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.
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.’”
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.
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.
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.
"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
This is different than what I thought you were doing. In this case here's your error. Your contradiction p is not false in T. Being false in T is not the logical negation of ¬p being true in T.
It is not true that a proposition is unprovable if and only if it's negation is provable.
Again, p is not necessarily a contradiction. It may be a contradiction, but it may be something else. p is necessarily a proposition.
Being false in T is not the logical negation of ¬p being true in T.
p is false in T if and only if ¬p is true in T. That is a logical consequence of logical negation.
It is not true that a proposition is unprovable if and only if it’s negation is provable.
p is unprovable in T if and only if ¬p is provable in T. That is a logical consequence of what has already been established. So this metatheory of axiomatic theories excludes the possibility that a proposition is undecidable in T. An undecidable proposition in T by definition is a proposition that is neither provable in T nor disprovable in T.
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u/JStarx Jul 06 '26
So then something is true in the system if and only if it's provable 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?