I mean that the unknown truth "p & ~Kp" of NonO cannot possibly become a known truth. If that is impossible from the outset, then so is knowability. — Luke

No it isn't. There are some things which are unknown truths which can become known, e.g. the number of coins in a jar.

To borrow Andrew M's example:

Suppose there are 163 coins in the jar and no-one knows there is.

It's thus true that there's 163 coins in the jar and no-one knows there is.

That true statement is unknowable. Why? Because anyone coming to know that there's 163 coins in the jar (say, by counting) would render the statement false (since the second conjunct would be false). The statement doesn't change from an unknown truth to a known truth. It changes from an unknown truth to a known falsity. — Luke

These are two different propositions:

1. There are 163 coins in the jar
2. There are 163 coins in the jar and no-one knows there is

It is possible that both propositions are true. It is possible that neither proposition is known to be true. It is possible to know the first proposition. It is not possible to know the second proposition. Therefore, the knowability principle is false.

I mean that the unknown truth "p & ~Kp" of NonO cannot possibly become a known truth. If that is impossible from the outset, then so is knowability.
— Luke

No it isn't. There are some things which are unknown truths which can become known, e.g. the number of coins in a jar. — Michael

Presumably, the unknown truth of the number of coins in a jar is not expressed as "p & ~Kp", since this is unknowable. So how would you express the unknown truth about the number of coins in a jar?

Presumably, the unknown truth of the number of coins in a jar is not expressed as "p & ~Kp", since this is unknowable. So how would you express the unknown truth about the number of coins in a jar? — Luke

1. p
2. p ∧ ¬Kp

Assume p is true. Both 1 and 2 are true. Neither 1 nor 2 are known to be true. 1 can be known to be true. 2 can't be known to be true.

1. p
2. p ∧ ¬Kp

Assume p is true. Both 1 and 2 are true. Neither 1 nor 2 are known to be true. 1 can be known to be true. 2 can't be known to be true. — Michael

1. does not express that it is unknown
2. expresses that it is unknown, but it is unknowable.

Therefore, the number of coins in the jar remains unknowable.

Therefore, the number of coins in the jar remains unknowable. — Luke

It isn't. We can count the coins and then we will know how many coins are in the jar.

1. does not express that it is unknown — Luke

Which is why it is possible to know it.

2. expresses that it is unknown, but it is unknowable. — Luke

Which is why the knowability principle is wrong.

Therefore, the number of coins in the jar remains unknowable.
— Luke

It isn't. We can count the coins and then we will know how many coins are in the jar. — Michael

Then this should be able to be expressed in the argument. If it cannot be expressed in the argument, then it is not a failure of the knowability principle, but a failure of logic. Otherwise, accept the logic and the number of coins in the jar is unknowable.

1. does not express that it is unknown
— Luke

Which is why it is possible to know it. — Michael

I asked how you would express (in logical notation) that it was unknown.

I asked how you would express (in logical notation) that it was unknown. — Luke

2 does that.

Then this should be able to be expressed in the argument. If it cannot be expressed in the argument, then it is not a failure of the knowability principle, but a failure of logic. Otherwise, accept the logic and the number of coins in the jar is unknowable. — Luke

I don't understand what you're asking for here. The argument simply shows that if you take the knowability principle and the non-omniscience principle as premises then it follows that the non-omniscience principle is false. It is then up to the reader to decide whether to accept that the non-omniscience principle is false or to reject the knowability principle.

So why can't you just accept that the knowability principle is wrong? Some truths are, in fact, unknowable.

I asked how you would express (in logical notation) that it was unknown.
— Luke

2 does that. — Michael

2 (when expressed as "p & ~Kp") is unknowable, which means that so is the number of coins in the jar.

I don't understand what you're asking for here. The argument simple shows that if you take the knowability principle and the non-omniscience principle as premises then it in fact follows that the non-omniscience principle is false. It is then up to the reader to decide whether to accept that the non-omniscience principle is false or to reject the knowability principle.

So why can't you just accept that the knowability principle is wrong? Some truths are, in fact, unknowable. — Michael

It is not possible to know any proposition of the form "p & ~Kp", which means that all unknown truths (expressed in this way, at least) are unknowable. An unknown truth cannot become a known truth, and vice versa. The result of the argument is therefore that all (known) truths are known and all unknown truths are unknowable, and never the twain shall meet. The conclusion is not a failure of KP, but a failure of logic.

On the one hand, you want me to accept the argument's implication that there is at least one unknowable truth, and that therefore KP must be rejected.

On the other hand, you do not accept the argument's implication that we cannot come to know mundane unknown truths such as the number of coins in a jar.

2 (when expressed as "p & ~Kp") is unknowable, which means that so is the number of coins in the jar. — Luke

That p ∧ ¬Kp is unknowable isn't that p is unknowable. The number of coins in the jar is p. We can know p.

It is not possible to know any proposition of the form "p & ~Kp", which means that all unknown truths (expressed in this way, at least) are unknowable. — Luke

This is where you have a fundamental misunderstanding that I don't know how to explain to you. Maybe like this?

a) p
b) a is not known to be true

Both a and b are true. Neither a nor b are known to be true. It is possible to know a but not possible to know b.

On the other hand, you do not accept the argument's implication that we cannot come to know mundane unknown truths such as the number of coins in a jar. — Luke

No it doesn't.

Look, smarter people than both of us have addressed Fitch's knowability paradox. None of them have argued that it somehow entails that all truths are unknowable; instead they accept that it shows that either the knowability principle is false or that every truth is known. Their solution to the problem (where they want to keep some form of the knowability principle) is to change the knowability principle. See Tennant's and Dummett's responses as detailed here.

That p ∧ ¬Kp is unknowable isn't that p is unknowable. — Michael

I'm asking you how else "p is unknown" could be expressed in logical notation - other than as "p ∧ ¬Kp", and other than as your mere assurance outside of logical notation that p is unknown.

It is not possible to know any proposition of the form "p & ~Kp", which means that all unknown truths (expressed in this way, at least) are unknowable.
— Luke

This is where you have a fundamental misunderstanding that I don't know how to explain to you. Maybe like this?

a) p
b) a is not known to be true

Both a and b are true. Neither a nor b are known to be true. It is possible to know a but not possible to know b. — Michael

I understand the conjunction. I don't see how this contradicts what I said.

I'm asking you how else "p is unknown" could be expressed in logical notation - other than as "p ∧ ¬Kp", and other than as your mere assurance outside of logical notation that p is unknown. — Luke

p ∧ ¬Kp is how you express it.

The problem is that you seem to go from "p ∧ ¬Kp" is unknowable to "p" is unknowable. And that just doesn't follow.

I'm asking you how else "p is unknown" could be expressed in logical notation - other than as "p ∧ ¬Kp", and other than as your mere assurance outside of logical notation that p is unknown.
— Luke

p ∧ ¬Kp is how you express it.

The problem is that you seem to go from "p ∧ ¬Kp" is unknowable to "p" is unknowable. And that just doesn't follow. — Michael

Please tell me where I am going wrong here:

The unknown truth that is the number of coins in the jar is expressed as: p ∧ ¬Kp

It is impossible to know the unknown truth: p ∧ ¬Kp

Therefore, it is impossible to know the unknown truth that is the number of coins in the jar.

Please tell me where I am going wrong here:

The unknown truth that is the number of coins in the jar is expressed as: p ∧ ¬Kp

It is impossible to know the unknown truth: p ∧ ¬Kp

Therefore, it is impossible to know the unknown truth that is the number of coins in the jar. — Luke

Here are two propositions:

1. the cat is on the mat
2. the cat is on the mat and the mat was bought from Ikea

Both are true, and even though the first proposition doesn't express it, the mat was bought from Ikea (as explained by the second proposition). And it's possible that I (eventually) know that the cat is on the mat but not that the mat was bought from Ikea (so I know the first but not the second).

Similarly:

3. the cat is on the mat
4. the cat is on the mat and nobody knows that the cat is on the mat

Both are true. And even though the third proposition doesn't express it, nobody knows that the cat is on the mat (as explained by the fourth proposition).

The issue is that it is possible to (eventually) know 3 but it isn't possible to (eventually) know 4.

Both are true. And even though the third proposition doesn't express it, nobody knows that the cat is on the mat (as explained by the fourth proposition). — Michael

Are you saying that we can change the expression of the unknown truth in Fitch’s proof to “p” instead of “p & ~Kp”?

You seem to be saying that the truth of the statement "It's true that there's milk in the fridge and no-one knows there is" is unknowable, which seems reasonable, since I don't know there's milk in the fridge unless I open it but then if I do that someone knows there is milk in the fridge. But when I open the fridge I know (excluding weirdness like the milk coming to be there only when I looked) that the statement was true before I looked. — Janus

:up:

So, again, there seems to be a time element involved.

If I go down the 'weirdness' rabbit hole and say that when I look and see the milk I still don't know that the milk had been there prior to my looking, then all bets are off. — Janus

Yes. While you're down the rabbit hole, be sure to check out the quantum superposition version: |milk in the fridge> + |no milk in the fridge>. :-)

Aye, there's the rub. If a truth is knowable, then it can come to be known; that is, it can change from being unknown to being known. However, as you note, the statement "p & ~Kp" does not (and cannot) change from being unknown to being known. — Luke

:up:

Therefore, the starting suppositions make it impossible for an unknown truth to become a known truth. — Luke

No, whether a statement is unknowable or not is conditional on the content of the statement. As @Michael points out, unknown truths that don't mention that they're unknown can be known.

But if "p & ~Kp" cannot possibly change from being unknown to being known, then of course it is unknowable: it's a rigged game from the outset. — Luke

Of course the statement is intentionally constructed to give that result. But it has real consequences for "any theory committed to the thesis that all truths are knowable" (from SEP).

No, whether a statement is unknowable or not is conditional on the content of the statement. As Michael is pointing out, regular statements that don't mention that they're not known can be known. — Andrew M

So is there a way to express an unknown truth in logical notation without mentioning that it is unknown?

So is there a way to express an unknown truth in logical notation without mentioning that it is unknown? — Luke

Sure, just don't mention it's unknown. So instead of "p & ~Kp", that would be "p". With the milk example, that would be "there's milk in the fridge". It's also true that it's initially unknown but since the statement doesn't mention that, its truth status doesn't change when someone comes to know it.

Sure, just don't mention it's unknown. So instead of "p & ~Kp", that would be "p". — Andrew M

How does that express that it is unknown?

How does that express that it is unknown? — Luke

It doesn't. That information is part of the context. The statement doesn't mention it. It also doesn't mention a host of other things, such as whether it's lite or full cream milk, whether it's in Alice's fridge or Bob's fridge, and so on.

Are you saying that we can change the expression of the unknown truth in Fitch’s proof to “p” instead of “p & ~Kp”? — Luke

I'm sorry but I just don't know how to fix your confusion. I've tried my best.

It doesn't. That information is part of the context. The statement doesn't mention it. It also doesn't mention a host of other things, such as whether it's lite or full cream milk, whether it's in Alice's fridge or Bob's fridge, and so on. — Andrew M

Then we can simply express the unknown truth in Fitch’s proof as “p” and the problem goes away: there are no unknowable truths.

EDIT: Does Fitch’s proof allow for some unknown truths to be expressed as “p” and others to be expressed as “p & ~Kp”?

Then we can simply express the unknown truth in Fitch’s proof as “p” and the problem goes away: there are no unknowable truths. — Luke

No, that's just changing the subject. There are unknowable truths regardless of whether there's a proof about them.

EDIT: Does Fitch’s proof allow for some unknown truths to be expressed as “p” and others to be expressed as “p & ~Kp”? — Luke

That a proposition is true is expressed by "p". That a proposition is unknown is expressed by "~Kp". If those two ideas need to be expressed together, then the conjunction symbol is used. "p" by itself implies nothing about whether the proposition is known or unknown, but it is nonetheless one or the other.

No, that's just changing the subject. There are unknowable truths regardless of whether there's a proof about them. — Andrew M

You want to disregard Fitch's proof, but I'm the one changing the subject?

Either an unknown truth is expressed as “p & ~Kp” and it follows that we must reject KP because some/all unknown truths are unknowable, or else an unknown truth is expressed as "p" and it follows that we need not reject KP because all truths are knowable.

The result of Fitch's proof is that some truths are unknowable. However, if one of its suppositions brackets off and excludes those unknown truths that do not mention they are unknown, then that leaves most unknown truths as knowable.

You want to say, in essence, that Fitch's proof affects only those unknown truths that mention they are unknown. Fine. There are some unknown truths which are unknowable, and it is only those unknown truths which mention that they are unknown. But unless it is necessary for an unknown truth to mention that it is unknown, then all truths are knowable.

Is there a reason why an unknown truth must mention that it is unknown, or can any unknown truth be expressed as "p"? If we can simply re-express the unknown truths of Fitch's proof such that they do not mention that they are unknown, then all truths are knowable. If this re-expression is possible, then knowing these truths is possible.

I get it now. Unknown truths can either mention they are unknown or not mention they are unknown. Only the former are unknowable. Since there is at least one unknowable truth then we must reject KP.

However, my point is that we can safely ignore these unknowable truths since they can be re-written without self-reference; the unknown truths on which they are based can be re-written such that they do not mention they are unknown. If the only unknowable truths are those that mention they are unknown, then there is no loss of information or knowledge which comes from expressing these unknown truths as “p” instead of “p & ~Kp”.

I get it now. Unknown truths can either mention they are unknown or not mention they are unknown. Only the former are unknowable. Since there is at least one unknowable truth then we must reject KP. — Luke

Yes.

However, my point is that we can safely ignore these unknowable truths since they can be re-written without self-reference; the unknown truths on which they are based can be re-written such that they do not mention they are unknown. If the only unknowable truths are those that mention they are unknown, then there is no loss of information or knowledge which comes from expressing these unknown truths as “p” instead of “p & ~Kp”. — Luke

Regardless of the symbols you use to express the proposition, it is impossible to know that the cat is on the mat and that nobody knows that the cat is on the mat.

What’s the issue with just accepting that some truths are unknowable?

But this would only apply to a small subset of unknown truths, those that follow the form "no one knows that X" where X is a true proposition.

And while this set of truths cannot be known, the past tense version "no one knew that X was true," can be known. I don't see this as a huge problem.

Indeed, I'm not even sure if this problem holds for an eternalists view of time in the first place. If all moments in time are real, then the issue is simply that the knowledge of the truth of the "known one knows that X is true" proposition has to occur simultaneously with the discovery of X being true. But the addition of a past tense would really just be an artifact of our languages' inability to transcend the present. . After all, if the past is real then there is a reality where "no one knows that X is true" is still true and someone in the future can have knowledge of this past truth. If this holds, the truthmaker of the the fact that "no one knows that X is true," still exists even after someone knows X.

This doesn't seem too dicey to me. There are plenty of good empirical reasons to accept eternalism (e.g., physics)

Not the best thread, this one.

Has someone claimed it's all quantum yet?

It'll happen.