
theorem Th5:
  for X being set, Y being non empty set, f being Function of X, Y
  holds {f"{y} where y is Element of Y : y in rng f} is a_partition of X
proof
  let X be set, Y be non empty set, f be Function of X, Y;
  set P = {f"{y} where y is Element of Y : y in rng f};
  for x being object holds x in X iff ex A being set st x in A & A in P
  proof
    let x be object;
    hereby
      assume A1: x in X;
      then A2: f.x in rng f & f.x in Y by FUNCT_2:4, FUNCT_2:5;
      reconsider A = f"{f.x} as set;
      take A;
      f.x in {f.x} by TARSKI:def 1;
      hence x in A by A1, FUNCT_2:38;
      thus A in P by A2;
    end;
    given A being set such that
      A3: x in A & A in P;
    consider y being Element of Y such that
      A4: f"{y} = A & y in rng f by A3;
    thus x in X by A3, A4;
  end;
  then A5: union P = X by TARSKI:def 4;
  A6: for A being Subset of X st A in P holds A <> {} &
    for B being Subset of X st B in P holds A = B or A misses B
  proof
    let A be Subset of X;
    assume A in P;
    then consider y1 being Element of Y such that
      A7: A = f"{y1} & y1 in rng f;
    consider x being object such that
      A8: x in X & y1 = f.x by A7, FUNCT_2:11;
    f.x in {y1} by A8, TARSKI:def 1;
    hence A <> {} by A7, A8, FUNCT_2:38;
    let B be Subset of X;
    assume B in P;
    then ex y2 being Element of Y st B = f"{y2} & y2 in rng f;
    hence thesis by A7, ZFMISC_1:11, FUNCT_1:71;
  end;
  P c= bool X
  proof
    let x be object;
    assume x in P;
    then ex y being Element of Y st f"{y} = x & y in rng f;
    hence thesis;
  end;
  hence thesis by A5, A6, EQREL_1:def 4;
end;
