It would make it seem much more like you are arguing in good faith, rather than doubling down on your position if you actually responded to what I am arguing here, (and others have argued) rather than playing a 'tit for tat' game of accusing me of poor reading comprehension, simply because I won't acquiesce to your stipulations. — Janus
I accuse you of poor reading comprehension because I
have been responding to your arguments, and you never seem to understand the responses. You are mostly putting forth things that I don't disagree with, as though they disprove the things that I am saying. So I'm not going to put forth counter-arguments to show that the things you're putting forth are wrong, because they're mostly not. They're just beside the point of anything that I was saying in this thread, not against it, and I'm trying to show you why that is.
For example:
tools marks are thought to be a sign strongly suggestive of artificial structures, and that conjecture is confirmed, although not logically proven, by countless examples drawn from experience — Janus
This is what I mean about you changing the focus of the discussion. First we were talking about how we could test
whether or not the Face was artificial. We could test that
using the implication of tool marks to artificiality, and we were discussing the right way to use that implication to test for artificiality. But now you're talking about how we could know
whether tool marks imply artificiality. That's a different thing. In a complete investigation that
is a further question that we could step back and ask too, but it's not the same question we were asking about at the start.
Anyway, on to the meat of things.
"If P then Q, Q, therefore P" is simply an invalid deduction. Confirmationism is an inductive, not an invalid deductive, thought process. — Janus
I have repeated over and over again that I'm
not simply talking about deductive vs inductive implications. I'm
talking about the direction of implication. Inferrring from "if P then probably Q", and "probably Q", to "probably P", all merely probabilistically, is
still not a good inference.
(I'm saying "good" instead of "valid" here so you don't think I mean non-probabilistic deduction; I've done that before as well. Also, probabilistic inference is not the same thing as induction, but let's leave that aside for now).
Meanwhile, inferring from "if Q then probably P, and "probably Q", to "probably P", even if it's all merely probabilistically,
is a good inference.
And that is the whole point of falsificationism, because "if Q then P" is logically equivalent to "if not P then not Q", as you already know.
(This is another place where you've been just talking past me. You speak as though you pointed out the equivalence of those to me, and that that defeated my point, when I already knew they were equivalent, and I in turn pointed out that switching from "if P then Q" to "if Q then P" makes
all the difference; I
never objected to "if Q then P", only "if P then Q", because the former is equivalent to falsification, and the latter is the only kind of confirmationism I'm against here).
So if you're affirming that the good direction of inference is from "if Q then (probably) P", and "(probably) Q", to "(probably) P", and you're not saying that the inference from "if P then (probably) Q", and "(probably) Q", to "(probably) P", is a good inference, then
you're completely agreeing with me, even if you think you're not.
the correct formulation for confirmationism is "if Q then P", — Janus
I just went to look up a quote about confirmationism in a reliable source, the Stanford Encyclopedia of Philosophy, to settle this merely nominal dispute once and for all, and I found something interesting: there are
two mutually contradictory things both called "confirmationism".
I learned confirmationism in school as synonymous with the hypothetico-deductive method, which to quote SEP means:
e HD-confirms h relative to k if and only if h ∧ k ⊨ e and k ⊭ e; — SEP
Where
e is some evidence,
h is some hypothesis,
k is some set of background assumptions, and
⊨ is a symbol for entailment, i.e. necessary implication.
So
on that hypothetico-deductivist confirmationist account, if the hypothesis (and background assumptions) entail the evidence (but the background assumptions alone don't), then seeing that evidence confirms the hypothesis. That is exactly
what I have been saying confirmation claims, except using "P" for the hypothesis+assumptions together, and "Q" for the evidence: that if P implies Q, and Q is the case, then P is the case.
But, it seems, Hempel's model, which I learned in school as something
against confirmationism and more a step toward falsificationism -- because it says exactly the opposite of that hypothetico-deductivism above, just like falsification does -- is apparently
also called "confirmationism", and I'm guessing that that's where you're coming from, Janus. Hempel's account says:
e confirms h relative to k if and only if e disconfirms ¬h relative to k. — SEP
Which is pretty much the same thing falsification says, except it doesn't call that "confirming" h, because falsificationism uses the term "confirmation" to refer to hypothetico-deductive confirmation. It just says that that's falsifying ¬h, which of course entails that h as well.
I get the feeling that maybe your point is that zero probability is not quite identical to impossibility, just very close. But even given that that's the case, it's not counter to my main thrust here, since as explained to Janus above, I'm not depending on these implications I'm loosely using here being completely strict absolute deductions; the same points I'm making apply even if they're taken only to be probabilistic relationships, as I'll explain more to Srap below right now.
Bayes theorem is:
P(H | E) = P(E | H) * P(H) / P(E)
Some simple algebra can rearrange that to:
P(H | E) * P(E) / P(E | H) = P(H)
Things like “P(X | Y)” are often phrased as “the probability of X given Y”, but that means the same thing as “the probability that X is true if Y is true” or “the probability that if Y is true then X is true” or “the probability that Y implies X”. “X given Y” = “X if Y” = “if Y then X” = “X implies Y”. It’s all the same thing. Just wrap a “the probability that” around any of those and you have what “P(X | Y)” means.
So we can rephrase the above formula as:
P(E implies H) * P(E) / P(H implies E) = P(H)
Meanwhile, the standard form of a falsificationist inference is:
~H implies ~E, and E, therefore H.
or equivalently:
E implies H, and E, therefore H.
If we want to make that probabilistic instead, using formal notation instead of just sticking “probably”s in there as I’ve been doing because I assumed we were all smart people who can read between the lines, we’d then have:
P(E implies H) * P(E) = P(H)
or if you like:
P(H | E) * P(E) = P(H)
The only difference between that and Bayes formula is that Bayes also has P(E | H) in the denominator on the left (of the rearranged presentation), i.e. the probability of the hypothesis is also inversely proportional to the probability that the hypothesis implies the evidence, i.e. the probability of the evidence given the hypothesis.
In other words, not only does Bayes theorem say that the more probably that the evidence implies the hypothesis, the more probable the hypothesis is true — which is exactly what falsification says, since “E implies H” is equivalent to “~H implies ~E” — additionally, the more probably that the hypothesis predicts the evidence, the
less probable that the hypothesis is true — the
exact opposite of what confirmationism would have you think.
That is to say, the standard form of a confirmationist inference:
H implies E, and E, therefore (probably) H.
rendered into proper probabilistic notation would be:
P(H implies E) * P(E) = P(H)
or if you like:
P(E | H) * P(E) = P(H)
where the first term there is exactly what Bayes theorem has the inverse of, in addition to also having the converse (“P(E implies H)” or "P(H | E)").
Let me repeat all that. Falsification rendered probabilistically:
P(H | E) * P(E) = P(H)
Bayes theorem, algebraically rearranged:
P(H | E) * P(E) / P(E | H) = P(H)
Versus confirmationism rendered probabilistically:
P(E | H) * P(E) = P(H)
So if anything, Bayes theorem is doubling down on falsificationism vs confirmationism, besides just rendering the whole thing into a probabilistic form.