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

theorem Th61:
  for n,a,k being Integer holds
  (n <> 0 implies (a + n * k) div n = (a div n) + k) &
  (a + n * k) mod n = a mod n
proof
  let n,a,k be Integer;
  thus
A1: now
    assume
A2: n <> 0;
    thus (a + n * k) div n = [\ (a + n * k)/n /] by INT_1:def 9
      .= [\ (a + n * k) * n" /] by XCMPLX_0:def 9
      .= [\ a * n" + (n * n") * k /]
      .= [\ a * n" + 1 * k /] by A2,XCMPLX_0:def 7
      .= [\ a * n" /] + k by INT_1:28
      .= [\ a/n /] + k by XCMPLX_0:def 9
      .= (a div n) + k by INT_1:def 9;
  end;
  per cases;
  suppose
A3: n <> 0;
    hence (a + n * k) mod n = (a + n * k) - ((a div n) + k) * n by A1,
INT_1:def 10
      .= a - (a div n) * n
      .= a mod n by A3,INT_1:def 10;
  end;
  suppose
    n = 0;
    hence thesis;
  end;
end;
