...if Smith is rational and understands the conventions of formal logic, and believes that p, and entertains the proposition (p V q) in light of his belief that p, we should be surprised indeed if he does not acknowledge that he believes that (p V q) is true.
We're told S believes p, hence S believes (p v q).
If someone objects, what you gonna do? — Banno
Each of these propositions is entailed by (f). Imagine that Smith realizes the entailment of each of these propositions he has constructed by (f), and proceeds to accept (g), (h), and (i) on the basis of (f). Smith has correctly inferred (g), (h), and (i) from a proposition for which be has strong evidence. Smith is therefore completely justified in believing each of these three propositions.
Exactly.An intersting notion. So Moore might believe he has a hand, and yet doubt it. Or Moore might know he has a hand, and yet doubt it. But not Moore might be certain he has a hand, and yet doubt it.
That might be right. — Banno
This gives a weighted disjunction, (p(99%) v ~p(1%)). And that does not lead to (p v q). It's so simple it seems to be invisible to everyone, but as soon as it is possible that ~p, the damaging disjunction (p v q) cannot be made at all. — unenlightened
IF p, then (p v q). That's valid, sound, true and contentless
— unenlightened
It's only sound if p is true. — Srap Tasmaner
The conclusion of an inference merits no more or less credence than what you grant your premises. If you're uncertain about your premises, then you should be just that uncertain about your conclusions. — Srap Tasmaner
It often seems professional epistemologists count it their duty to construct and assault straw men. Consider their collective abuse of the moldy old straw man they call "the skeptic".What irks me about Gettier is that he appears to be assaulting a straw man. Who is it that believes knowledge is exactly justified true belief? — Banno
No, it's always sound, because it already has your 'if' incorporated. I put it in capitals so you would notice. It is contentless because it does not claim that p is true. S wrongly makes the claim, that p is true, and then uses this formula to arrive illegitimately at (p v q). — unenlightened
What here do you disagree with? — Srap Tasmaner
I'll just note that science, probability, and induction/abduction are what S does to arrive at his belief p. No quarrel with him there. — unenlightened
Dretske and Nozick focused on a form of skepticism that combines K with the assumption that we do not know that skeptical hypotheses are false. For example, I do not know not-biv: I am not a brain in a vat on a planet far from earth being deceiving by alien scientists. On the strength of these assumptions, skeptics argue that we do not know all sorts of commonsense claims that entail the falsity of skeptical hypotheses. For example, since not-biv is entailed by h, I am in San Antonio, skeptics may argue as follows:
(1) K is true; i.e., if, while knowing p, S believes q because S knows that p entails q, then S knows q.
(2) h entails not-biv.
(3) So if I know h and I believe not-biv because I know it is entailed by h then I know not-biv.
(4) But I do not know not-biv.
(5) Hence I do not know h.
Dretske and Nozick were well aware that this argument can be turned on its head, as follows:
(1) K is true; i.e., if, while knowing p, S believes q because S knows that p entails q, then S knows q.
(2) h entails not-biv.
(3) So if I know h and I believe not-biv because I know it is entailed by h then I know not-biv.
(4)′ I do know h.
(5)′ Hence I do know not-biv.
Turning tables on the skeptic in this way was roughly Moore's (1959) antiskeptical strategy. (Tendentiously, some writers now call this strategy dogmatism). However, instead of K, Moore presupposed the truth of a stronger principle:
PK: If, while knowing p, S believes q because S knows that q is entailed by S's knowing p, then S knows q.
Unlike K, PK underwrites Moore's famous argument: Moore knows he is standing; his knowing that he is standing entails that he is not dreaming; therefore, he knows (or rather is in a position to know) that he is not dreaming.
According to your calculus, does (p(99%) V ~p(1%)) imply ((p V q)99%)? Or how are your probabilistic weightings related to propositional logic? — Cabbage Farmer
Where do you derive this principle from? It isn't a law of logic. If I might use an analogy, the higher you want to build, the more secure you need to make your foundations. — unenlightened
We can extend the treatment to "certainty": So long as we mean mere practical certainty or a feeling of sureness, but not absolute theoretical certainty, certainty is compatible with doubt. — Cabbage Farmer
it just seems to be the case that minds like ours never or almost never attain absolute certainty. — Cabbage Farmer
Cabbage Farmer wrote:
The truth conditions of (p V q) are met as soon as the truth conditions of p are met. Or as soon as the truth conditions of q are met. Or as soon as the truth conditions of (p AND q) are met.
It seems this is the point you're neglecting, which has sent you off on a search for red herring.
Smith believes that p.
Smith does not merely believe that there are abstract inferential relations between any pair of propositions and their disjunction. He believes the truth condition for a particular disjunctive claim has been satisfied. He believes that Jones owns a Ford...
Cabbage Farmer wrote:
What I'm suggesting is that
(p AND (IF p THEN (p V q))
is already enough to give a truth condition for (p V q). Or in other words:
(p V q) is true if p.
...Smith entertains the proposition (p V q) and makes a rational judgment.
We're told that S has evidence of p, that S knows that p ∨ q follows from p, and so that S has evidence of p ∨ q. We're also told that S, recognising that he has evidence of p ∨ q, believes p ∨ q. — Michael
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