Originally posted by vistesd
Okay, DT—after the contradiction, the symbolic logic gets beyond me. Can you explain in lay terms why you can’t connect this to an omniscience (which, as I recall, had to do with the distribution of justification versus the distribution of knowledge—which I also did not really comprehend)?
EDIT: I am re-reading LJ's argument & your replies . . .
EDIT ...[text shortened]... the possibility, which I do not see as leading to a contradiction (but i might be missing much).
I've defined omniscience sufficiently widely that I can say that there is an omniscient entity if and only if all propositions that are true are known. The converse of this is that if and only if there is not an omniscience then there exists some proposition which is both true and unknown.
We have:
(1) ¬O <-> ∃P (P & ¬K(P))
The statement there exists is the same as a string of disjunctions. Propositions are sentences in some language and therefore form a countable set, so we can enumerate them, P1, P2, P3, ... (*)
This means we can write:
(2) ¬O <-> (P1 & ¬K(P1)) or (P2 & ¬K(P2)) or (P3 & ¬K(P3)) or ...
Now, hoping to derive a contradiction I looked at K(¬O), it is known that there is not an omniscience:
(3) K(¬O) <-> K(∃P P & ¬K(P))
which is the same as:
(4) K(¬O) <-> K[(P1 & ¬K(P1)) or (P2 & ¬K(P2)) or (P3 & ¬K(P3)) or ...]
For clarity I'll replace P1 & ¬K(P1) with C1 and so forth, so we have:
(5) K(¬O) <-> K(C1 or C2 or C3 or ...)
Now, there are two problems, one is that in getting this far as I've assumed an axiom called K which assumes that knowledge is preserved by entailment. It is not clear that that is the case. However that is not the serious flaw, the killer is that my argument relies on this:
(6) K(C1 or C2 or C3 or ...) -> K(C1) or K(C2) or K(C3) or ...
but knowledge won't distribute in that way. It is not true that if I know C1 or C2, I either know C1 or know C2, because I may not know which one is true.
My point in reply to twhitehead (which is the post you replied to) is as follows:
Suppose a proposition is known to be either true or false but it's truth value is not known. If the proposition is true it is not known to be true, and if false then it's converse is true and not known to be. So we have that:
(P & ¬K(P)) or (¬P & ¬K(¬P))
Let's make C = P & ¬K(P), and C* = ¬P & ¬K(¬P). I'm using * to mean dual as there is a sort of duality between the two propositions. So we have:
C or C*
and given that we can claim to know this we have:
K(C or C*)
but we cannot get from this to:
K(C) or K(C*)
which twhitehead needs for his objection to work (although he just denied he was claiming that so I don't know what he's arguing).
Regarding edit 2, I think you are right as believing a proposition does not entail said proposition is true. So where K(P) -> P of necessity we have ¬(B(P) -> P), in other words a belief isn't stopped from being a belief by the object of the belief not existing or not having the property that it is believed to have. Compare a belief and a knowledge claim that "all swans are white.", the agent who believes that is mistaken due to the existence of
Cygnus atratus, however it would not stop their belief from being a belief. Their "knowledge" that "All swans are white." would never have been knowledge because the proposition is required to be true.
The logic for the proof of the unknowability of some propositions requires that K(P) -> P, but for belief we do not have B(P) (meaning some agent believes P) implying P. So we cannot get between:
B(P) & B(¬B(P))
and
B(P) & ¬B(P).
Note that B(P) means that
any agent believes the proposition, so that B(¬B(P)) is not incoherent, agent 1 can believe P and agent 2 not believe P and not believe that anyone else could believe it. In the case of a single agent they might be mad.
(*) In case anyone thinks that you can use a diagonalisation argument to dispute this, it won't work as one cannot construct a grammatical sentence by making a list of grammatical sentences and choosing the first letter so that it is different from the first letter of the first sentence, the second letter different from the second letter of the second sentence and so forth. The result will not automatically obey grammatical rules, which is what that type of argument requires.