
theorem Th2:
  for X being set holds 2 c= card X iff ex x,y being object st x in X
  & y in X & x<>y
proof
  let X be set;
  thus 2 c= card X implies ex x,y being object st x in X & y in X & x<>y
  proof
    assume 2 c= card X;
    then card 2 c= card X;
    then consider f being Function such that
A1: f is one-to-one and
A2: dom f = 2 and
A3: rng f c= X by CARD_1:10;
    take x=f.0;
    take y=f.1;
A4: 0 in dom f by A2,CARD_1:50,TARSKI:def 2;
    then f.0 in rng f by FUNCT_1:def 3;
    hence x in X by A3;
A5: 1 in dom f by A2,CARD_1:50,TARSKI:def 2;
    then f.1 in rng f by FUNCT_1:def 3;
    hence y in X by A3;
    thus thesis by A1,A4,A5,FUNCT_1:def 4;
  end;
  given x,y being object such that
A6: x in X & y in X and
A7: x<>y;
  {x,y} c= X
  by A6,TARSKI:def 2;
  then card {x,y} c= card X by CARD_1:11;
  hence thesis by A7,CARD_2:57;
end;
