Proof: (a) In the case of equal probabilities Let H(N,…1N)=A(N) By use condition 3), it is then easy to have A(MN=HS(IMN,., IMN Hs(1M,…,1M)+,,(1M)(1N,…,1N) Then A(M+AN a(N2=2A(N), A(S)=CA(S), A(t3=BA(t) For any given B, it is always possible to find a proper a such that a+1Proof: (a) In the case of equal probabilities S S H (1/N, …, 1/N) = A(N) By use condition (3), it is then easy to have A(MN) = H (1/MN, …, 1/MN) S S M i=1 = A(M) + A(N) Then A(N ) = 2 A(N), A(S ) = A(S), A(t ) = A(t) 2 a a b b Let = H (1/M, …, 1/M) + (1/M) H (1/N, …, 1/N) For any given b, it is always possible to find a proper a such that S t < S a b a+1 (*)