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