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I'm having a little bit of a logical fuss right now. Yesterday, I was happy with a statement like

$$ ((p \implies q) \wedge (\lnot p \implies q)) \implies q $$ because $p \vee \lnot p = \top$. (Let's stick to standard propositional logic, and work in $\sf ZFC$). Now if $p=\sf CH$, then the above statement makes me think twice. It is well known that $\sf CH$ is independent of $\sf ZFC$, so it feels weird to say $\sf{CH} \vee \lnot\sf{CH}=\top$.

I guess my question is this: Is it true that one of $\sf CH$ and $\lnot \sf CH$ is true in $\sf ZFC$, and that the statement of independence just tells us that we can't know which? Or is it actually the case that neither $\sf CH$ nor its negation are true in $\sf ZFC$ since there are models of $\sf ZFC$ where $\sf CH$ holds, as well as models where $\lnot \sf CH$ holds?

Willie Wong
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nullUser
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  • I removed the tag (independence) since the word has multiple meanings (within logic, probability, elementary set theory as applied to the physical sciences). – Willie Wong Aug 14 '13 at 07:53

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$\mathsf{CH}\lor\neg\mathsf{CH}$ is a tautology, so it is true outright in any system in which it can be stated. $\mathsf{ZFC}$ does not imply either $\mathsf{CH}$ or $\neg\mathsf{CH}$; there are models of $\mathsf{ZFC}+\mathsf{CH}$ and models of $\mathsf{ZFC}+\neg\mathsf{CH}$. (And I've actually seen a theorem $q$ proved in just this way: it was shown that $(\mathsf{CH}\to q)\land(\neg\mathsf{CH}\to q)$, from which $q$ followed trivially.)

Brian M. Scott
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    But if $\sf{CH} \vee \lnot \sf{CH}$ is true in $\sf ZFC$, then one of $\sf{CH},\lnot \sf{CH}$ is true in $\sf ZFC$. If $\sf{CH}$ is true, then how could there be a model of $\sf {ZFC} + \lnot CH$? Whatever made $\sf CH$ true in $\sf ZFC$ is still there, so why does it not remain true by when we add more axioms? – nullUser Aug 13 '13 at 20:22
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    @nullUser: Your first sentence is false. It is not the case that one of $\mathsf{CH}$ and $\neg\mathsf{CH}$ is true in $\mathsf{ZFC}$, though it is trivially true that their disjunction is. – Brian M. Scott Aug 13 '13 at 20:23
  • True independence means that each can be taken as an axiom, just like geometry has axioms ("straight lines that are parallel never meet") that seem as though they must be true or false. – abiessu Aug 13 '13 at 20:23
  • @Brian Wait, you're telling me that $p \vee q$ is not the same as $p$ or $q$? – nullUser Aug 13 '13 at 20:24
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    @nullUser: For example, let $P$ be the parallel postulate in Euclid, and let $\Sigma$ be the other axioms of Euclidean geometry. From $\Sigma$ we can deduce $P\lor\neg P$, simply because we can deduce any tautology, but we know that both $\Sigma\cup{P}$ (Euclidean geometry) and $\Sigma\cup{\neg P}$ (non-Euclidean geometry) are consistent. – Brian M. Scott Aug 13 '13 at 20:25
  • @nullUser: It appears that your understanding of English '$p$ or $q$' is different from the meaning of $p\lor q$ in formal logic. – Brian M. Scott Aug 13 '13 at 20:26
  • My understanding of $\vee$ was given by the usual truth table given the truth values of $p,q$. But in this case $p$ does not have a truth value? This is just confusing me more... Could you suggest an introductory reference on formal logic? – nullUser Aug 13 '13 at 20:32
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    @nullUser: Any given model of $\mathsf{ZFC}$ satisfies exactly one of $\mathsf{CH}$ and $\neg\mathsf{CH}$, so in any given model exactly one of them is true. The problem is simply that the axioms of $\mathsf{ZFC}$ are not strong enough to imply imply $\mathsf{CH}$ and are also not strong enough to imply $\neg\mathsf{CH}$. But we can still safely say that in any model of $\mathsf{ZFC}$ the statement $\mathsf{CH}\lor\neg\mathsf{CH}$ will be true! – Brian M. Scott Aug 13 '13 at 20:37
  • @BrianM.Scott If you assume the principle of bivalence, and CH* comes as an atomic proposition, and if (CH* ∨¬ CH) is true, then one of {CH, ¬ CH} holds true. In other words, in Polish notation, ...

    A C ApNp p C ApNp Np is a tautology. So do you deny CH as an atomic proposition?

     – Doug Spoonwood Aug 14 '13 at 16:27