 To see the rationale for (2), recall our discussion of confirmation theory  (§1.2).  According to our definition of confirmation, evidence confirms a hypothesis just in case \(p(H\mid E)&gt;p(H)\), which Bayes’ theorem tells us is equivalent to \(p(E\mid H) &gt; p(E)\). Likewise, \(E\) disconfirms \(\neg H\) just in case \(p(E) &gt; p(E\mid \neg H)\). Now, we can prove that if \(p(E\mid H) &gt; p(E)\), then \(p(E) &gt; p(E\mid \neg H)\). So if \(E\) confirms \(H\), it disconfirms \(\neg H\), which amounts to \(E\) supporting \(H\) over \(\neg H\).