let b be Ordinal; :: thesis: for f being Function holds
( ( f is b -limited & not f is empty & f is segmental ) iff ex a being Ordinal st
( dom f = b \ a & a in b ) )

let f be Function; :: thesis: ( ( f is b -limited & not f is empty & f is segmental ) iff ex a being Ordinal st
( dom f = b \ a & a in b ) )

thus ( f is b -limited & not f is empty & f is segmental implies ex a being Ordinal st
( dom f = b \ a & a in b ) ) :: thesis: ( ex a being Ordinal st
( dom f = b \ a & a in b ) implies ( f is b -limited & not f is empty & f is segmental ) )
proof
assume A1: b = sup (dom f) ; :: according to EXCHSORT:def 3 :: thesis: ( f is empty or not f is segmental or ex a being Ordinal st
( dom f = b \ a & a in b ) )

assume A2: not f is empty ; :: thesis: ( not f is segmental or ex a being Ordinal st
( dom f = b \ a & a in b ) )

given c, d being Ordinal such that A3: dom f = c \ d ; :: according to EXCHSORT:def 1 :: thesis: ex a being Ordinal st
( dom f = b \ a & a in b )

set x = the Element of c \ d;
take a = d; :: thesis: ( dom f = b \ a & a in b )
A4: b = c by A2, A1, A3, Th6;
thus dom f = b \ a by A2, A1, A3, Th6; :: thesis: a in b
( a c= the Element of c \ d & the Element of c \ d in b ) by A2, A3, A4, ORDINAL6:5;
hence a in b by ORDINAL1:12; :: thesis: verum
end;
given a being Ordinal such that A5: ( dom f = b \ a & a in b ) ; :: thesis: ( f is b -limited & not f is empty & f is segmental )
a in dom f by A5, ORDINAL6:5;
hence b = sup (dom f) by A5, Th6; :: according to EXCHSORT:def 3 :: thesis: ( not f is empty & f is segmental )
thus not f is empty by A5, ORDINAL6:5; :: thesis: f is segmental
take b ; :: according to EXCHSORT:def 1 :: thesis: ex b being Ordinal st proj1 f = b \ b
take a ; :: thesis: proj1 f = b \ a
thus proj1 f = b \ a by A5; :: thesis: verum