Past exam of the philosophy of science supplementary subject of the University of Oxford / 2017 / 5 / b

Let $R(x)$ mean that $x$ is a raven and $B(x)$ that it is black. The hypothesisis 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.^[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.