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 mod n = l mod n
proof
  assume that
A1: k mod n = 0;
  per cases;
  suppose
A2: n = 0;
    hence k + l mod n = 0 by Def2
      .= l mod n by A2,Def2;
  end;
  suppose
A3: n <> 0;
    then consider m being Nat such that
A4: k = n * m + 0 and 0 < n by A1,Def2;
    consider t being Nat such that
A5: l = n * t + (l mod n) and
A6: l mod n < n by A3,Def2;
    k + l = n * (m + t) + (l mod n) by A4,A5;
    hence thesis by A6,Def2;
  end;
end;
