reserve A,B for Ordinal,
  K,M,N for Cardinal,
  x,x1,x2,y,y1,y2,z,u for object,X,Y,Z,X1,X2, Y1,Y2 for set,
  f,g for Function;
reserve m,n for Nat;
reserve x1,x2,x3,x4,x5,x6,x7,x8 for object;
reserve A,B,C for Ordinal,
  K,L,M,N for Cardinal,
  x,y,y1,y2,z,u for object,X,Y,Z,Z1,Z2 for set,
  n for Nat,
  f,f1,g,h for Function,
  Q,R for Relation;
reserve n,k for Nat;

theorem
  card X +` card Y = card X & card Y in card X implies card (X \ Y) = card X
proof
  assume that
A1: card X +` card Y = card X and
A2: card Y in card X;
A3: card X c= card (X \/ Y) by CARD_1:11,XBOOLE_1:7;
  card (X \/ Y) c= card X by A1,Th33;
  then
A4: card (X \/ Y) = card X by A3;
  (X \ Y) \/ Y = X \/ Y by XBOOLE_1:39;
  then
A5: card X = card (X \ Y) +` card Y by A4,Th34,XBOOLE_1:79;
  then
A6: card (X \ Y) c= card X by Th93;
A7: now
    assume not card X is finite;
    then
A8: card X +` card X = card X by Th74;
    assume not thesis;
    then card (X \ Y) in card X by A6,CARD_1:3;
    then card X in card X by A2,A5,A8,Th95;
    hence contradiction;
  end;
  now
    assume card X is finite;
    then reconsider X,Y as finite set by A2,CARD_3:92;
    card Y = card card Y;
    then card (card X + card Y) = card card X by A1,Th37;
    then card X + card Y = card X + 0;
    then Y = {};
    hence thesis;
  end;
  hence thesis by A7;
end;
