let X, Y be set ; ( ( X c= [:X,Y:] or X c= [:Y,X:] ) implies X = {} )
assume A1:
( X c= [:X,Y:] or X c= [:Y,X:] )
; X = {}
assume A2:
X <> {}
; contradiction
now consider z being
set such that A19:
z in X
by A2, XBOOLE_0:7;
z in X \/ (union X)
by A19, XBOOLE_0:def 3;
then consider Y1 being
set such that A20:
Y1 in X \/ (union X)
and A21:
for
x being
set holds
( not
x in X \/ (union X) or not
x in Y1 )
by TARSKI:2;
assume A22:
X c= [:X,Y:]
;
contradictionthen
Y1 in union X
by A20, XBOOLE_0:def 3;
then consider Y2 being
set such that A26:
Y1 in Y2
and A27:
Y2 in X
by TARSKI:def 4;
Y2 in [:X,Y:]
by A22, A27, TARSKI:def 3;
then consider x,
y being
set such that A28:
x in X
and
y in Y
and A29:
Y2 = [x,y]
by Def2;
(
Y1 = {x} or
Y1 = {x,y} )
by A26, A29, TARSKI:def 2;
then A30:
x in Y1
by TARSKI:def 1, TARSKI:def 2;
x in X \/ (union X)
by A28, XBOOLE_0:def 3;
hence
contradiction
by A21, A30;
verum end;
hence
contradiction
by A1, A3; verum