# Proving Expected Utility Theorem

I am struggling to understand the proof of the second step in the Expected Utility Theorem, particularly the part that deals with preferences over weighted sums of lotteries. The statement I am trying to prove is:

For the best possible lottery $$L_h$$ and the worst possible lottery $$L_l$$, the compound lottery $$bL_h + (1-b)L_l$$ is preferred over $$aL_h + (1-a)L_l$$ if and only if $$b > a$$.

I've attempted to prove it by contradiction but have encountered difficulties in rigorously showing how having 'more' of $$L_h$$ in one lottery over another implies it should be preferred. I of course understand the intuition, but I am struggling with the formal proof using the independence axiom.

We have that $$L_h \succ L_l$$.
Then as $$L_h \succ L_l$$ we have by IIA that for $$b \in ]0,1[$$ $$b L_h + (1-b) L_l \succ b L_l + (1-b) L_l = L_l.$$ Next as $$b L_h + (1-b) L_l \succeq L_l$$ we have again by IIA that for all all $$q \in ]0,1[$$ $$q(b L_h + (1-b) L_l) + (1-q) (b L_h + (1-b) L_l) \succ q L_l + (1-q)(b L_h + (1-b) L_l)$$ Simplifying gives: $$b L_h + (1-b) L_l \succ (1-q) b L_h + [(1-b) + bq] L_l$$ Now let $$q = \frac{b- a}{b} \in ]0,1[$$ (because $$b > a$$). This gives: $$b L_h + (1-b) L_l \succ a L_h + (1-a) L_l.$$