let X be set ; ( X c= [:X,X:] implies X = {} )
assume A1:
X c= [:X,X:]
; X = {}
assume
X <> {}
; contradiction
then consider z being set such that
A2:
z in X
by XBOOLE_0:7;
z in X \/ (union X)
by A2, XBOOLE_0:def 3;
then consider Y being set such that
A3:
Y in X \/ (union X)
and
A4:
for x being set holds
( not x in X \/ (union X) or not x in Y )
by TARSKI:2;
then
Y in union X
by A3, XBOOLE_0:def 3;
then consider Z being set such that
A8:
Y in Z
and
A9:
Z in X
by TARSKI:def 4;
Z in [:X,X:]
by A1, A9, TARSKI:def 3;
then consider x, y being set such that
A10:
x in X
and
y in X
and
A11:
Z = [x,y]
by Def2;
( Y = {x} or Y = {x,y} )
by A8, A11, TARSKI:def 2;
then A12:
x in Y
by TARSKI:def 1, TARSKI:def 2;
x in X \/ (union X)
by A10, XBOOLE_0:def 3;
hence
contradiction
by A4, A12; verum