Question 2.18

I have only been able to make progress on this question by assuming a modified version of G5 stated in terms of preference instead of indifference:-

G5*. If (p:g) and (p:h) are in G and g#h then (p:g)#(p:h).

I hope my adapted notation is not too baffling!

Look upon the p, g and h as vectors.
: denotes the circle operation in the text. # denotes the preference ordering.
I shall also use t instead of alpha.

Put p=(t,1-t),g=((p:a),(r:a)) and
h=((q:a),(r:a)).

We have (p:a)#(q:a) and (r:a)#(r:a) so
(t:(p:a),(1-t):(r:a))#(t:(q:a),(1-t):(r:a)) by G5*.

The simple gamble induced by
(t:(p:a),(1-t):(r:a)) is
(tp+((1-t)r:a) and by G6 these are indifferent. Similarly for the expression in q. The result follows.

But with G5 alone....?