Charlie: Your proposal to remove condition 3 works when T is a finite theory, but you cannot do that when T is infinite. Indeed, we use 3 for the infinite extension form Tn to T∞. I suspect that you cannot remove condition 3.
Is the idea that the proof necessary to use 1 is of infinite length, and you want your logic to be finitary? Hm. This seems odd, because P(s|T∞) is in some sense already a function with an infinitely long argument. How do you feel about using 2 in the form of P(T∞|T0)=P(Tn|T0)⋅[someprobability], therefore P(T∞|T0)≤P(Tn|T0), which has the same amount of argument as P(s|T∞)? I’m confused about at least one thing.
Also, is there some reason you prefer not to reply using the button below the comment?
Charlie: Your proposal to remove condition 3 works when T is a finite theory, but you cannot do that when T is infinite. Indeed, we use 3 for the infinite extension form Tn to T∞. I suspect that you cannot remove condition 3.
Is the idea that the proof necessary to use 1 is of infinite length, and you want your logic to be finitary? Hm. This seems odd, because P(s|T∞) is in some sense already a function with an infinitely long argument. How do you feel about using 2 in the form of P(T∞|T0)=P(Tn|T0)⋅[someprobability], therefore P(T∞|T0)≤P(Tn|T0), which has the same amount of argument as P(s|T∞)? I’m confused about at least one thing.
Also, is there some reason you prefer not to reply using the button below the comment?