[FOM] "Leibniz's Law"

Jeffrey Ketland ketland at ketland.fsnet.co.uk
Sat Jun 7 19:12:32 EDT 2003

The following seems to be a probabilistic version of Leibniz. Using the
probability axioms, we get:

     If F is not opaque and Pr(a=b) = 1, then Pr(Fa) = Pr(Fb).

Set Pr(a=b) = 1 and assume F is not opaque. So, Fa implies a=b -> Fb.
So, Pr(Fa) <= Pr(~(a=b) or Fb)
We can use: Pr(X or Y) = Pr(X) + Pr(Y) - Pr(X & Y)
   Pr(Fa) <= Pr(~(a=b) or Fb)
               <= 1 - Pr(a=b) + Pr(Fb) - Pr(~(a=b) & Fb)
               <= Pr(Fb) - Pr(~(a=b) & Fb)
Also, Pr(~(a=b)) = 0. And ~(a=b) & Fb implies ~(a=b).
So, Pr(~(a=b & Fb) <= Pr(~(a=b)) <= 0 So, Pr(~(a=b) & Fb) = 0.
So, Pr(Fa) <= Pr(Fb).
By a similar argument, Pr(Fb) <= Pr(Fa).
So, Pr(Fa) = Pr(Fb).

Is that right?
If so, then (assuming you're probabilistically coherent) if you assign
distinct probabilities to "Bacon wrote M" and "Shakespeare wrote M", you
must assign a probability < 1 to "Bacon = Shakespeare".

I don't see how to prove that if a = b then Pr(Fa) = Pr(Fb). One would
expect it to hold on an objective interpretation of probability, like
Popper's propensity view.

--- Jeff

Jeffrey Ketland
School of Philosophy
University of Leeds
j.j.ketland at leeds.ac.uk
ketland at ketland.fsnet.co.uk

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