In a message dated 5/4/99 2:13:30 PM Eastern Daylight Time,
trybulec at math.uwb.edu.pl writes:
> One cannot prove it anyway
>
> (x,y) U {x} = {{x},{x,y}} U {x} = {{x},{x,y},x} =/= (x,y)
> by regularity.
>
Sorry, of course, the right "theorem" is (x, y) U {{x}} = (x, y), because
(x,y) = { {x}, {x, y} }.
V. Makarov