civic rapids
There are 11 premises (you probably only need 4-6 of them tho) [P(~AB) is short for P(~A&B)]
- P(ABC) ≈ P(BC)
- P(AB) ≈ P(B)
- P(~A~BC) ≈ P(~BC)
- P(~A~B) ≈ P(~B)
- P(~A~BC) ≈ P(~B)
- P(ABC) << P(B)
- P(~A~B~C) << P(~B)
- P(A~BC) << P(A~B)
- P(~ABC) ≈ P(~AB)
- P(~A~BC) ≈ P(~A~B)
- P(A~C) ≈ P(A)
we want to prove ¬(P(B | ¬A & C) ≈ 1). It's very important that these are approximate premises. This for example, means that you cannot use P(C | ~A) ≈ 1 to mean if ~A then C: that makes an assumption that you would need to support, i.e. P(C | ~A) ≈ 1 ≠ P(C | ~A) = 1.
Countermodel
Ω=[0,1] with the standard measure
A=[0, 1-ε²-ε³]
B=[ε⁴, 1-ε³]
C=[1-ε, 1]
These are the hardest to satisfy:
P(A|B&C) = (1-ε²-ε³)/(1-ε²) ≈ 1
P(~A&~C|~B) = 0
P(C|A&~B) = 0
P(C|~A&B) = 1
The full check:
- P(ABC) ≈ P(BC): ABC=[1-ε, 1-ε²-ε³], BC=[1-ε, 1-ε³] ✓
- P(AB) ≈ P(B): AB=[ε⁴, 1-ε²-ε³], B=[ε⁴, 1-ε³] ✓
- P(~A~BC) ≈ P(~BC): ~A~BC=[1-ε²-ε³, 1], ~BC=[1-ε³, 1] ✓
- P(~A~B) ≈ P(~B): ~A~B=[1-ε²-ε³, 1], ~B=[0, ε⁴]∪[1-ε³, 1] ✓
- P(~A~BC) ≈ P(~B): - same as above -
- P(ABC) << P(B): ABC=[1-ε, 1-ε²-ε³], B=[ε⁴, 1-ε³] ✓
- P(~A~B~C) << P(~B): ~A~B~C=∅, ~B≠∅ ✓
- P(A~BC) << P(A~B): A~BC=∅, A~B≠∅ ✓
- P(~ABC) ≈ P(~AB): ~ABC=~AB=[1-ε²-ε³, 1-ε³] ✓
- P(~A~BC) ≈ P(~A~B): ~A~BC=~A~B=[1-ε³, 1] ✓
- P(A~C) ≈ P(A): A~C=[0, 1-ε], A=[0, 1-ε²-ε³] ✓
anticonclusion:
~ABC=[1-ε²-ε³, 1-ε³], ~AC=[1-ε²-ε³, 1]
=> P(~ABC) ≈ P(~AC)
=> P(B | ~AC) ≈ 1
additional premise
however, we found an additional premise, namely: P( ~A | ~B~C) >> P( A | ~B~C), which contradicts the countermodel:
P( ~A | ~B~C) >> P( A | ~B~C) means (by def. conditional probability) that P(~A~B~C) >> P(A~B~C): ~A~B~C=∅, A~B~C=[0,ε⁴] ❌
I'm super stuck... So wondering if anyone here can find a proof with this additional premise: a proof would solve the Boltzmann brain paradox or is there are countermodel? Curious to hear your thoughts <3
noble carbon