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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 e. 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 e. dom F <-> X e. A ) )
5	4	biimpar	\|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ X e. A ) -> X e. dom F )
6	1	fpr2a	\|- ( ( ( R Fr A /\ R Po A /\ R Se A ) /\ X e. 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 e. A ) -> ( F ` X ) = ( X G ( F \|` Pred ( R , A , X ) ) ) )