theorem lfnauxeq3 (F: set) (k _n1 _n2: nat): $ _n1 = _n2 -> lfnaux F k _n1 = lfnaux F k _n2 $;
_n1 = _n2 -> _n1 = _n2
_n1 = _n2 -> lfnaux F k _n1 = lfnaux F k _n2