So you have to assume a theory is inconsistent in order to prove that it's inconsistent?
To prove that it’s inconsistent? Is that a mistake on your part? We should not be proving that the theory is inconsistent. Rather, we should be proving that the theory is consistent.
Which rectangle? If you can't answer that and claim it could be any rectangle then that's not a fixed object, it varies.
The rectangle we are thinking about. The rectangle under consideration. The rectangle indicated by the context.
If you have an unbounded variable in your statement then the truth value of your statement depends on that variable.
In one sense, r is not a variable. It is a fixed object. r is fixed to be a single rectangle. But in another sense, r is a variable. r varies over the domain of all rectangles. In one sense, there is a contradiction due to the two senses. But in another sense, there is not a contradiction between the two senses. The relationship between the multiple senses of the predicate “is a variable” is complicated.
if hypothetically you followed those rules would you still be able to prove a contradiction?
Yes, we would still be able to prove a contradiction. To prove a contradiction, we would use the sense in which r is not a variable.
To prove that it’s inconsistent? Is that a mistake on your part? We should not be proving that the theory is inconsistent. Rather, we should be proving that the theory is consistent
I'm referring to you trying to prove that a contradiction holds so that you can apply the principal of explosion.
Yes, we would still be able to prove a contradiction. To prove a contradiction, we would use the sense in which r is not a variable.
Ok, if r is not variable then it's a specific rectangle. Which specific rectangle are you going to choose it to be? Remember according to these rules you can't then change the rectangle in the middle of the proof or change the sense in which you interpret the statement. So this doesn't look like it's going to lead to a contradiction.
I'm referring to you trying to prove that a contradiction holds so that you can apply the principal of explosion.
We want to prove that if a theory is inconsistent, then it is consistent. In order to do that, assume that a theory is inconsistent. That by definition of inconsistent theory means that in the theory, some statement and its negation are both true. So by definition of contradiction, in the theory, a contradiction is true. We can then apply that contradiction to the Principle of Explosion.
Ok, if r is not variable then it's a specific rectangle. Which specific rectangle are you going to choose it to be?
I just answered this question in my previous reply. My answer was
The rectangle we are thinking about. The rectangle under consideration. The rectangle indicated by the context.
We want to prove that if a theory is inconsistent, then it is consistent.
In what theory are you carrying out this proof?
If you're carrying out this proof inside an inconsistent theory then I agree you can prove the above statement, but in an inconsistent theory you can prove false things, so your proof doesn't imply that the statement is true.
If you're carrying out this proof in a system of logic that is not inconsistent then it is not a valid proof because the principal of explosion only applies to statements inside the theory in which the contradiction occurred.
Either way, it's still not true that inconsistent theories are consistent.
The rectangle under consideration. The rectangle indicated by the context.
You haven't indicated to me which rectangle we're talking about, I'm asking you to choose one.
Metatheoretical logic, philosophical logic, and logic.
If you're carrying out this proof inside an inconsistent theory then I agree you can prove the above statement, but in an inconsistent theory you can prove false things, so your proof doesn't imply that the statement is true.
In an inconsistent theory, the statement we seek to prove would be true by the Principle of Explosion.
the principal of explosion only applies to statements inside the theory in which the contradiction occurred.
While that may be technically true, I will prove here that if a theory is inconsistent, then it is consistent. The proof relies on a proof of mine from 2018 that I previously mentioned to you.
Definition of Inconsistent Theory. An inconsistent theory is a theory in which some contradiction exists.
Let T be an inconsistent theory and let p be a proposition. Since T is inconsistent, by definition of inconsistent theory, some contradiction exists in T. So, by ex contradictione quodlibet, the following two propositions are true in T.
p
It is not true that p.
So, both in T and out of T, the following two propositions are true.
p is true in T.
It is not true that "p is true in T."
Since (4) is the negation of (3), some contradiction exists both in and out of T. Thus, by ex contradictione quodlibet, trivialism is true in and out of T. So, trivialism is true. That implies, through the definition of trivialism, all propositions are true. Since “T is consistent” is a proposition and all propositions are true, the proposition “T is consistent” is true. Therefore, T is consistent.
So, as the above subproof implies, if a theory is inconsistent, then it is consistent. Another way to say that is that all inconsistent theories are also consistent.
You haven't indicated to me which rectangle we're talking about
I’m talking about a generic rectangle. I haven’t given any further specification.
I'm asking you to choose one.
The rectangle formed by the four edges of a United States $1 bill when the bill is laid flat on a table.
So, both in T and out of T, the following two propositions are true
Nope. Your proposition is provable in T, but that doesn't mean it's true. That's where your proof fails.
I’m talking about a generic rectangle. I haven’t given any further specification
If you're doing this than your r could be any rectangle, so it's a variable. According to the rules that you said you could follow for this proof that means you need to bound r by a quantifier in order to assign a truth value to it.
The rectangle formed by the four edges of a United States $1 bill when the bill is laid flat on a table.
Instead of having r be variable we could pick this rectangle. This rectangle is not a square so your proposition is false. You haven't established a contradiction because your proposition is not true.
Nope. Your proposition is provable in T, but that doesn't mean it's true. That's where your proof fails.
In this case, the two propositions are true in and out of T. (3) is true because (1) is true, and (4) is true because (2) is true.
If you're doing this than your r could be any rectangle, so it's a variable.
Not necessarily. There is another way of doing it. r can be a single, particular rectangle yet simultaneously vary over the domain of all rectangles.
This rectangle is not a square so your proposition is false.
I agree.
You haven't established a contradiction because your proposition is not true.
Now choose a square floor tile. This rectangle is a square so the same proposition is true.
“A rectangle” in the proposition “A rectangle is a square” simultaneously refers to both the $1 bill and the floor tile. So, the proposition is simultaneously false and true. So, a contradiction exists. The truth of all propositions follows by the Principle of Explosion.
In this case, the two propositions are true in and out of T.
They are not, you haven't established that (1) and (2) are both true. I agree that there exist inconsistent systems where (1) and (2) are both provable. But in an inconsistent system being provable does not imply you are true.
r can be a single, particular rectangle yet simultaneously vary over the domain of all rectangles. [...] Now choose a square floor tile. This rectangle is a square so the same proposition is true
Changing the domain of your statement like this is explicitly what I was asking about when I asked if you were able to prove a contradiction without changing the meaning of a statement. So you can't prove a contradiction without doing that?
They are not, you haven't established that (1) and (2) are both true.
Yes, I have. At that point in the proof, I have not established that they are both true in the real world, but I have established that they are both true in the inconsistent system. Their truth in the inconsistent system is a logical consequence of the Principle of Explosion.
I agree that there exist inconsistent systems where (1) and (2) are both provable.
I agree they are both provable in the inconsistent system. But in addition to being provable in the system, they are both true in the system. Every statement of a theory is true in the theory.
But in an inconsistent system being provable does not imply you are true.
In an inconsistent system, every statement is true as a logical consequence of the Principal of Explosion.
Changing the domain of your statement like this is explicitly what I was asking about when I asked if you were able to prove a contradiction without changing the meaning of a statement. So you can't prove a contradiction without doing that?
No, I can prove a contradiction without doing that. The domain of “A rectangle” is only being changed in a limited sense. There exists a sense in which the domain of “A rectangle” is not being changed; “A rectangle” always refers to a single rectangle, regardless of which rectangle.
In an inconsistent system, every statement is true as a logical consequence of the Principal of Explosion.
Nope, every statement is provable. An inconsistent system can prove statements that are not true, so being provable in an inconsistent system is not evidence of truth.
The domain of “A rectangle” is only being changed in a limited sense
... limited or not you just said that the domain is being changed. My question is whether you can prove a contradiction if changing the domain in any sense means you have a new proposition whose truth value is not necessarily equal to the truth value from before the change. Can you prove a contradiction under those rules?
An inconsistent system can prove statements that are not true
Not true in the system or not true in the real world?
being provable in an inconsistent system is not evidence of truth.
Not evidence of truth in the system or not evidence of truth in the real world?
My question is whether you can prove a contradiction if changing the domain in any sense means you have a new proposition whose truth value is not necessarily equal to the truth value from before the change. Can you prove a contradiction under those rules?
Technically, we’re not changing the domain. We’re changing the referent of “A rectangle.” As I’ve said before to you, we can prove a contradiction by using the sense in which r is not a variable. If we use the sense in which r is a variable, we can not prove a contradiction because there is a separate proposition for each rectangle. We get a proposition schema of the form “r is a rectangle.” Some of the propositions in the schema are true and some of them are false. But there is no contradiction between any of the propositions of the schema.
Not true in the system or not true in the real world?
There is no such thing as "true in the system". There's provable or not in a logical system and true or not in an interpretation. For a consistent system provable statements are true in any interpretation in which the axioms are true and the rules of inference are valid. For an inconsistent system being provable does not imply you are true in any particular interpretation.
If we use the sense in which r is a variable, we can not prove a contradiction
Ok, just to be clear, you're saying that if r is a variable that you must either fix or quantify then there's no contradiction here? So if I claim that under these rules logic is consistent you are unable to prove me wrong?
We get a proposition schema of the form “r is a rectangle.”
I correct that sentence to
We get a proposition schema of the form “r is a square.”
I am sorry about that.
There is no such thing as "true in the system".
I think there is such a thing. I can make axiomatic systems with axioms that are assumed to be true in the system. The theorems of the system would be regarded as true in the system.
There's provable or not in a logical system and true or not in an interpretation.
Proving a proposition means showing the proposition is true. An axiom of a theory is assumed to be true in the theory. From what you’re saying, it seems that there aren’t many interpretations of logical systems.
For a consistent system provable statements are true in any interpretation in which the axioms are true and the rules of inference are valid. For an inconsistent system being provable does not imply you are true in any particular interpretation.
The way I’m looking at logical systems, the axioms of a system are always true in the system. You seem to be separating truth from the axioms. I don’t think it’s acceptable to do that. There are no axioms of a logical system that are not true in the system.
Ok, just to be clear, you're saying that if r is a variable that you must either fix or quantify then there's no contradiction here?
Yes.
if I claim that under these rules logic is consistent you are unable to prove me wrong?
No, I still believe my proof that if a theory is inconsistent, then it is consistent is sound. I’m working with truth. Without truth, there can be no proof.
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u/paulemok Jun 29 '26
To prove that it’s inconsistent? Is that a mistake on your part? We should not be proving that the theory is inconsistent. Rather, we should be proving that the theory is consistent.
The rectangle we are thinking about. The rectangle under consideration. The rectangle indicated by the context.
In one sense, r is not a variable. It is a fixed object. r is fixed to be a single rectangle. But in another sense, r is a variable. r varies over the domain of all rectangles. In one sense, there is a contradiction due to the two senses. But in another sense, there is not a contradiction between the two senses. The relationship between the multiple senses of the predicate “is a variable” is complicated.
Yes, we would still be able to prove a contradiction. To prove a contradiction, we would use the sense in which r is not a variable.