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 ).
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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).
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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.
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