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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?

DBK
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goblin GONE
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1 Answers1

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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.

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Hunan Rostomyan
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  • 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. – miracle173 Aug 30 '13 at 07:11
  • 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 ⇒ Q). – miracle173 Aug 30 '13 at 07:12
  • 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 Θ ⋃ Γ have to be consistent. Why should they? Consistent here means that they do not imply a contradiction. – miracle173 Aug 30 '13 at 07:14
  • @miracle173, user18921, thank you very much for the criticisms. – Hunan Rostomyan Aug 31 '13 at 04:41
  • 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? – miracle173 Sep 02 '13 at 15:04
  • @miracle173 _Theories_ are sets of _sentences_. We can say that theory T is falsifiable if and only if some sentence S in T is falsifiable. Definitions (1-2) are really just for sentences, but with that equivalence, everything said of sentences will also apply to the theories they're members of. As for references, these definitions above are just the general way of thinking about the logic of falsifiability (see the SEP article on Popper, section 3; and anything on the topic by logical empiricists). – Hunan Rostomyan Sep 02 '13 at 17:39
  • If theories P and Q are sets of sentences, how do you define the set *P # Q*, where *#* is an operation like *and*, *or*, *->*? I supposed by *P # Q = { p #q, where p from P and q from Q}*. But then simple idendities like *P= P and P* are not valid anymore. So how do you define *P # Q*? – miracle173 Sep 02 '13 at 21:42
  • When I said "everything said of sentences will also apply to the theories they're members of" I wasn't speaking carefully; what I wanted to say is that certain things true of the sentences can tell us things about the theories they're members of. For example, if s in P is falsifiable then P is falsifiable. But then if s in P is not falsifiable we can't conclude that P is not falsifiable. To keep things simple I just wanted to avoid talking of theories _and_ sentences. Let me know if you still want to ask about the meaning of sentences of form P # Q where # is a connective and P,Q are sets. – Hunan Rostomyan Sep 02 '13 at 22:01
  • without a definition of the operations on *theories* the proof does not make any sense because terms like *(P -> Q)* don't have any meaning. – miracle173 Sep 02 '13 at 23:16
  • In my answer I'm only _really_ talking about sentences, assuming that what's being said can be extended to theories as well. With the assumption that theories are _sets_ of sentences, you're right, the sentential operations on sets are obviously not defined. Two solutions: (i) define a function [X] from sets to sentences (consisting of a conjunction of members of X) and interpret (P # Q) as ([P] # [Q]), or (ii) interpret 'and' and 'not' for sets as set-intersection and complement. I'd go with (i), but (ii) might also work (have to work it out to be sure). – Hunan Rostomyan Sep 02 '13 at 23:43
  • @HunanRostomyan In your disproof for Claim2, you demostrate a single case in which PvQ is not *falsified*, but *falsified* isn't the same as *falsafiable*. It seems that the claim, "(PvQ) is falsifiable", is clearly true if P and Q are each independently falsifiable, for the simple reason given by the OP: if you falsify P and you falsify Q, then PvQ is *by definition* falsified. Am I missing something here? Claim2 seems intuitively true, and your disproof seems tangential and unrelated. Where am I confused? – Ken Bellows Sep 03 '13 at 12:22
  • @KenB Thanks Ken, but here is the idea behind it: P v Q is _falsifiable_ iff there exists a model where some world satisfies ~P and ~Q. When we know that P and Q are falsifiable, all we know is that (a) there exists a model where some world satisfies ~P, and (b) there exists a model where some world satisfies ~Q. From (a-b) we cannot make the conclusion that (c) there exists a model where some world satisfies ~P and ~Q. I briefly considered the possibility of constructing the model of (c) by taking the union of the (a-b) models, but what guarantees that the combined model is consistent?! – Hunan Rostomyan Sep 03 '13 at 16:39
  • Honest question in an attempt to learn more about a topic on which I am less than an expert: Your given definition of falsifiability seems specific to a particular system of modal logic. Which logical system are you working under here? Or perhaps, who established this definition, and where could I read more? Intuitively, I don't agree with it, but I imagine that I might if I understood it better. How is it useful to an argument to say something is falsifiable only if I *decide* or *imagine* that there exists a world where it's false? – Ken Bellows Sep 03 '13 at 16:55
  • Falsifiability, like satisfiability, is about _possibilities_, so a natural background logic for the logic of falsifiability is modal logic. I have in mind **K**, but I'm not against the application of weaker or stronger logics either. (This particular definition, (1-2) above, as far as I remember, traces back to Hempel or Ayer.) To say that something is falsifiable is exactly the same as to say that the negation of it is satisfiable, i.e., if you can conceive of a world in a model that makes its negation true. – Hunan Rostomyan Sep 03 '13 at 17:06
  • @KenB How do you feel about the fact that the falsifiability of P v Q entails the falsifiability of both P and Q? (This is the converse of Claim 2.) Does it sound right to you, intuitively? (It's a fact, on this view, because a world that satisfies ~P and ~Q satisfies ~P and satisfies ~Q, so falsifies both P and Q.) – Hunan Rostomyan Sep 03 '13 at 17:11
  • @HunanRostomyan Certainly ~(PvQ) entails (~P)^(~Q); that's just DeMorgan's law. I have no problem with that. But if I'm understanding you correctly, the counterexample would be a situation in which ~P and ~Q contradict each other or are otherwise related such that no world could ever possibly exist in which ~P^~Q is true? Is that accurate? – Ken Bellows Sep 03 '13 at 17:26
  • It's a bit more involved than that. Look at (a, b, c) a few comments above. All we can say is that ~P is satisfied on some world w and ~Q on some world v, possibly w != v. Claim 2 says that from this fact we can infer that some world u satisfies _both_ ~Q and ~P. But it's possible that a falsifier for P v Q exists, but the argument squeezed into Claim 2 is simply _invalid_: because the existence of a falsifier for P v Q doesn't follow from the existence of falsifiers for P and for Q. I can't think of an argument that would vindicate Claim 2, so if you come up with one, please let me know. – Hunan Rostomyan Sep 03 '13 at 17:35
  • @KenB I have updated my answer; thanks for your comments. – Hunan Rostomyan Sep 03 '13 at 20:14