reserve X,X1,X2,Y,Y1,Y2 for set, p,x,x1,x2,y,y1,y2,z,z1,z2 for object;
reserve f,g,g1,g2,h for Function,
  R,S for Relation;

theorem
  f is one-to-one iff for y ex x st f"{y} c= {x}
proof
  (for y ex x st f"{y} c= {x}) iff for y st y in rng f ex x st f"{y} = {x}
  proof
    thus (for y ex x st f"{y} c= {x}) implies for y st y in rng f ex x st f"{y
    } = {x}
    proof
      assume
A1:   for y ex x st f"{y} c= {x};
      let y;
      consider x such that
A2:   f"{y} c= {x} by A1;
      assume y in rng f;
      then consider x1 being object such that
A3:   x1 in dom f and
A4:   y = f.x1 by Def3;
      take x;
      f.x1 in {y} by A4,TARSKI:def 1;
      then f"{y} <> {} by A3,Def7;
      hence thesis by A2,ZFMISC_1:33;
    end;
    assume
A5: for y st y in rng f ex x st f"{y} = {x};
    let y;
A6: now
      set x = the set;
      assume
A7:   not y in rng f;
      take x;
      rng f misses {y} by A7,ZFMISC_1:50;
      then f"{y} = {} by RELAT_1:138;
      hence f"{y} c= {x};
    end;
    now
      assume y in rng f;
      then consider x such that
A8:   f"{y} = {x} by A5;
      take x;
      thus f"{y} c= {x} by A8;
    end;
    hence thesis by A6;
  end;
  hence thesis by Th73;
end;
