Metamath Proof Explorer


Theorem fpr3

Description: Law of founded partial recursion, 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 )