reason
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- Apr 26, 2007
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For Evan,
Note: c(e) refers to the logical content of e (not including tautologies).
Therefore, h = h1 + e
We can now replace any instance of h with h1 + e.
However, if we break apart h1 + e, we can isolate h1: that part of h which is not equal to e. We can then ask whether h1 is supported by e,
Since h1 is derived from c(h) minus c(e), then the answer is no.
In other words, that portion of h which goes beyond e is not made more probable by e. The apparent increase in probability of all of h arises from treating hypotheses as indivisible elements -- a fallacy of composition. What probabilities you get, and what supports what, just depends on how you say it.
Edit: I haven't actually tried writing this out in this form before, so I probably (heh) made some mistakes, knowing me.
Note: c(e) refers to the logical content of e (not including tautologies).
If p(h|e) is greater than p(h), then p(e|h) = 1.
If p(e|h) = 1, then c(e) is a subset of c(h).
If c(h1) = c(h) - c(e), then c(h1) + c(e) = c(h)
If p(e|h) = 1, then c(e) is a subset of c(h).
If c(h1) = c(h) - c(e), then c(h1) + c(e) = c(h)
Therefore, h = h1 + e
We can now replace any instance of h with h1 + e.
If p(h1 + e|e) is greater than p(h1 + e), then p(e|h1 + e) = 1.
However, if we break apart h1 + e, we can isolate h1: that part of h which is not equal to e. We can then ask whether h1 is supported by e,
Is p(e|h1) = 1?
Since h1 is derived from c(h) minus c(e), then the answer is no.
In other words, that portion of h which goes beyond e is not made more probable by e. The apparent increase in probability of all of h arises from treating hypotheses as indivisible elements -- a fallacy of composition. What probabilities you get, and what supports what, just depends on how you say it.
Edit: I haven't actually tried writing this out in this form before, so I probably (heh) made some mistakes, knowing me.