let f be Function; :: thesis: ( f is Tree-yielding iff for x being object st x in dom f holds
f . x is Tree )
thus ( f is Tree-yielding implies for x being object st x in dom f holds
f . x is Tree ) :: thesis: ( ( for x being object st x in dom f holds
f . x is Tree ) implies f is Tree-yielding )

proof

assume A1: for x being object st x in rng f holds
x is Tree ; :: according to TREES_3:def 3,TREES_3:def 9 :: thesis: for x being object st x in dom f holds
f . x is Tree
let x be object ; :: thesis: ( x in dom f implies f . x is Tree )
assume x in dom f ; :: thesis: f . x is Tree
then f . x in rng f by FUNCT_1:def 3;
hence f . x is Tree by A1; :: thesis: verum

end;

assume A2: for x being object st x in dom f holds
f . x is Tree ; :: thesis: f is Tree-yielding
let x be object ; :: according to TREES_3:def 3,TREES_3:def 9 :: thesis: ( x in rng f implies x is Tree )
assume x in rng f ; :: thesis: x is Tree
then ex y being object st
( y in dom f & x = f . y ) by FUNCT_1:def 3;
hence x is Tree by A2; :: thesis: verum