Metamath Proof Explorer


Theorem fpr2

Description: Law of well-founded recursion over a partial order, part two. Now we establish the value of F within A . (Contributed by Scott Fenton, 11-Sep-2023) (Proof shortened by Scott Fenton, 18-Nov-2024)

Ref Expression
Hypothesis fprr.1 F = frecs R A G
Assertion fpr2 R Fr A R Po A R Se A X A F X = X G F Pred R A X

Proof

Step Hyp Ref Expression
1 fprr.1 F = frecs R A G
2 1 fpr1 R Fr A R Po A R Se A F Fn A
3 2 fndmd R Fr A R Po A R Se A dom F = A
4 3 eleq2d R Fr A R Po A R Se A X dom F X A
5 4 biimpar R Fr A R Po A R Se A X A X dom F
6 1 fpr2a R Fr A R Po A R Se A X dom F F X = X G F Pred R A X
7 5 6 syldan R Fr A R Po A R Se A X A F X = X G F Pred R A X