Theorem fpr3

Description: Law of well-founded recursion over a partial order, part three. Finally, we show that F is unique. We do this by showing that any function H with the same properties we proved of F in fpr1 and fpr2 is identical to F . (Contributed by Scott Fenton, 11-Sep-2023)

	Ref	Expression
Hypothesis	fprr.1	|- F = frecs ( R , A , G )
Assertion	fpr3	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) ) -> F = H )

Proof

Step	Hyp	Ref	Expression
1		fprr.1	|- F = frecs ( R , A , G )
2		simpl	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) ) -> ( R Fr A /\ R Po A /\ R Se A ) )
3	1	fpr1	|- ( ( R Fr A /\ R Po A /\ R Se A ) -> F Fn A )
4	1	fpr2	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ z e. A ) -> ( F ` z ) = ( z G ( F |` Pred ( R , A , z ) ) ) )
5	4	ralrimiva	|- ( ( R Fr A /\ R Po A /\ R Se A ) -> A. z e. A ( F ` z ) = ( z G ( F |` Pred ( R , A , z ) ) ) )
6	3 5	jca	|- ( ( R Fr A /\ R Po A /\ R Se A ) -> ( F Fn A /\ A. z e. A ( F ` z ) = ( z G ( F |` Pred ( R , A , z ) ) ) ) )
7	6	adantr	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) ) -> ( F Fn A /\ A. z e. A ( F ` z ) = ( z G ( F |` Pred ( R , A , z ) ) ) ) )
8		simpr	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) ) -> ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) )
9		fpr3g	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( F Fn A /\ A. z e. A ( F ` z ) = ( z G ( F |` Pred ( R , A , z ) ) ) ) /\ ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) ) -> F = H )
10	2 7 8 9	syl3anc	|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ ( H Fn A /\ A. z e. A ( H ` z ) = ( z G ( H |` Pred ( R , A , z ) ) ) ) ) -> F = H )