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

theorem Th66:
  for n,a,b being Integer holds (a + b) mod n = ((a mod n) + (b mod n)) mod n
proof
  let n,a,b be Integer;
  per cases;
  suppose
A1: n = 0;
    hence (a + b) mod n = 0 by INT_1:def 10
      .= ((a mod n) + (b mod n)) mod n by A1,INT_1:def 10;
  end;
  suppose
    n <> 0;
    then
    a mod n + (a div n) * n = (a - (a div n) * n) + (a div n) * n & b mod
    n + (b div n) * n = (b - (b div n) * n) + (b div n) * n by INT_1:def 10;
    then (a + b) - ((a mod n) + (b mod n)) = ((a div n) + (b div n)) * n;
    then n divides (a + b) - ((a mod n) + (b mod n)) by INT_1:def 3;
    then a+b,(a mod n)+(b mod n) are_congruent_mod n by INT_2:15;
    hence thesis by Th64;
  end;
end;
