
theorem
Sierp36 190,10
proof
  thus Sum digits(190,10) = 10 by Th53;
  190=19*10;
  hence 10 divides 190 by INT_1:def 3;
  let m be Nat;
  assume A1: Sum digits(m,10) = 10 & 10 divides m;
  then consider j being Nat such that
  A2: m=10*j by NAT_D:def 3;
  assume m < 190;
  then 10*j < 10*19 by A2;
  then j < 18+1 by XREAL_1:64;
  then j <= 18 by NAT_1:9;
  then j=0 or ... or j=18;
  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) = 1 by A2,Th17;
    hence contradiction by A1;
  end;
  suppose j=2;
    then Sum digits(m,10) = 2 by A2,Th19;
    hence contradiction by A1;
  end;
  suppose j=3;
    then Sum digits(m,10) = 3 by A2,Th21;
    hence contradiction by A1;
  end;
  suppose j=4;
    then Sum digits(m,10) = 4 by A2,Th23;
    hence contradiction by A1;
  end;
  suppose j=5;
    then Sum digits(m,10) = 5 by A2,Th25;
    hence contradiction by A1;
  end;
  suppose j=6;
    then Sum digits(m,10) = 6 by A2,Th27;
    hence contradiction by A1;
  end;
  suppose j=7;
    then Sum digits(m,10) = 7 by A2,Th29;
    hence contradiction by A1;
  end;
  suppose j=8;
    then Sum digits(m,10) = 8 by A2,Th31;
    hence contradiction by A1;
  end;
  suppose j=9;
    then Sum digits(m,10) = 9 by A2,Th33;
    hence contradiction by A1;
  end;
  suppose j=10;
    then Sum digits(m,10) = 1 by A2,Th35;
    hence contradiction by A1;
  end;
  suppose j=11;
    then Sum digits(m,10) = 2 by A2,Th37;
    hence contradiction by A1;
  end;
  suppose j=12;
    then Sum digits(m,10) = 3 by A2,Th39;
    hence contradiction by A1;
  end;
  suppose j=13;
    then Sum digits(m,10) = 4 by A2,Th41;
    hence contradiction by A1;
  end;
  suppose j=14;
    then Sum digits(m,10) = 5 by A2,Th43;
    hence contradiction by A1;
  end;
  suppose j=15;
    then Sum digits(m,10) = 6 by A2,Th45;
    hence contradiction by A1;
  end;
  suppose j=16;
    then Sum digits(m,10) = 7 by A2,Th47;
    hence contradiction by A1;
  end;
  suppose j=17;
    then Sum digits(m,10) = 8 by A2,Th49;
    hence contradiction by A1;
  end;
  suppose j=18;
    then Sum digits(m,10) = 9 by A2,Th51;
    hence contradiction by A1;
  end;
end;
