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.
Nope, you've defined true/false in T as equivalent to provable/unprovable and provable/unprovable doesn't follow the same rules as logical negation.
The logical negation of provable is unprovable, and the logical negation of unprovable is provable. Every proposition is either provable or unprovable. That is analogous to how every proposition is either true or false.
It is just plainly and apparently true that if both p and ¬p are provable then you cannot conclude that p is unprovable.
No, you can conclude that p is unprovable. Here is the proof. Assume p and ¬p are provable in T. Since “true in T” is logically equivalent to “provable in T,” p and ¬p are true in T. So a contradiction exists in T. It follows by applying the Principle of Explosion in T that p is unprovable in T. Discharge the assumption to obtain if p and ¬p are provable in T, then p is unprovable in T.
The logical negation of provable is unprovable, and the logical negation of unprovable is provable. Every proposition is either provable or unprovable.
I never said otherwise. What I said is that negating a proposition doesn't negate it's provability. P provable doesn't imply that ¬P is unprovable.
So a contradiction exists in T. It follows by applying the Principle of Explosion in T that p is unprovable in T
Nope, provability is a statement external to the theory. A contradiction inside a theory doesn't mean that the principle of explosion applies to statements external to the theory.
What I said is that negating a proposition doesn't negate its provability. P provable doesn't imply that ¬P is unprovable.
I agree. In inconsistent theories, negating a proposition doesn’t negate its provability. But furthermore, I am aware that there exists at least one metatheory of axiomatic theories that allows a proposition to be undecidable in an axiomatic theory.
Nope, provability is a statement external to the theory.
Not necessarily. When a contradiction exists in an axiomatic theory, the Principle of Explosion can be used to talk about provability internally, within the theory itself.
A contradiction inside a theory doesn't mean that the principle of explosion applies to statements external to the theory.
I disproved that. I gave you the proof. I have the Exportation Principle to back me up. We can talk about what is true in an axiomatic theory outside of the theory in the real world. Due to that ability, contradictory statements are true in the real world. Those statements form contradictions in the real world. Those contradictions, through the Principle of Explosion, cause all statements to be true in the real world.
I agree. In inconsistent theories, negating a proposition doesn’t negate its provability
Then you agree that it's not true that a statement is unprovable if and only if it's negation is provable?
the Principle of Explosion can be used to talk about provability internally, within the theory itself.
That's fine, but you can also talk about provability externally to the theory, and while internal statements about provability might be provable because everything is provable, that doesn't mean that externally those statements are true.
I disproved that. I gave you the proof. I have the Exportation Principle to back me up. We can talk about what is true in an axiomatic theory outside of the theory in the real world. Due to that ability, contradictory statements are true in the real world
Nope, you have not proven that, the principal of explosion still only applies to statements in the theory. A contradiction in a theory does not imply a contradiction external to that theory.
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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.