
theorem
Sierp36 48,12
proof
  thus Sum digits(48,10) = 12 by Th99;
  48=4*12;
  hence 12 divides 48 by INT_1:def 3;
  let m be Nat;
  assume A1: Sum digits(m,10) = 12 & 12 divides m;
  then consider j being Nat such that
  A2: m=12*j by NAT_D:def 3;
  assume m < 48;
  then 12*j < 12*4 by A2;
  then j < 3+1 by XREAL_1:64;
  then j <= 3 by NAT_1:9;
  then j=0 or ... or j=3;
  then per cases;
  suppose j=0;
    then Sum digits(m,10) = 0 by A2,Th6;
    hence contradiction by A1;
  end;
  suppose j=1;
    then Sum digits(m,10) = 3 by A2,Th93;
    hence contradiction by A1;
  end;
  suppose j=2;
    then Sum digits(m,10) = 6 by A2,Th95;
    hence contradiction by A1;
  end;
  suppose j=3;
    then Sum digits(m,10) = 9 by A2,Th97;
    hence contradiction by A1;
  end;
end;
