reserve a,b,p,k,l,m,n,s,h,i,j,t,i1,i2 for natural Number;

theorem
  k mod n = 0 implies (k + l) div n = (k div n) + (l div n)
proof
  assume
A1: k mod n = 0;
  per cases;
  suppose
A2: n <> 0;
    then
A3: k = n * (k div n) + 0 by A1,INT_1:59;
A4: ex t being Nat st l = n * t + (l mod n) & l mod n < n by A2,Def2;
    l = n * (l div n) + (l mod n) by A2,INT_1:59;
    then k + l = n * ((k div n) + (l div n)) + (l mod n) by A3;
    hence thesis by A4,Def1;
  end;
  suppose n = 0;
    hence thesis;
  end;
end;
