reserve k,m,n for Nat,
  i1,i2,i3 for Integer,
  e for set;

theorem Th3:
  m <> 0 implies (k mod (m*n)) mod n = k mod n
proof
  assume
A1: m <> 0;
  per cases;
  suppose
A2: n <> 0;
    reconsider k9 = k, m9 = m, n9 = n as Integer;
    m9*n9 <> 0 by A1,A2,XCMPLX_1:6;
    hence (k mod (m*n)) mod n = (k9 - (k9 div (m9*n9))*(m9*n9)) mod n9 by
INT_1:def 10
      .= (k9 + (-(m9*(k9 div m9*n9)))*n9) mod n9
      .= k mod n by NAT_D:61;
  end;
  suppose
A3: n = 0;
    hence (k mod (m*n)) mod n = 0 by NAT_D:def 2
      .= k mod n by A3,NAT_D:def 2;
  end;
end;
