Hopefully not taking away anyone’s fun here, but to reconcile Raven(x)->Black(x) but not vice versa, what this statement wants to say, letting P(R) and P(B) be the probabilities of raven and black, respectively, is P(R|B)=0 and P(B|R)=1, which gives us that
P(R|B) = 0
P(RB)/P(B) = 0
P(RB) = 0
and
P(B|R) = 1
P(BR)/P(R) = 1
P(BR) = P(R)
But of course this leads to a contradiction, so it can’t really be true that Black(x)-/->Raven(x), can it? Sure, because what is really meant by implies (-/->) is not P(B|R) = 0 but P(B|R)<1. But in logic we often forget this because anything with a probability less than 1 is assigned a truth value of false.
Logic has its value, since sometimes you want to prove something is true 100% of the time, but this is generally only possible in pure mathematics. If you try to do it elsewhere you’ll get exceptions (e.g. albino ravens). So leave logic to mathematicians; you should use Bayesian inference.
Hopefully not taking away anyone’s fun here, but to reconcile Raven(x)->Black(x) but not vice versa, what this statement wants to say, letting P(R) and P(B) be the probabilities of raven and black, respectively, is P(R|B)=0 and P(B|R)=1, which gives us that
P(R|B) = 0 P(RB)/P(B) = 0 P(RB) = 0
and
P(B|R) = 1 P(BR)/P(R) = 1 P(BR) = P(R)
But of course this leads to a contradiction, so it can’t really be true that Black(x)-/->Raven(x), can it? Sure, because what is really meant by implies (-/->) is not P(B|R) = 0 but P(B|R)<1. But in logic we often forget this because anything with a probability less than 1 is assigned a truth value of false.
Logic has its value, since sometimes you want to prove something is true 100% of the time, but this is generally only possible in pure mathematics. If you try to do it elsewhere you’ll get exceptions (e.g. albino ravens). So leave logic to mathematicians; you should use Bayesian inference.