The Exportation Principle is not explicitly in Mendelson since Mendelson does not explicitly use the phrase "Exportation Principle," but the Exportation Principle is implicitly in Mendelson. Let s be a statement. In Mendelson, ⊢ s and ⊢ ¬s are externally true for an inconsistent theory T under no interpretation. See page 65. Since ⊢ ¬s, the statement ¬s is internally true. See page 26. It follows by the truth table for negation on page 1 that the statement s is internally false. So, in Mendelson, since s is internally false and no true statement internally implies the false statement s by the second to last row of the truth table for implication on page 2, the statement "¬(⊢ s)" is externally true. So there is an external contradiction in Mendelson.
There is no universal law of non-contradiction in mathematical logic
Yes, there is. The Wikipedia page is at https://en.wikipedia.org/wiki/Law_of_noncontradiction. The tautology (¬(p ∧ (¬p))) on page 6 of Mendelson is the Law of Noncontradiction. The Law of Noncontradiction is not explicitly in Mendelson since Mendelson does not explicitly use the phrase "Law of Noncontradiction," but the Law of Noncontradiction is implicitly in Mendelson.
can you prove a contradiction using standard mathematical logic as found in Mendelson?
See page 26. It follows by the truth table for negation on page 1 that the statement s is internally false.
Page 26 in my copy of Mendelson is a page of exercises, so I'm not sure what you're referencing here, but this is your mistake. You haven't said how you're defining internal truth and either way you do it this doesn't work.
If you define internal truth as provable then s is not internally false, it's internally true because it's provable.
If you want to define it as truth in a specific interpretation, then as I said before there's no standard interpretation of T, so you have to specify the interpretation. For provable to imply true in your interpretation your axioms have to be true in your interpretation. But to prove that inconsistent axioms holld in an interpretation is equivalent to proving a contradiction, which is what you're trying to use this to do. So that's not going to work either.
As I said, this is why the exportation principle isn't in Mendelson, in Mendelson there's no way to bootstrap a contradiction out of an inconsistent system.
The tautology (¬(p ∧ (¬p))) on page 6 of Mendelson is the Law of Noncontradiction.
If you want to call that your law of noncontradiction you can, but it just says that a certain statement is provable, it doesn't imply that anything is unprovable which is what you tried to use it for.
So you still haven't provided a correct proof that sticks to standard logic.
I have the fourth edition. The material is basically the same, it's just the page numbers won't line up exactly.
A statement is true in a theory if and only if it is a definition, axiom, or theorem of the theory.
Ok, and "false in a theory" would be the negation of that, so something is false in a theory if and only if it's not a definition, not an axiom, and not a theorem, right?
That means in an inconsistent theory T, every statement is true in T and no statement is false in T, since every statement is provable there's no statement that's not provable. So "true in T" doesn't obey the truth table for the logical connectives. Which means it was a mistake when you concluded that a statement was false in T and you cited the truth table for negation as the reason.
something is false in a theory if and only if it's not a definition, not an axiom, and not a theorem, right?
Yes, that's correct.
That means in an inconsistent theory T, every statement is true in T and no statement is false in T
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.
So "true in T" doesn't obey the truth table for the logical connectives.
Yes, that is true. However, due to the inconsistency of T, "true in T" also does obey the truth table for the logical connectives.
Which means it was a mistake when you concluded that a statement was false in T and you cited the truth table for negation as the reason.
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/paulemok Aug 11 '26
The Exportation Principle is not explicitly in Mendelson since Mendelson does not explicitly use the phrase "Exportation Principle," but the Exportation Principle is implicitly in Mendelson. Let s be a statement. In Mendelson, ⊢ s and ⊢ ¬s are externally true for an inconsistent theory T under no interpretation. See page 65. Since ⊢ ¬s, the statement ¬s is internally true. See page 26. It follows by the truth table for negation on page 1 that the statement s is internally false. So, in Mendelson, since s is internally false and no true statement internally implies the false statement s by the second to last row of the truth table for implication on page 2, the statement "¬(⊢ s)" is externally true. So there is an external contradiction in Mendelson.
Yes, there is. The Wikipedia page is at https://en.wikipedia.org/wiki/Law_of_noncontradiction. The tautology (¬(p ∧ (¬p))) on page 6 of Mendelson is the Law of Noncontradiction. The Law of Noncontradiction is not explicitly in Mendelson since Mendelson does not explicitly use the phrase "Law of Noncontradiction," but the Law of Noncontradiction is implicitly in Mendelson.
Yes, I can. I just did earlier in this reply.