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.
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.