let X be non trivial set ; for x being Element of X ex y being set st
( y in X & x <> y )
let x be Element of X; ex y being set st
( y in X & x <> y )
ex y being set st
( y in X & y <> x )
proof
assume A1:
for
y being
set holds
( not
y in X or
y = x )
;
contradiction
X c= {x}
hence
contradiction
;
verum
end;
then consider y being set such that
A2:
( y in X & y <> x )
;
take
y
; ( y in X & x <> y )
thus
( y in X & x <> y )
by A2; verum