Metamath Proof Explorer


Theorem frrdmss

Description: Show without using the axiom of replacement that the domain of the well-founded recursion generator is a subclass of A . (Contributed by Scott Fenton, 18-Nov-2024)

Ref Expression
Hypothesis frrrel.1 F=frecsRAG
Assertion frrdmss domFA

Proof

Step Hyp Ref Expression
1 frrrel.1 F=frecsRAG
2 eqid f|xfFnxxAyxPredRAyxyxfy=yGfPredRAy=f|xfFnxxAyxPredRAyxyxfy=yGfPredRAy
3 2 1 frrlem7 domFA