Solution (source code)

= Solution

Let $R(x)$ mean that $x$ is a <raven> and $B(x)$ that it is black. The hypothesis
$$
H:\ \forall x\,[R(x)\Rightarrow B(x)]
$$
is <logically equivalent> to $\forall x\,[\neg B(x)\Rightarrow\neg R(x)]$. If the observed paper is both nonblack and a nonraven, it is a positive instance of the second formulation. Combining an unrestricted positive-instance rule with preservation of <scientific confirmation> under <logical equivalence> produces the <raven paradox>.<How Bayesian Confirmation Theory Handles the Paradox of the Ravens by Branden Fitelson and James Hawthorne>[opening derivation of the Nicod and equivalence conditions]{ref}

Whether the observation actually supplies evidence depends on what was already known and how the object was selected. In <Bayesian confirmation>, with background $K$, support means $P(H\mid E,K)>P(H\mid K)$. If the observer already knows the selected item is a nonraven and its colour is conditionally independent of $H$, then $P(E\mid H,K)=P(E\mid\neg H,K)$: the <likelihood ratio> is one and there is no confirmation. If all of $E$ was already known, learning it again also gives no update.

Under a different sampling scheme, the same classification can provide slight support. In the <finite-population model of raven confirmation>, fix $r$ <ravens> and $M$ nonblack objects. Compare no nonblack raven with exactly one. Sampling uniformly among the nonblack objects gives a nonraven with probabilities $1$ and $1-1/M$, hence likelihood ratio $(1-1/M)^{-1}>1$. If $M$ is large this is tiny evidence, not a proof. This example establishes possibility under explicit assumptions, not that every white-paper observation has that evidential role.

The appropriate answer is therefore conditional: mere logical instantiation does not settle the support supplied by observing this paper. State the background and sampling procedure. A familiar paper deliberately selected as a known nonraven normally supplies no new evidence about raven colour; a specified alternative sampling model may yield a small positive update.