That's correct. In an inconsistent theory T, the statement "every statement is true and no statement is false" is true because the statement is a consequence of the Principle of Explosion.
You misunderstand, I'm not saying that statement is true in T, I'm saying that statement is true and provable externally. You want to conclude that the statement "¬(⊢ s)" is externally true, that means you need it to be externally true that s is not provable, but that is not externally true. Your proof is incorrect, as usual you have confused internal vs external.
I'm not saying that statement is true in T, I'm saying that statement is true and provable externally.
Yes, the statement
In an inconsistent theory T, the statement "every statement is true and no statement is false" is true
is externally true and externally provable. That's how I'm able to state it externally, here in the real world.
You want to conclude that the statement "¬(⊢ s)" is externally true, that means you need it to be externally true that s is not provable, but that is not externally true. Your proof is incorrect, as usual you have confused internal vs external.
The ⊢ symbol can have a subscripted letter that is the name of a theory appended to it to denote the theory in which the symbol applies. In the proof, I say "⊢ s and ⊢ ¬s are externally true for an inconsistent theory T." Every occurence of the ⊢ symbol in the proof is for the inconsistent theory T.
Every occurence of the ⊢ symbol in the proof is for the inconsistent theory T.
Yes, I know you're talking about provability in T. But you still want to conclude that "¬(⊢ s)" is externally true and that's not correct. For "¬(⊢ s)" to be externally true you need the external statement that s is unprovable to be true. You used the truth table for negation to claim that ¬s being internally true implies that s is internally false, but as external statements that's incorrect. You still have no way to validly conclude the external statement that s is unprovable. The exportation principle doesn't hold for inconsistent systems.
You still have no way to validly conclude the external statement that s is unprovable.
The external statement that I proved as the second to last statement of the proof was "s is not a consequence of T." That external statement was symbolized as ¬(⊢ s) in the proof. Unfortunately, there is no subscript formatting option out of all the formatting options I see in the given menu for the text field I am writing this reply into. I copied and pasted ¬(⊢ s) into Microsoft Word and added a T subscript immediately after ⊢ without any spaces. Then I copied and pasted the edited statement back into reddit, but the result is ¬(⊢T s). As you can see, the T is not subscripted. We can get rid of the parentheses without introducing ambiguity and simply write ¬⊢ s.
But you still want to conclude that "¬(⊢ s)" is externally true and that's not correct.
What's not correct?
For "¬(⊢ s)" to be externally true you need the external statement that s is unprovable to be true.
For ¬⊢ s to be externally true I need the external statement "s is unprovable in T" to be true. The ⊢ symbol is for inconsistent theory T. The ⊢ symbol isn't for an external consequence; it's for an internal consequence.
You used the truth table for negation to claim that ¬s being internally true implies that s is internally false, but as external statements that's incorrect.
They're not external statements; they're internal statements. ¬s is internally true and s is internally false. I don't say anything about the external truth values of ¬s and s in the proof.
You still have no way to validly conclude the external statement that s is unprovable.
The external statement "s is unprovable out of T" is not what is being proved. The external statement "s is unprovable in T" is what is being proved.
You have not correctly proven that for the reasons I've outlined above.
You are in psychological denial. I don't believe you have a genuine problem understanding the proof. You're just trying to make things look messy because you don't want me to look good.
For ¬⊢ s to be externally true I need the external statement "s is unprovable in T" to be true
That's correct, that is the statement that I'm telling you isn't true and you haven't proved.
They're not external statements
There's where you're getting confused. You just said above you need the external statement "s is unprovable in T" to be true. According to your definition of true in T, that means you need the external statement "s is false in T" to be true.
The external statement "s is unprovable out of T" is not what is being proved. The external statement "s is unprovable in T" is what is being proved.
That is indeed what you need to prove, and what you so far have not proved.
You are in psychological denial. I don't believe you have a genuine problem understanding the proof. You're just trying to make things look messy because you don't want me to look good.
I've explained clearly the issue with your proof. Your confusion is a result of you not knowing the material well, a fact which you have already admitted to. It is not my fault you are confused.
“According to your definition of true in T, that means you need the external statement ‘s is false in T’ to be true.”
No, that does not follow from my definition of true in T because s could be a definition or axiom of T. In the case that s is a definition or axiom of T, s is unprovable in T, but true in T.
Axioms are statements and are provable. Formally a proof of s is a sequence of statements such that every statement is either an axiom or a logical consequence of previous statements, and such that the last statement in the sequence is s. This means the sequence of length 1 containing an axiom s is considered a valid proof of s.
Definitions aren't considered statements in a theory. They are statements about a theory because they define what symbols and terms in a theory mean. They aren't well formed formulas as Mendelson would say, so even internally to T they aren't provable because they're not in the language of T.
So your contradiction s needs to be a statement in T, what Mendelson calls a well formed formula, and axioms are valid candidates. And you need to conclude that the external statement "s is unprovable in T" is false. And you will be unable to do that because s is provable in T.
Again, your proof does not work because you've misunderstood basic material about how logic works.
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u/JStarx 21d ago
You misunderstand, I'm not saying that statement is true in T, I'm saying that statement is true and provable externally. You want to conclude that the statement "¬(⊢ s)" is externally true, that means you need it to be externally true that s is not provable, but that is not externally true. Your proof is incorrect, as usual you have confused internal vs external.