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.
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.