skyrms_axiom_1 - of the pair is equal to the sum of the...

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Our Axiom (1) Can’t Be Derived from Skyrms’s Six “Rules” Branden Fitelson 02/10/07 The argument has three steps: Step 1. Skyrms’s six rules can be derived from Axioms (2) and (3). You will show this on your first problem set. Step 2. Axiom (1) can’t be derived from Axioms (2) and (3). To show this, it is sufficient to give an interpretation on which Axioms (2) and (3) are true but Axiom (1) is false. Consider an interpretation with a Boolean algebra composed of the propositions B = { p, p,p ∨∼ p,p p } . [Convince yourself that B is a Boolean algebra!]. And, with a function Pr ( · ) defined over B as follows: Pr (p) = 2 Pr ( p) = - 1 Pr (p ∨∼ p) = 1 Pr (p p) = 0 Axiom (2): This interpretation contains one tautology, p ∨ ∼ p , and the interpretation assigns it a Pr-value of 1. So Axiom (2) is true on this inter- pretation. Axiom (3): There are four pairs of mutually exclusive propositions on this interpretation: { p, p } , { p,p p } , {∼ p,p p } , and { p ∨∼ p,p p } . You can convince yourself that in each case the Pr-value of the disjunction
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Unformatted text preview: of the pair is equal to the sum of the Pr-values of the disjuncts. So Axiom 3 is true on this interpretation. Axiom (1): It is not the case that on this interpretation the Pr-value of every proposition is non-negative. So Axiom (1) is false on this interpre-tation. Step 3. Derivation is transitive. Given that Skyrmss six rules can be derived from our Axioms (2) and (3), then if Axiom (1) could be derived from the six rules, it could be derived from Axioms (2) and (3). But we just showed that Axiom (1) cant be derived from Axioms (2) and (3). Therefore Axiom (1) cant be derived from Skyrmss six rules. QED . An interesting exercise: Convince yourself that Kolmogorovs axioms are all independent. This wil require showing for each of the three axioms that it cant be derived from the other two....
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