Agreed. People find material implication and inclusive disjunction counterintuitive, and then mistake their objections to them for objections to arguments that use them.

Believing that a disjunction follows from a belief is not equivalent to believing the disjunction.

That's the problem.

My argument shows it and dissolves any and all issues with believing disjunction by virtue of showing the difference... clearly.

Un wrote:...

But since there is in fact no connection between p and q, there is no justification for saying it.

The justification for Smith saying it, is the fact that there is no connection. "Either Jones owns a Ford or Brown is in Barcelona" is Smith's certainty being put on display. The problem, of course, is that Gettier's account is inadequate for properly representing Smith's believing the disjunction.

This changes with Gettier though. Gettier knows Jones does not own a Ford. Gettier knows Brown is in Barcelona. Gettier also knows that Smith believes the disjunction because Smith believes Jones owns a Ford. If we fill out my solution with Gettier's belief we arrive at a different disjunction(s) than Smith. None of which are problematic.

Michael wrote:

If I believe that my child is eating cake, and if I believe that three children are eating cake, then I believe that the statement "my child is one of the three eating cake" is true. This seems perfectly reasonable.

Nothing at all unreasonable there.

I would just note something that is not like Gettier Case II. Michael is talking about combining two separate beliefs of his own. Case II only uses one belief of Smith's.

The similarity is that what's being derived has two different sets of truth conditions being combined into one. That is notable, I think, because of the historical notion of "proposition". A disjunction is called a "proposition", as is a conjunction. However, a disjunction consists of two separate propositions, each with it's own set of truth conditions, and in Case II, one of which need not be believed by the speaker/author.

Propositions are not equivalent to belief.

Nothing at all unreasonable there.

I would just note something that is not like Gettier Case II. Michael is talking about combining two separate beliefs of his own. Case II only uses one belief of Smith's.

The similarity is that what's being derived has two different sets of truth conditions being combined into one. That is notable, I think, because of the historical notion of "proposition". A disjunction is called a "proposition", as is a conjunction. However, a disjunction consists of two separate propositions, each with it's own set of truth conditions, and in Case II, one of which need not be believed by the speaker/author.

Propositions are not equivalent to belief. — creativesoul

If I believe that the statement "Jones owns a Ford" is true and written in this book, and if I believe that "Brown is in Barcelona" is also written in this book, then I believe that the statement "one of the two statements written in this book" is true.

How is that any different to saying "I believe that this or that statement is true"? Or "I believe that 'Jones owns a Ford' is true or 'Brown is in Barcelona' is true"? Or "I believe that Jones owns a Ford or Brown is in Barcelona"?

Two kids, Bill and Ted. One of them is yours. If it's not Ted, it must be Bill, and if it's not Bill, it must be Ted.

The justification for Smith saying it, is the fact that there is no connection. "Either Jones owns a Ford or Brown is in Barcelona" is Smith's certainty being put on display. The problem, of course, is that Gettier's account is inadequate for properly representing Smith's believing the disjunction. — creativesoul

Exactly so. which is to say it is rhetorical. It only follows and then trivially if he is right about Jones.

This changes with Gettier though. Gettier knows Jones does not own a Ford. Gettier knows Brown is in Barcelona. Gettier also knows that Smith believes the disjunction because Smith believes Jones owns a Ford. If we fill out my solution with Gettier's belief we arrive at a different disjunction(s) than Smith. None of which are problematic. — creativesoul

Yes, Gettier has authorial infallibility. He knows what he knows absolutely in his invented world, and he knows in this instance that Smith does not know what he rationally believes. But since Smith is defined to be rational, he knows that his rational belief p can be false and he knows that its falsehood does not imply the the truth of q. So he cannot believe (p v q) precisely because that implies that if perchance he is wrong about p, then q would be true. It is impossible for him to assent to this because even Gettier tells us, for realism's sake, that he forms several disjunctions mutually contradictory under ~p which is infallibly true, unknown to Smith.

Two kids, Bill and Ted. One of them is yours. If it's not Ted, it must be Bill, and if it's not Bill, it must be Ted. — unenlightened

Sure, but we're talking about belief. If I believe that one of the kids in the phone booth is mine I don't necessarily believe that if it's not Bill then it's Ted. I might believe that it's definitely Bill and definitely not Ted. But it's still the case that I believe that one of the kids in the phone booth is mine.

Sure, but we're talking about belief. — Michael

So you believe it's Bill, and you don't believe it's Ted. But should it turn out that Bill is not yours, because your wife had an affair, you do not thereby conclude you have fathered Ted.

p1. (B v T)
p2. ~B
c1 T

p1. (B v T)
p3. ~T
c2. B

I think this demonstrates the implications I have indicated, and the necessity of including falsehood in any expression of a disjunction. To believe the disjunction is to believe " if it's not Bill, it must be Ted." And that is unbelievable. But note that when you paraphrase the disjunction informally as "one of the kids in the phone booth is mine", you contrive to avoid the implication by avoiding the disjunction. That's cheating.

So you don't believe that "either Donald Trump is the President or Hillary Clinton is the President" is true?

I believe that it's true, even though I don't believe that if it's not Donald Trump then it's Hillary Clinton.

Yes. I believe Trump is, and Clinton is not. Or to use your informal locution, I believe that one of Trump and Clinton (namely Trump) is president.

But what you present here is a perfectly valid disjunction that I would have assented to before knowing the result of the election "Either Trump or Clinton will be President". You need to do better than keep asking these rhetorical and irrelevant questions. The disjunction is valid because if either one of them was not elected, the other would have been, in other words there is a real connection between the two sides of the disjunction. Now do try and address the argument a little.

How is it any different?

If I believe that Donald Trump is the President then I will believe that "Donald Trump is the President or Hillary Clinton is the President" is true.

If I believe that Bill is my child then I will believe that "Bill is my child or Ted is my child" is true.

If I believe that Jones owns a Ford then I will believe that "Jones owns a Ford or Brown is in Barcelona" is true.

If believe that p then I will believe that "p ∨ q" is true.

But let's continue with this example, as you seem to be OK with it.

I am justified in believing that Donald Trump is the President. Therefore I am justified in believing that "Donald Trump is the President or Hillary Clinton is the President" is true. Donald Trump isn't the President but Hillary Clinton is the President. "Donald Trump is the President or Hillary Clinton is the President" is true. I have a justified true belief.

But let's continue with this example, as you seem to be OK with it. — Michael

Let's not. Let's look at the formal implications of a disjunction that I laid out and you ignored.

Let's not. Let's look at the formal implications of a disjunction that I laid out and you ignored. — unenlightened

The formal implication of a disjunction is that if "p" is true then "p ∨ q" is true. Therefore the rational person who believes that "p" is true will believe that "p ∨ q" is true.

This just doesn't make sense:

1. I believe that "p" is true
2. I believe that if "p" is true then "p ∨ q" is true
3. I don't believe that "p ∨ q" is true

Because if you don't believe that "p ∨ q" is true then you believe that both "p" and "q" are false, but it's been established that you believe that "p" is true.

That's an implication of p, that I am questioning. Look at the implications of the disjunction that I put to you.

p1. (B v T)
p2. ~B
c1 T

p1. (B v T)
p3. ~T
c2. B — unenlightened

p1. (B v T)
p2. ~B
c1 T

p1. (B v T)
p3. ~T
c2. B — unenlightened

Yes. So if I believe p1 and p2 then I will believe c1, and if I believe p1 and p3 then I will believe c2. What's the problem?

That's an implication of p, that I am questioning. — unenlightened

But it's basic logic, so I don't see how you can question it. If "p" is true then "p ∨ q" is true. Because "p ∨ q" is true if "p" is true (or if "q" is true).

You seem to be pushing for a logic where "p" is true but "p ∨ q" isn't. I don't know how you can expect that to work. It's alien to every truth table I know.

Quite simply, what you're saying is illogical.

What's the problem? — Michael

I think it's called "Gettier" ;) . I'll try and formalise things as best I can.

On your side, we have:

a. If p then (p v q).

And on my side, we have:

b. If (p v q) then (if ~p then q).

The problem with putting these together is that (a) is conditional on p and (b) is a conditional on ~p.

This is not a problem under infallibility since both conditions cannot be met. But Gettier specifies that Smith believes p, but ~p. And this brings both conditionals into play at once. So we now have

c. If S believes p, then S believes (p v q)

d. If S believes (p v q) then S believes (if ~p then q)

This leads to:

e. If S believes p, then S believes that (if ~p then q).

Now (e) states that if S believes p, then S believes that if ~p, absolutely any and every proposition is true, including contradictions. I think it vastly overstates anyone's confidence in their beliefs. I am pretty damn sure that Trump is the president, but not so sure as to believe that if I am wrong about that, a thousand devils inhabit my left knee and also do not. After all, someone might have just assassinated the fruitcake (hope springs eternal). In other words, the formalisms of infallibility do not transfer to fallible beliefs.

I think we both want to escape (e) and our difference is more or less whether it is better to reject (c) or (d).

Edit. It occurs to me that the explosion at (e) is the equivalent of the formalism that from (p & ~p) anything follows. And that is what Gettier has set up for us - Smith believes p but ~p.

I think the situation is this:

1. Smith believes that if "p" is true then "p ∨ q" is true
2. Smith believes that if "p" is false then "p ∨ q" might be false, and so doesn't believe that if "p" is false then "q" is true

I don't know how to formally reconcile these.

Here's an attempt:

1. p
2. p ⊨ p ∨ q
3. p ∨ q
4. p ∨ q ⊨ ¬p → q
5. ¬p → q

6. p ⊨ ¬p → q

6 is the principle of explosion, a valid rule of inference. Therefore, if Smith believes that "p" is true then it is correct to believe that if "p" is false then "q" is true.

I think what you're bringing up is the paradox of material implication.

Yes, I'm with you.

So now we bracket that...

S believes {
1. p
2. p ⊨ p ∨ q
3. p ∨ q
4. p ∨ q ⊨ ¬p → q
5. ¬p → q
6. p ⊨ ¬p → q) }

7. ¬p (Gettier's stipulation)

And we have arrived at my (e), which is the explosion of belief. This is too volatile a situation to be tolerated. One false belief plus formal logic lead to the total collapse of knowledge.

I don't know about anyone else, but I am going to forbid S from moving from 1 to 2 unless p is a tautology, on the grounds that all other beliefs are not certain and therefore have the form "Believably p, but (&) conceivably ¬p". Now if S takes my advice and substitutes this formula for 1, then 2 does not follow from 'conceivably ¬p', and the bomb is defused.

Then what if we use this form:

1. One or both of "Jones owns a Ford" and "Brown is in Barcelona" is true

If it helps, this proposition can be presented to Smith ahead of time. It is only later that he is presented with evidence that Jones owns a Ford. Does he believe that 1 is true? I say he does.

Incidentally, the only mentions I can find of rejecting disjunction introduction are paraconsistent logics.

There's also relevance logic that denies disjunctive syllogism, which is what I was arguing for earlier. In relevance logic, p entails p ∨ q but p ∨ q doesn't entail ¬p → q.

1. One or both of "Jones owns a Ford" and "Brown is in Barcelona" is false — Michael

No that won't do. Firstly, it is more like your 3 than your 1. And secondly, they could both have been true.

Incidentally, the only mentions I can find of rejecting disjunction introduction are paraconsistent logics. — Michael

That makes perfect sense to me, because what I have done is to divide the claim into two parallel but opposed logical realms, the believable and the conceivable. So the claim "S believes p" is not expressed by S as "p". Instead, because Gettier has had the decency after all this time to let him know that he can have justified false beliefs, S says

1a. " Believably p, but conceivably ¬p." In this way he runs 2 arguments in superposition :

2a. Believably (p v q) but conceivably ¬(p v q)

And so on. And the two lines of argument remain in superposition until either Jones or Gettier confirms absolutely that either his belief or his unbelieved contrary conception is uniquely true, at which point he sensibly discards whichever line starts with the untrue premis.

No that won't do. Firstly, it is more like your 3 than your 1. And secondly, they could both have been true. — unenlightened

Sorry, meant "is true" not "is false".

2b. Believably (p v q) but conceivably (¬p v q)

That might be more fun.

This might actually be a better account of Smith's beliefs:

Because Smith believes p and because p entails p ∨ q, Smith believes p ∧ (p ∨ q).
p ∧ (p ∨ q) doesn't entail ¬p → q, so Smith doesn't believe ¬p → q.

Or to make it stronger, because Smith believes p ∧ ¬q and because p ∧ ¬q entails p ∨ q, Smith believes p ∧ ¬q ∧ (p ∨ q).
p ∧ ¬q ∧ (p ∨ q) doesn't entail ¬p → q, so Smith doesn't believe ¬p → q.

p ∧ (p ∨ q) doesn't entail ¬p → q, — Michael

?

Are you using "entail" in some special sense?

1a. " Believably p, but conceivably ¬p." — unenlightened

Isn't this what "probably p" already says? Why do this superposition analysis at all? Do belief and conception vary freely, or is there some relation there? If you just stick with probability, it's direct: as p seems less probable, ¬p seems more probable. Isn't that more sensible?

2b. Believably (p v q) but conceivably (¬p v q) — unenlightened

That says "Believably ¬p→q but conceivably p→q", but again manages to lose the connection between them. For any q, either p implies it or ¬p does, and if one doesn't, then the other definitely does.

Are we trying to reinvent "or" here?

The whole point of Gettier's Case II is that it's a bizarre coincidence that Smith's belief is true, and true for reasons that have nothing to do with his reasons for holding that belief.

It just doesn't matter if his holding that belief is also bizarre. There are two elements to a coincidence. Of course his holding that belief is bizarre! It's an abuse of logic. But if, for whatever peculiar and idiosyncratic reasons, he is inclined to form such a belief, it will be true and justified but not knowledge. You either accept that, and scrap JTB, or you block the supposed justification.

ADDED: Or I guess you could say that JTB "almost always" works, or "usually" works, or works for "normal" cases -- @Fafner is making a related argument elsewhere.