theorem Th37:
  n0,m0 are_coprime implies sigma(n0*m0) = sigma(n0) * sigma m0
proof
  assume
A1: n0,m0 are_coprime;
A2: Sigma 1 is multiplicative by Th36;
  thus sigma(n0*m0) = (Sigma 1).(n0*m0) by Def3
    .= ((Sigma 1).n0) * (Sigma 1).m0 by A1,A2
    .= sigma(1,n0) * (Sigma 1).m0 by Def3
    .= sigma(n0) * sigma m0 by Def3;
end;
