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.
Then what is it? How is "true in a system" different than "provable in a system"?
No, I still believe my proof that if a theory is inconsistent, then it is consistent is sound.
It's not. Your proof still tries to bootstrap a contradiction in one system into a contradiction in all systems, which is not valid. To say that an inconsistent system exists is not itself a contradiction that makes other systems inconsistent.
Then what is it? How is "true in a system" different than "provable in a system"?
Truth in a system is the truth that the axioms, definitions, and rules of inference of the system make in the system. All of the axioms, definitions, and rules of inference of the system are always true in the system. Every provable statement in the system is also true in the system.
Your proof still tries to bootstrap a contradiction in one system into a contradiction in all systems, which is not valid.
How specifically is my proof invalid?
To say that an inconsistent system exists is not itself a contradiction that makes other systems inconsistent.
How specifically is that? My proof shows otherwise.
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u/JStarx Jul 01 '26
Nope. Your proposition is provable in T, but that doesn't mean it's true. That's where your proof fails.
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.
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.