即证明\((A*A)*A=A*(A*A)\)
因为\(A*A=B\)，所以即证明\(B*A=A*B\)
设\(C_1=B*A\)，那么有$$C_1[i][j]=min_{1≤k≤p}(B[i,k]+A[k,j])$$
\[=min_{1≤k≤p}(min_{1≤l≤p}(A[i][l]+A[l][k])+A[k,j])
\]
\[=min_{1≤k≤p}(min_{1≤l≤p}(A[i][l]+A[l][k]+A[k,j]))
\]
\[=min_{1≤k≤p,1≤l≤p}(A[i][l]+A[l][k]+A[k,j])
\]
设\(C_2=A*B\)，同理有$$C_2[i][j]=min_{1≤k≤p,1≤l≤p}(A[i][k]+A[k][l]+A[l,j])$$
显然\(C_1=C_2\)