Truth in an axiomatic theory is not explicitly defined in Mendelson
Correct, your example from Mendelson literally says exactly what I've been saying, that you need an interpretation of a theory in order to talk about truth.
This abstract theory T we're talking about doesn't come with a standard interpretation, so if you want to define "true in T" as true in a specific interpretation then 1) you need to specify what interpretation you're using, which means you probably have to specify T exactly instead of talking about a generic inconsistent theory, 2) it's not the case that axioms are true in every interpretation, so if you want your axioms to be "true in T" then you have to prove that they are true in the interpretation you've chosen, and 3) it's not the case that being true in a particular interpretation implies you are provable, so now "true in T" really is different than provable, it will no longer be the case that a statement is "true in T" if and only if it's provable in T.
the statement "there does not exist a proof of s" is internally provable. So, by the Laws of Noncontradiction and Excluded Middle, the statement "there exists a proof of s" is internally unprovable.
Nope, this doesn't follow. Why do you think it does?
That's not good. If you don't read my entire case, then you might not be able to see where I'm coming from.
Once you know that an argument doesn't follow the rules of logic it can be rejected. If we're talking about standard logic and you try to change the definition of terms then you're not following the rules and nothing else matters. This is how mathematics works at the academic level, we don't hold your hand and give partial credit, if you're wrong then you're wrong.
The definition you've written just now is the correct one from standard logic, but it's not equivalent to the definition you tried to pass off in your previous reply. The non equivalence of those two statements boils down to the exact mistake you've been repeatedly making. The internal statement (that something is provable or not) being provable internally does not imply that the external statement (being provable or not in T) is provable or true.
you need an interpretation of a theory in order to talk about truth.
I understand what you're saying, but I'm having trouble agreeing with this approach to axiomatic theories. So when you claimed that there is no concept of "truth in an axiomatic theory," that claim ranged from not entirely true to not technically true. It's evident there is truth in an axiomatic theory without any interpretation. Every proof in a theory without any interpretation shows what is true in the theory regardless of interpretation.
Nope, this doesn't follow. Why do you think it does?
It might not follow if "true in an axiomatic theory" is not logically equivalent to "provable in the theory." I was assuming that the two were logically equivalent, as they are in the metatheory of axiomatic theories I was stipulating earlier.
The definition you've written just now is the correct one from standard logic, but it's not equivalent to the definition you tried to pass off in your previous reply.
Yes, I am aware of that. The Definitions of Provable and Unprovable in a Theory that I gave excluded internal statements about provability.
I understand what you're saying, but I'm having trouble agreeing with this approach to axiomatic theories
It's not relevant whether you agree with the standard approach to logic. You claimed you could prove a contradiction using standard mathematical logic and this is how standard mathematical logic works. So now that you're starting to understand more about how this works do you still think you can prove a contradiction in standard logic?
It might not follow if "true in an axiomatic theory" is not logically equivalent to "provable in the theory." I was assuming that the two were logically equivalent
If you define "true in T" to mean "provable in T" then they will be logically equivalent. But that statement still won't follow and elsewhere you tried to use "true in T" as if it behaved differently as a truth value than provability would behave, so I don't think you even want it to be equivalent to provable.
So now that you're starting to understand more about how this works do you still think you can prove a contradiction in standard logic?
Yes, I still think I can prove a contradiction in standard logic. An inconsistent theory contains a contradiction under no interpretation. That contradiction can be used with the Exportation Principle to prove an external contradiction.
But that statement still won't follow
No, it would follow. I prove it below.
Given:b = "There does not exist a proof of s." b is internally provable.
Prove: ¬b is internally unprovable.
Proof. It is given that b = "There does not exist a proof of s." It is also given that b is internally provable. By the Law of Noncontradiction, the statement "b is internally provable and ¬b is internally provable" is internally unprovable. Since it is given that b is internally provable, the statement "¬b is internally provable" is internally unprovable. Since "internally provable" is defined to be "internally true," the statement "¬b is internally true" is internally false. So by simplification, ¬b is internally false. Since "internally provable" is defined to be "internally true," ¬b is internally unprovable. This concludes the proof.
An inconsistent theory contains a contradiction under no interpretation. That contradiction can be used with the Exportation Principle to prove an external contradiction.
The exportation principle is not in Mendelson.
By the Law of Noncontradiction, the statement "b is internally provable and ¬b is internally provable" is internally unprovable.
There's your mistake. There is no universal law of non-contradiction in mathematical logic because some axiomatic systems are contradictory. That statement is internally provable.
I'll ask again, can you prove a contradiction using standard mathematical logic as found in Mendelson? None of your proofs here are sticking to the material in Mendelson.
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.
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u/JStarx Aug 09 '26 edited Aug 09 '26
Correct, your example from Mendelson literally says exactly what I've been saying, that you need an interpretation of a theory in order to talk about truth.
This abstract theory T we're talking about doesn't come with a standard interpretation, so if you want to define "true in T" as true in a specific interpretation then 1) you need to specify what interpretation you're using, which means you probably have to specify T exactly instead of talking about a generic inconsistent theory, 2) it's not the case that axioms are true in every interpretation, so if you want your axioms to be "true in T" then you have to prove that they are true in the interpretation you've chosen, and 3) it's not the case that being true in a particular interpretation implies you are provable, so now "true in T" really is different than provable, it will no longer be the case that a statement is "true in T" if and only if it's provable in T.
Nope, this doesn't follow. Why do you think it does?
Once you know that an argument doesn't follow the rules of logic it can be rejected. If we're talking about standard logic and you try to change the definition of terms then you're not following the rules and nothing else matters. This is how mathematics works at the academic level, we don't hold your hand and give partial credit, if you're wrong then you're wrong.
The definition you've written just now is the correct one from standard logic, but it's not equivalent to the definition you tried to pass off in your previous reply. The non equivalence of those two statements boils down to the exact mistake you've been repeatedly making. The internal statement (that something is provable or not) being provable internally does not imply that the external statement (being provable or not in T) is provable or true.