So, I could instead think of iteratively eliminating inconsistent sets of sentences from the measure M, and looking at the limit of M(X)/M(all). I will assume for convenience that we eliminate one inconsistency at a time (so, remove mass from one set of sentences at a time). Let’s call this M{d}, so M{d+1} has one more inconsistent set removed from the measure (which is to say, set to measure 0).
Each set eliminated may take some mass away from the numerator, and will definitely take some mass away from the denominator. (Yet the numerator will always remain less than or equal to the denominator.)
If we remove a set of sentences which does not include X and has no overlap with any of the other sets of sentences removed so far, then the mass of both the numerator and the denominator get multiplied by 1 - .5^n, where n is the number of sentences in the set. To make this more general, for M{0}(Y) we can think of ~Y as having already been removed from its mass to begin; therefore, we can say that for any M{d}(Y), if d+1 is based on an inconsistent set S not involving anything removed from M{d}(Y) previously, then M{d+1}(Y) = M{d}(Y) * (1 - .5^|S|).
If S does contain a sentence which has been involved in inconsistency previously, we remove a strictly smaller fraction. If a subset of S has already been removed, then nothing gets removed at all. But if a set S being removed from M{d}(Y) contains only Y and some other sentences which have never appeared before, then M{d+1}(Y) = M{d}(Y) * (1 - .5^|S-1|).
Let’s name the ratio M(X)/M(all) at depth d as R(d). Considering R(d)/R(d+1), we know its limit must be in the range [0,1], because M(X) cannot be more than M(all) (so R(d) cannot diverge upward). If the limit of R(d) is greater than zero, the limit of R(d)/R(d+1) must be 1.
Assume X and ~X are both consistent.
In some ordering of logical depth, we have that: infinitely often, we eliminate a set of the form {Z, Z->~X, X} where Z and Z->~X have both never been seen before in an elimination set. Thus, M{d+1}(X) = .75 M{d}(X).
M{d}(all) = M{d}(X) + M{d}(~X). Nothing will be removed from M{d}(~X). Since X and ~X are consistent, both have some measure. Therefore, M{d+1}(all) = .75 M{d}(X) + M{d}(~X) = M{d}(all) ( .75 R(d) + M{d}(~X)/M{d}(all)) = M{d}(all) ( .75 R(d) + 1 - R(d)) = M{d}(all) * (1 - .25 R(d)).
I didn’t follow but the conclusion doesn’t sound right to me. Your argument looks like it should apply to any language and proof system. So if I introduce “X” as a logical symbol that isn’t constrained by any axioms or inference rules, why would its probability go to either 0 or 1? It seems to converge to 1⁄2 since for any consistent truth assignment with X=T we have a symmetric consistent truth assignment with X=F.
Yea, I suspect that’s right: so if we introduce X, it must go to either zero or 1, but either option violates the symmetry between X and ~X. Therefore, it must not converge. But this needs to be formalized more before I’m sure.
To convey my argument without as much of the math:
Suppose that P(X) is 1⁄2 at some stage. Then there will be inconsistent sets yet to remove which will take it at least C away from 1⁄2, where C is a constant that does not go down as the process continues.
The intuition behind this is that removing an inconsistent sentence which has not appeared in any of the inconsistencies removed so far, reduces mass by 1⁄2. Thus, the mass is changing significantly, all the time. Now to make this into a proof we need to show that P(X) changes significantly no matter how far into the process we go; IE, we need to show that a significantly different amount of mass can be removed from the ‘top’ and the ‘bottom’ (in the fraction M(X) / M(all) at finite depth).
The inconsistency {Y, Y->X, ~X} is supposed to achieve this: it only removes mass from the bottom, but there are infinitely many sets like this (we can make Y arbitrarily complex), and each of them reduces the bottom portion by the same fraction without touching the top. Specifically, the bottom becomes 7/8ths of its size (if I’ve done it right), so P(x) becomes roughly .57.
The fraction can re-adjust by decreasing the top in some other way, but this doesn’t allow convergence. No matter how many times the fraction reaches .5 again, there’s a new Y which can be used to force it to .57.
Ok, I see.
So, I could instead think of iteratively eliminating inconsistent sets of sentences from the measure M, and looking at the limit of M(X)/M(all). I will assume for convenience that we eliminate one inconsistency at a time (so, remove mass from one set of sentences at a time). Let’s call this M{d}, so M{d+1} has one more inconsistent set removed from the measure (which is to say, set to measure 0).
Each set eliminated may take some mass away from the numerator, and will definitely take some mass away from the denominator. (Yet the numerator will always remain less than or equal to the denominator.)
If we remove a set of sentences which does not include X and has no overlap with any of the other sets of sentences removed so far, then the mass of both the numerator and the denominator get multiplied by 1 - .5^n, where n is the number of sentences in the set. To make this more general, for M{0}(Y) we can think of ~Y as having already been removed from its mass to begin; therefore, we can say that for any M{d}(Y), if d+1 is based on an inconsistent set S not involving anything removed from M{d}(Y) previously, then M{d+1}(Y) = M{d}(Y) * (1 - .5^|S|).
If S does contain a sentence which has been involved in inconsistency previously, we remove a strictly smaller fraction. If a subset of S has already been removed, then nothing gets removed at all. But if a set S being removed from M{d}(Y) contains only Y and some other sentences which have never appeared before, then M{d+1}(Y) = M{d}(Y) * (1 - .5^|S-1|).
Let’s name the ratio M(X)/M(all) at depth d as R(d). Considering R(d)/R(d+1), we know its limit must be in the range [0,1], because M(X) cannot be more than M(all) (so R(d) cannot diverge upward). If the limit of R(d) is greater than zero, the limit of R(d)/R(d+1) must be 1.
Assume X and ~X are both consistent.
In some ordering of logical depth, we have that: infinitely often, we eliminate a set of the form {Z, Z->~X, X} where Z and Z->~X have both never been seen before in an elimination set. Thus, M{d+1}(X) = .75 M{d}(X).
M{d}(all) = M{d}(X) + M{d}(~X). Nothing will be removed from M{d}(~X). Since X and ~X are consistent, both have some measure. Therefore, M{d+1}(all) = .75 M{d}(X) + M{d}(~X) = M{d}(all) ( .75 R(d) + M{d}(~X)/M{d}(all)) = M{d}(all) ( .75 R(d) + 1 - R(d)) = M{d}(all) * (1 - .25 R(d)).
Thus, infinitely often, R(d+1) = {.75 M{d}(X)} / {M{d}(all) * (1 - .25 R(d))} = .75 R(d) / (1 - .25 R(d))
Let c be R(d+1) - R(d).
So, infinitely often, we have
R(d) + c = .75 R(d) / (1 - .25 R(d))
c = .75 R(d) / (1 - .25 R(d)) - R(d)
If c goes to zero, R(d) must go to:
0 = .75 R(d) / (1 - .25 R(d)) - R(d)
R(d) = .75 R(d) / (1 - .25 R(d))
1 = .75 / (1 - .25 R(d))
1 - .25 R(d) = .75
.25 = .25 R(d)
R(d) = 1
So we see that if the probability converges, it must go to either 0 or 1! Unless I’ve made some mistake.
I didn’t follow but the conclusion doesn’t sound right to me. Your argument looks like it should apply to any language and proof system. So if I introduce “X” as a logical symbol that isn’t constrained by any axioms or inference rules, why would its probability go to either 0 or 1? It seems to converge to 1⁄2 since for any consistent truth assignment with X=T we have a symmetric consistent truth assignment with X=F.
Yea, I suspect that’s right: so if we introduce X, it must go to either zero or 1, but either option violates the symmetry between X and ~X. Therefore, it must not converge. But this needs to be formalized more before I’m sure.
I think it will converge to 1⁄2 because the symmetry applies at each level of depth, not just at the limit (at least approximately)
To convey my argument without as much of the math:
Suppose that P(X) is 1⁄2 at some stage. Then there will be inconsistent sets yet to remove which will take it at least C away from 1⁄2, where C is a constant that does not go down as the process continues.
The intuition behind this is that removing an inconsistent sentence which has not appeared in any of the inconsistencies removed so far, reduces mass by 1⁄2. Thus, the mass is changing significantly, all the time. Now to make this into a proof we need to show that P(X) changes significantly no matter how far into the process we go; IE, we need to show that a significantly different amount of mass can be removed from the ‘top’ and the ‘bottom’ (in the fraction M(X) / M(all) at finite depth).
The inconsistency {Y, Y->X, ~X} is supposed to achieve this: it only removes mass from the bottom, but there are infinitely many sets like this (we can make Y arbitrarily complex), and each of them reduces the bottom portion by the same fraction without touching the top. Specifically, the bottom becomes 7/8ths of its size (if I’ve done it right), so P(x) becomes roughly .57.
The fraction can re-adjust by decreasing the top in some other way, but this doesn’t allow convergence. No matter how many times the fraction reaches .5 again, there’s a new Y which can be used to force it to .57.