Theorem tfrlem4

Description: Lemma for transfinite recursion. A is the class of all "acceptable" functions, and F is their union. First we show that an acceptable function is in fact a function. (Contributed by NM, 9-Apr-1995)

	Ref	Expression
Hypothesis	tfrlem.1	|- A = { f | E. x e. On ( f Fn x /\ A. y e. x ( f ` y ) = ( F ` ( f |` y ) ) ) }
Assertion	tfrlem4	|- ( g e. A -> Fun g )

Proof

Step	Hyp	Ref	Expression
1		tfrlem.1	|- A = { f | E. x e. On ( f Fn x /\ A. y e. x ( f ` y ) = ( F ` ( f |` y ) ) ) }
2	1	tfrlem3	|- A = { g | E. z e. On ( g Fn z /\ A. w e. z ( g ` w ) = ( F ` ( g |` w ) ) ) }
3	2	eqabri	|- ( g e. A <-> E. z e. On ( g Fn z /\ A. w e. z ( g ` w ) = ( F ` ( g |` w ) ) ) )
4		fnfun	|- ( g Fn z -> Fun g )
5	4	adantr	|- ( ( g Fn z /\ A. w e. z ( g ` w ) = ( F ` ( g |` w ) ) ) -> Fun g )
6	5	rexlimivw	|- ( E. z e. On ( g Fn z /\ A. w e. z ( g ` w ) = ( F ` ( g |` w ) ) ) -> Fun g )
7	3 6	sylbi	|- ( g e. A -> Fun g )