If there’s a short proof of ϕ from ψ and a short proof of ψ from ϕ and they both have relatively long disproofs, then counterfacting on ϕ, ψ should have a high value, and counterfacting on ψ, ϕ should have a high value.
The way to read ⊢ is that the stuff on the left is your collection of axioms (A is a finite collection of axioms and A,ϕ just means we’re using the stuff in A as well as the statement ϕ as our axioms), and it proves some statement.
For the first formulation of the value of a statement, the value would be 1 if adding ϕ doesn’t provide any help in deriving a contradiction from A. Or, put another way, the shortest way of proving ¬ϕ, assuming A as your axioms, is to derive ⊥ and use principle of explosion. It’s “independent” of A, in a sense.
There’s a technicality for “equivalent” statements. We’re considering “equivalent” as “propositionally equivalent given A” (Ie, it’s possible to prove an iff statement with only the statements in A and boolean algebra alone. For example, ¬(ϕ∨ψ)=¬ϕ∧¬ψ is a statement provable with only boolean algebra alone. If you can prove the iff but you can’t do it with boolean algebra alone, it doesn’t count as equivalent. Unless ϕ is propositionally equivalent to ψ, then ϕ∨ψ is not equivalent to ϕ, (because maybe ϕ is false and ψ is true) which renders the equality you wrote wrong, as well as making the last paragraph incoherent.
In classical probability theory, P(ϕ)+P(ψ)=P(ϕ∨ψ) holds iff P(ϕ∧ψ) is 0. Ie, if it’s impossible for both things to happen, the probability of “one of the two things happen” is the same as the sum of the probabilities for event 1 and event 2.
In our thing, we only guarantee equality for when ϕ=¬ψ (assuming A). This is because VA(ϕ∨ψ)=VA(ϕ∨¬ϕ)=VA(⊤)=VA(⊤∧A)=VA(A)=1. (first two = by propositonally equivalent statements getting the same value, the third = by ⊤ being propositionally equivalent to A assuming A, fourth = by ⊤∧A being propositionally equivalent to A, final = by unitarity. Equality may hold in some other cases, but you don’t have a guarantee of such, even if the two events are disjoint, which is a major difference from standard probability theory.
The last paragraph is confused, as previously stated. Also, there’s a law of boolean algebra that ϕ∨(ψ∧Z) is the same as (ψ∨ϕ)∧(ψ∨Z). Also, the intuition is wrong, VA(ϕ∧ψ) should be less than VA(ϕ), because “probability of event 1 happens” is greater than “probability that event 1 and event 2 happens”.
Highlighting something and pressing ctrl-4 turns it to LaTeX.
If there’s a short proof of ϕ from ψ and a short proof of ψ from ϕ and they both have relatively long disproofs, then counterfacting on ϕ, ψ should have a high value, and counterfacting on ψ, ϕ should have a high value.
The way to read ⊢ is that the stuff on the left is your collection of axioms (A is a finite collection of axioms and A,ϕ just means we’re using the stuff in A as well as the statement ϕ as our axioms), and it proves some statement.
For the first formulation of the value of a statement, the value would be 1 if adding ϕ doesn’t provide any help in deriving a contradiction from A. Or, put another way, the shortest way of proving ¬ϕ, assuming A as your axioms, is to derive ⊥ and use principle of explosion. It’s “independent” of A, in a sense.
There’s a technicality for “equivalent” statements. We’re considering “equivalent” as “propositionally equivalent given A” (Ie, it’s possible to prove an iff statement with only the statements in A and boolean algebra alone. For example, ¬(ϕ∨ψ)=¬ϕ∧¬ψ is a statement provable with only boolean algebra alone. If you can prove the iff but you can’t do it with boolean algebra alone, it doesn’t count as equivalent. Unless ϕ is propositionally equivalent to ψ, then ϕ∨ψ is not equivalent to ϕ, (because maybe ϕ is false and ψ is true) which renders the equality you wrote wrong, as well as making the last paragraph incoherent.
In classical probability theory, P(ϕ)+P(ψ)=P(ϕ∨ψ) holds iff P(ϕ∧ψ) is 0. Ie, if it’s impossible for both things to happen, the probability of “one of the two things happen” is the same as the sum of the probabilities for event 1 and event 2.
In our thing, we only guarantee equality for when ϕ=¬ψ (assuming A). This is because VA(ϕ∨ψ)=VA(ϕ∨¬ϕ)=VA(⊤)=VA(⊤∧A)=VA(A)=1. (first two = by propositonally equivalent statements getting the same value, the third = by ⊤ being propositionally equivalent to A assuming A, fourth = by ⊤∧A being propositionally equivalent to A, final = by unitarity. Equality may hold in some other cases, but you don’t have a guarantee of such, even if the two events are disjoint, which is a major difference from standard probability theory.
The last paragraph is confused, as previously stated. Also, there’s a law of boolean algebra that ϕ∨(ψ∧Z) is the same as (ψ∨ϕ)∧(ψ∨Z). Also, the intuition is wrong, VA(ϕ∧ψ) should be less than VA(ϕ), because “probability of event 1 happens” is greater than “probability that event 1 and event 2 happens”.
Highlighting something and pressing ctrl-4 turns it to LaTeX.