For all A, B, C, D and p, if A ≤ B and C ≤ D, then pA + (1-p)C ≤ pB + (1-p)D.
If we now take C and D to be the same lottery, we get independence, as long as C ≤ C. Now, given completeness, C ≤ C is always true (because at least one of C=C, CC must be true, and thus we can always get C ≤ C, -- switching C with C if needed!).
So we don’t need consistency, we need a weak form of completeness, in which every lottery can be at least compared with itself.
Mathematically:
Independence + other 3 axioms ⇒ Independence II
Independence II ⇒ Independence
Hence: ~Independence ⇒ ~Independence II
My theorem implies: ~Independence II ⇒ You can be money pumped
Hence: ~Independence ⇒ You can be money pumped
Note, Independence II does not imply Independence, without using at least the consistency axiom.
The contrapositive of independence II is:
For all A, B, C, D and p, if A ≤ B and C ≤ D, then pA + (1-p)C ≤ pB + (1-p)D.
If we now take C and D to be the same lottery, we get independence, as long as C ≤ C. Now, given completeness, C ≤ C is always true (because at least one of C=C, CC must be true, and thus we can always get C ≤ C, -- switching C with C if needed!).
So we don’t need consistency, we need a weak form of completeness, in which every lottery can be at least compared with itself.
Transitivity and Continuity are unnecessary, however.
That is my reading of it too. I know Stuart is putting forward analytic results here, I was concerned that this one was not correctly represented.