It’s interesting that this is basically the opposite of the Gaifman condition—clearly there are conflicting intuitions about what makes a ‘good’ conditional logical probability.
On the open problem; In order to prove 3 from 2, all you need is that P(s∧T|R)=P(s|R) when s proves T − 3 follows from 2 if you do that substitution, and then divide by P(T|R), which is less than or equal 1 (this may assume an extra commonsense axiom that probabilities are positive).
Now consider applying rule 1 to P(s∧T|R), T proven by s. R proves that only one of s∧T,¬s∧T,¬s∧¬T is true, and also proves that only one of s,¬s∧T,¬s∧¬T is true. Thus 3 is derivable from 1 and 2.
It’s interesting that this is basically the opposite of the Gaifman condition—clearly there are conflicting intuitions about what makes a ‘good’ conditional logical probability.
On the open problem; In order to prove 3 from 2, all you need is that P(s∧T|R)=P(s|R) when s proves T − 3 follows from 2 if you do that substitution, and then divide by P(T|R), which is less than or equal 1 (this may assume an extra commonsense axiom that probabilities are positive).
Now consider applying rule 1 to P(s∧T|R), T proven by s. R proves that only one of s∧T,¬s∧T,¬s∧¬T is true, and also proves that only one of s,¬s∧T,¬s∧¬T is true. Thus 3 is derivable from 1 and 2.