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 = frecs R A G
Assertion frrdmss dom F A

Proof

Step Hyp Ref Expression
1 frrrel.1 F = frecs R A G
2 eqid f | x f Fn x x A y x Pred R A y x y x f y = y G f Pred R A y = f | x f Fn x x A y x Pred R A y x y x f y = y G f Pred R A y
3 2 1 frrlem7 dom F A