with only two worlds w and v s.t. w satisfies ¬ P and v satisfies (¬ P ∧ ¬ Q). Here, P is falsified in all worlds, Q is falsified by v, but (P → Q) ≡ (¬ P ∨ Q) is not falsified by any of the worlds, because neither world satisfies both (P and ¬ Q).                                                                   ■

Facts 1 and 3 help establish two of the three claims you made in paragraph 1. Your last claim, which came second in your first paragraph, is here turned into a question (Problem 2). I think it's going to be settled in the negative, but that remains to be seen. If you find the answer, please leave a comment.

Addendum. I'd like to offer two further observations that will help directly address the critique:

Fact 4. Tautologies are not falsifiable. (and that's a good thing!)

Proof. Take an arbitrary tautology s. If s is falsifiable, then (by definitions 1–2 above) there exists a set of observation sentences Γ such that ¬ s is a logical consequence of Γ. But Since s is a tautology, ¬ s is a contradiction, and therefore Γ implies a contradiction, i.e., it is inconsistent. But Γ is a set of observation sentences, so it cannot be inconsistent. Therefore: s is not falsifiable. And since s was arbitrary, we've shown that no tautology is falsifiable.                                                   ■

Fact 5. Contradictions are falsifiable. (not that anyone was unclear about this one)

Proof. Take an arbitrary contradiction s. Since s is a contradiction, ¬ s is a tautology, so it's a logical consequence of any sentence whatsoever. Take an arbitrary set of observation sentences Γ. From the previous two sentences we know that: ¬ s is a logical consequence of Γ. Since ¬ s is a logical consequence of a set of observation sentences (namely Γ), we know that s is falsifiable. And since s was arbitrary, we've shown that all contradictions are falsifiable.                                                     ■

That Popper's criterion doesn't conflict with Facts 4 and 5 is a good thing. It's okay if tautologies are not falsifiable, and it's okay to say that they're not "scientific." Is "2 + 2 = 4" scientific? No, because to settle it we don't need to appeal to observation at all. Only empirical or synthetic statements can be falsified (and thus be "scientific"). Popper's proposed criterion of demarcation attempts to tell good, "scientific" synthetic statements apart from bad, "unscientific" synthetic statements.

Updates

• Sept. 3, 2013 Today's discussion with Ken B convinced me that my attempts to settle Problem 2 negatively have failed, so I propose it as an open question. For solutions please leave a comment.
• Aug. 30, 2013 Today's revision was necessitated by miracle173's important criticisms. The OP had already pointed in this direction (10 days ago!), so many thanks to both for their criticisms.
• Aug. 29, 2013 Thanks to the anonymous editor for the corrections and improvements.

If you find errors or have suggestions, please leave a comment or simply edit the post.

• 1) In Fact2 you say nothing about the possibility that Gamma and Sigma may contradict. In the proof of Fact3 is is essential for you that the finite set of observations does not contradit. Commented Aug 30, 2013 at 7:11
• 2) In a comment above you say that "Fact 3 shows that a world that falsifies P and Q cannot falsify (P --> Q)". But Fact 3 says more. If there is a world that falsifies P and another world that falsifies Q then there may be another world that falsifies (P &Implies; Q). Commented Aug 30, 2013 at 7:12
• 3) Your proof of Fact3 looks strange to me. You derive it from "if P is falsifiable then Not P is not falsifiable". Is this sentence true? For me it is not evident that &Theta; &Union; &Gamma; have to be consistent. Why should they? Consistent here means that they do not imply a contradiction. Commented Aug 30, 2013 at 7:14
• @miracle173, user18921, thank you very much for the criticisms. Commented Aug 31, 2013 at 4:41
• This formal processing of falsifiability is new to me and there are some things unclear to me, e.g. what is exactly meant by a '' theory'' and therefore I don't understand precisely neither the definition of ''falsifiable'' nor expressions like ''if P then Q''. In your proof it seems to me that P, Q ... are simple sentences. Is it possible that you cite some references accessible in the internet? Commented Sep 2, 2013 at 15:04