Considering presumptions (1), (2), and you may (3), how does the newest argument toward first end go?
See today, very first, that the suggestion \(P\) gets in just towards the earliest together with third ones site, and you can subsequently, the insights off both of these premise is very easily secure
Fundamentally, to ascertain another conclusion-that is, one according to our history education in addition to offer \(P\) its apt to be than simply not too Goodness doesn’t occur-Rowe requires just one extra presumption:
\[ \tag <5>\Pr(P \mid k) = [\Pr(\negt G\mid k)\times \Pr(P \mid \negt G \amp k)] + [\Pr(G\mid k)\times \Pr(P \mid G \amp k)] \]
\[ \tag <6>\Pr(P \mid k) = [\Pr(\negt G\mid k) \times 1] + [\Pr(G\mid k)\times \Pr(P \mid G \amp k)] \]
\tag <8>&\Pr(P \mid k) \\ \notag &= \Pr(\negt G\mid k) + [[1 – \Pr(\negt G \mid k)]\times \Pr(P \mid G \amp k)] \\ \notag &= \Pr(\negt G\mid k) + \Pr(P \mid G \amp k) – [\Pr(\negt G \mid k)\times \Pr(P \mid G \amp k)] \\ \end
\]
\tag <9>&\Pr(P \mid k) – \Pr(P \mid G \amp k) \\ \notag &= \Pr(\negt G\mid k) – [\Pr(\negt G \mid k)\times \Pr(P \mid G \amp k)] \\ \notag &= \Pr(\negt G\mid k)\times [1 – \Pr(P \mid G \amp k)] \end
\]
However because off assumption (2) you will find one \(\Pr(\negt G \middle k) \gt 0\), while in look at assumption (3) you will find one \(\Pr(P \middle G \amplifier k) \lt step one\), which means that one to \([1 – \Pr(P \middle G \amplifier k)] \gt 0\), so that it following employs out-of (9) you to
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