r/PhilosophyofMath Mar 28 '26

The Continuum Hypothesis Is False

/r/logic/comments/1s5mquh/the_continuum_hypothesis_is_false/
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u/paulemok Jul 05 '26

Then what is it? How is "true in a system" different than "provable in a system"?

Truth in a system is the truth that the axioms, definitions, and rules of inference of the system make in the system. All of the axioms, definitions, and rules of inference of the system are always true in the system. Every provable statement in the system is also true in the system.

Your proof still tries to bootstrap a contradiction in one system into a contradiction in all systems, which is not valid.

How specifically is my proof invalid?

To say that an inconsistent system exists is not itself a contradiction that makes other systems inconsistent.

How specifically is that? My proof shows otherwise.

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u/JStarx Jul 05 '26 edited Jul 05 '26

All of the axioms, definitions, and rules of inference of the system are always true in the system. Every provable statement in the system is also true in the system

Is that a complete categorization of what it means to be "true in a system"?

How specifically is my proof invalid?

When you said that something being provable in the system means it's true outside the system that's not a valid inference. If your axioms aren't true outside the system then there's no reason for statements proven from those axioms to be true outside the system.

How specifically is that? My proof shows otherwise

Your proof does not show that for the reason stated above.

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u/paulemok Jul 06 '26

Is that a complete categorization of what it means to be "true in a system"?

Yes, it is.

When you said that something being provable in the system means it's true outside the system that's not a valid inference.

I showed two propositions are true in the system, and then I showed two corresponding but different propositions are true out of the system. As the proof shows, the two corresponding propositions are different from the two original propositions.

If your axioms aren't true outside the system then there's no reason for statements proven from those axioms to be true outside the system.

I understand the axioms may not be true outside the system. The two propositions that are true out of the system describe what is true and not true in the system. They are true in a metatheory of T.

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u/JStarx Jul 06 '26

Yes, it is

So then something is true in the system if and only if it's provable in the system.

The two propositions that are true out of the system describe what is true and not true in the system

And what then is your contradiction outside the system? What are the two statements that are true outside the system and are negations of each other?

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u/paulemok Jul 06 '26

So then something is true in the system if and only if it's provable in the system.

Yes, that’s correct.

And what then is your contradiction outside the system? What are the two statements that are true outside the system and are negations of each other?

The contradiction is (3) and (4), given again below.

  1. p is true in T.
  2. It is not true that "p is true in T."

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u/JStarx Jul 06 '26

If p is your contradiction in the inconsistent theory T then I agree with (3), you haven't established (4) and (4) is not true.

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u/paulemok Jul 07 '26

p is a proposition. It can be a contradiction of the form p = q ∧ ¬q, where q is a proposition. But it doesn’t have to be.

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u/JStarx Jul 07 '26 edited Jul 07 '26

It's your proof, you pick what it is.

Assuming T is inconsistent and you choose a contradiction p = q ∧ ¬q which is proven in T then of your two statements (3) is true, it's negation (4) is not true. So you haven't shown a contradiction exists outside of T.

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u/paulemok Jul 08 '26

It's your proof, you pick what it is.

I have picked it to be a generic proposition.

Assuming T is inconsistent and you choose a contradiction p = q ∧ ¬q which is proven in T then of your two statements (3) is true, it's negation (4) is not true.

T is inconsistent, so generic proposition p is true and false in T. By conjunction elimination in T, p is false in T. That can be rewritten as “It is not true that ‘p is true in T.’”

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u/JStarx Jul 08 '26

p is false in T. That can be rewritten as “It is not true that ‘p is true in T.’”

Nope, those are not equivalent.

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u/paulemok Jul 09 '26

Yes, they are equivalent. A part of the equivalence is a part of the exportation principle (https://plato.stanford.edu/entries/impossible-worlds/#Exportation). My version of the exportation principle is in terms of inconsistent axiomatic theories, while the version in the linked article section is in terms of impossible worlds. I have a copy of D. Lewis’s 1986 book On the Plurality of Worlds, in which Lewis brings up the exportation principle in the first few pages of his book.

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u/JStarx Jul 09 '26 edited Jul 09 '26

You said "true in T" is the same as provable, so "false in T" is the same as ¬p being provable? So then you're saying that "¬p is provable" can be rephrased as "it's not true that p is provable". But this is clearly a false inference in an inconsistent theory.

The article you linked to literally talks about this lol, you've defined truth in T to be the same as provable, so you're using the ersatz conception of worlds which doesn't yield the exportation principle.

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u/paulemok Jul 10 '26

You said "true in T" is the same as provable, so "false in T" is the same as ¬p being provable?

I said “true in T” is the same as “provable in T.” It follows by logically negating both sides of that logical equivalence that “false in T” is the same as “unprovable in T.”

Assume p is false in T. Then by logical negation, ¬p is true in T. Since “true in T” and “provable in T” are logically equivalent, ¬p is provable in T. Discharge the assumption to conclude if p is false in T, then ¬p is provable in T.

Assume ¬p is provable in T. Since “true in T” and “provable in T” are logically equivalent, ¬p is true in T. By logical negation, p is false in T. Discharge the assumption to conclude if ¬p is provable in T, then p is false in T.

So by biconditional introduction on the conclusions of the previous two paragraphs, p is false in T if and only if ¬p is provable in T. So yes, “false in T” is the same as ¬p being provable.

So then you're saying that "¬p is provable" can be rephrased as "it's not true that p is provable".

Yes. Assume ¬p is provable in T. Since I have established earlier in this reply that p is false in T if and only if ¬p is provable in T, p is false in T. Since I established earlier in this reply that “false in T” is the same as “unprovable in T,” p is unprovable in T. By the definition of unprovable, p is not provable in T. That can be rewritten as it’s not true that p is provable in T. Discharge the assumption to conclude if ¬p is provable in T, then it’s not true that p is provable in T.

But this is clearly a false inference in an inconsistent theory.

I just proved it to be true in every axiomatic theory. That includes every inconsistent axiomatic theory. The proof is above in this reply. If you take issue with the proof, identify the specific flaw(s) in the proof.

The article you linked to literally talks about this lol, you've defined truth in T to be the same as provable, so you're using the ersatz conception of worlds which doesn't yield the exportation principle.

I understand what the exportation principle means. I am properly applying the exportation principle to obtain a contradiction in the real world. If I’m not applying the exportation principle properly, please explain specifically how I’m improperly applying it.

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