So we can all agree that if someone doesn't believe in p then he believes that p is false, and vice versa. — MathematicalPhysicist

No.

I don't believe that you had Weetabix for breakfast and I don't believe that you didn't have Weetabix for breakfast.

Poor comedian...

I'm not trying to be funny. I'm pointing out that not believing that something is true is not the same as believing that it's false.

I have a brother and his name is Andrew.

Do you believe that the above is true? Do you believe that the above is false? It's appropriate to not believe either, especially if you have no evidence either way.

$~B(a v b v c) ⊢ B~(a v b v c) ⊢ B(~a&~b&~c) ⊢B(~a)&B(~b)&B(~c) ⊢ B(~a) ⊢ ~B(a)$

Is a bit counter intuitive.

Bob: 'I don't believe in anything you said'
Alice: 'Why?'
Bob: 'Because the first thing you said was wrong'

Also counter models. In a universe of discourse with at least one non-agent ~B can't be equivalent to B~ as ~B is true for non-agents but B never is.

that would be translated to ~B(~p), not B(~p) — Mr Phil O'Sophy

I know. I'm pointing out that ¬Bp doesn't entail B¬p. It's possible for ¬Bp ∧ ¬B¬p.

that would be translated to ~B(~p), not B(~p) — Mr Phil O'Sophy

You can just substitute ¬p=q.

¬B(p) ∧ ¬B(¬p) → ¬(¬B(p) ↔ B(¬p))

it would be:
1. I don't believe you had Weetabix
is equivalent to
2. I believe you didn't have Weetabix. — Mr Phil O'Sophy

And that's wrong, because they're not equivalent, as I tried to explain. I don't believe that you had Weetabix and I don't believe that you didn't have Weetabix. 1) is true but 2) is false.

No... I wanted to show that not believing in a disjunction implies believing in the negation of each disjunct. This is very strange behaviour, and is implied by the OP.

1. There are <10000 asteroids around Saturn.
2. There are 10000 asteroids around Saturn.
3. There are >10000 asteroids around Saturn.

I believe in (1 v 2 v 3). I don't believe in (1 v 2), that doesn't mean I believe that ~1 &~2, since that commits me to 3. I suspend belief in each disjunct while believing the disjunction as a whole is true (it's a tautology).

Also, collapsing B~ and ~B kinda removes the point of it being a modal operator...

I have a brother.

Do you believe that the above is true or false? Do you understand that it is possible (even appropriate) to believe neither, i.e. to withhold judgement?

1. I believe you don't have a brother.
2. I don't believe you have a brother.

are both not saying that as far as i'm concerned, you don't have a brother? — Mr Phil O'Sophy

Not if we're being strict about the logic. It is possible to not believe that p is true and to not believe that p is false. This is the position we (should) have when we have no evidence either way.

Do you believe that I have a brother? Do you believe that I don't have a brother?

As that does leave it open to an ambiguity I guess, but I don't think when people say that in isolation that they mean 3.
'I neither believe nor disbelieve that you have a brother.'
— Mr Phil O'Sophy
. I think its more likely that if they mean 3. they say it, whereas if they say three, they imply 1, unless otherwise stated. — Mr Phil O'Sophy

Sure, but I was addressing the OP which uses formal logical syntax. My ordinary language response was just an attempt to show why it's wrong.

Another crazy thing occurs if you introduce quantification.

~B(Ex~P(x)) <-> B(~Ex~P(x)) <-> (B(AxP(x))

"The agent withholds belief about whether there exists an entity without property P"
iff
"The agent believes every entity has property P"

I lack belief in some things which are not gazoompas. Therefore I believe everything is a gazoompa.

I think we can see that lack of belief and disbelief are different now.

~B(Ex~P(x)) — fdrake

Wouldn't that mean you lack the belief that there is some thing that is not a gazoompa? If you want to say you lack belief in some thing, shouldn't you put ∃ before the B?

They're different.

~B(Ex~P(x))
~ExB(~P(x))

Agent withholds belief that there is an X which is not P.
There is no X which A does not believe is P.

But shouldn't the first one be Ex~B(~P(x))?

~B(Ex~P(x)) <-> B(~Ex~P(x)) <-> (B(AxP(x))

Do you agree with that holding when B~ <=> ~B (abusing notation)?

Yea, that looks correct to me, but the translation to English is wrong imo.

How would you translate it?

~B(Ex~P(x)) = not believing there is any x with the property P.

not believing there is any x without the property p?

Ah, yes, my bad.

Yeah. I agree that it's a more precise translation. My motivation for translating the ~B as withholding belief or suspending belief was to highlight the contrast between it and B~; both can be translated as 'doesn't believe' and it depends on the context which means which. Equating the two is an error made tempting by 'does not believe' meaning both in ordinary language. There are good reasons to maintain that they are distinct despite this, however.

The problem though is only this part:

I lack belief in some things which are not gazoompas. — fdrake

(Should be "I lack the belief that there are things that are not gazoompas")

Going through my comments there's a load of mistakes so they were probably terrible to read through (sorry :p).

Yeah this is fair enough. I've been imprecise a few times in the thread, which is certainly a shame since it's a thread about logic; and no one is less sympathetic to imprecision (rightly, probably) than logicians.

There are four possibilities:

• B(p)
• ~B(p)
• B(~p)
• ~B(~p)

The first two contradict each other, as do the last two.

The first and the third contradict each other; but not the second and the last. One can consistently hold that

~B(p) & ~B(~p)

if one chooses not to have an opinion about (p). That's the point of view of an agnostic when asked if they believe in god, for example.

So while ~B(p) implies ~B(~p), it does not imply B(~p).

Hence the OP is incorrect in saying ~B(p) <=> B(~p).

Then may ask the following question.
You don't believe I ate W and you don't believe I didn't eat W, then what do you believe: did I eat or didn't I?
We have ~Bp & ~B~p, the question is do we have BpvB~p?
How can someone believe something and also believe its negation at the same time?
What is the semantics of 'belief'?

Can someone both believe in God and believe there's no God?
Besides me of course.... :=)

You don't believe I ate W and you don't believe I didn't eat W, then what do you believe: did I eat or didn't I? — MathematicalPhysicist

What colour do you believe the shirt I am wearing right now is?

We have ~Bp & ~B~p, the question is do we have BpvB~p? — MathematicalPhysicist

No.

How can someone believe something and also believe its negation at the same time? — MathematicalPhysicist

They can't.