Theorem fprfung

Description: A "function" defined by well-founded recursion is indeed a function when the relationship is a partial order. Avoids the axiom of replacement. (Contributed by Scott Fenton, 18-Nov-2024)

	Ref	Expression
Hypothesis	fprfung.1	|- F = frecs ( R , A , G )
Assertion	fprfung	|- ( ( R Fr A /\ R Po A /\ R Se A ) -> Fun F )

Proof

Step	Hyp	Ref	Expression
1		fprfung.1	|- F = frecs ( R , A , G )
2		eqid	|- { f | E. x ( f Fn x /\ ( x C_ A /\ A. y e. x Pred ( R , A , y ) C_ x ) /\ A. y e. x ( f ` y ) = ( y G ( f |` Pred ( R , A , y ) ) ) ) } = { f | E. x ( f Fn x /\ ( x C_ A /\ A. y e. x Pred ( R , A , y ) C_ x ) /\ A. y e. x ( f ` y ) = ( y G ( f |` Pred ( R , A , y ) ) ) ) }
3	2 1	fprlem1	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( g e. { f | E. x ( f Fn x /\ ( x C_ A /\ A. y e. x Pred ( R , A , y ) C_ x ) /\ A. y e. x ( f ` y ) = ( y G ( f |` Pred ( R , A , y ) ) ) ) } /\ h e. { f | E. x ( f Fn x /\ ( x C_ A /\ A. y e. x Pred ( R , A , y ) C_ x ) /\ A. y e. x ( f ` y ) = ( y G ( f |` Pred ( R , A , y ) ) ) ) } ) ) -> ( ( x g u /\ x h v ) -> u = v ) )
4	2 1 3	frrlem9	|- ( ( R Fr A /\ R Po A /\ R Se A ) -> Fun F )