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

theorem Th62:
  for n being natural Number st n > 0
  for a being Integer holds a mod n >= 0 & a mod n < n
proof
  let n be natural Number;
  assume
A1: n > 0;
  let a be Integer;
  now
    a div n = [\ a/n /] by INT_1:def 9;
    then a div n <= a/n by INT_1:def 6;
    then (a div n) * n <= a/n * n by XREAL_1:64;
    then (a div n) * n <= (a * n") * n by XCMPLX_0:def 9;
    then (a div n) * n <= a * (n" * n);
    then (a div n) * n <= a * 1 by A1,XCMPLX_0:def 7;
    then (a div n) * n - (a div n) * n <= a - (a div n) * n by XREAL_1:9;
    hence 0 <= a mod n by INT_1:def 10;
    assume a mod n >= n;
    then a - (a div n) * n >= n by A1,INT_1:def 10;
    then (a + -(a div n) * n) + (a div n) * n >= n + (a div n) * n by XREAL_1:6
;
    then a - n >= (n + (a div n) * n) - n by XREAL_1:9;
    then (a - n) * n" >= ((a div n) * n) * n" by XREAL_1:64;
    then (a - n) * n" >= (a div n) * (n * n");
    then a * n" - n * n" >= (a div n) * 1 by A1,XCMPLX_0:def 7;
    then a * n" - 1 >= (a div n) by A1,XCMPLX_0:def 7;
    then
A2: a/n - 1 >= (a div n) by XCMPLX_0:def 9;
    a div n = [\ a/n /] by INT_1:def 9;
    hence contradiction by A2,INT_1:def 6;
  end;
  hence thesis;
end;
