
theorem Th7:
  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
A2: n <> 0;
    then
A3: b mod n + (b div n) * n = (b - (b div n) * n) + (b div n) * n by
INT_1:def 10;
    a mod n + (a div n) * n = (a - (a div n) * n) + (a div n) * n by A2,
INT_1:def 10;
    then (a - b) - ((a mod n) - (b mod n)) = ((a div n) - (b div n)) * n by A3;
    then n divides (a - b) - ((a mod n) - (b mod n)) by INT_1:def 3;
    hence thesis by NAT_D:64,INT_2:15;
  end;
end;
