Proof: Assume we have a preference ordering that satisfies A1, A2, A3, A5, and B2, and consider lotteries A,B∈L, and p∈[0,1], with A⪰B. Let C=pA+(1−p)B. It suffices to show C⪰B. Assume not, for the purpose of contradiction. Then (by axiom A1), B≻C. Thus by axiom B2 there exists a D∈L such that B≻D≻C. By axiom A3, we may assume D=qB+(1−q)C for some q∈[0,1]. Observe that C=rA+(1−r)D where r=pq1−p+pq∈[0,1]. r is positive, since otherwise C=D≻C. Thus, we can apply A5 to get that since D⪰rA+(1−r)D, we have D⪰A. Thus D⪰A⪰B≻D, a contradiction.
Claim: A1, A2, A3, A5, and B2 imply A4.
Proof: Assume we have a preference ordering that satisfies A1, A2, A3, A5, and B2, and consider lotteries A,B∈L, and p∈[0,1], with A⪰B. Let C=pA+(1−p)B. It suffices to show C⪰B. Assume not, for the purpose of contradiction. Then (by axiom A1), B≻C. Thus by axiom B2 there exists a D∈L such that B≻D≻C. By axiom A3, we may assume D=qB+(1−q)C for some q∈[0,1]. Observe that C=rA+(1−r)D where r=pq1−p+pq∈[0,1]. r is positive, since otherwise C=D≻C. Thus, we can apply A5 to get that since D⪰rA+(1−r)D, we have D⪰A. Thus D⪰A⪰B≻D, a contradiction.