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 𝐹𝐴

Proof

Step Hyp Ref Expression
1 frrrel.1 𝐹 = frecs ( 𝑅 , 𝐴 , 𝐺 )
2 eqid { 𝑓 ∣ ∃ 𝑥 ( 𝑓 Fn 𝑥 ∧ ( 𝑥𝐴 ∧ ∀ 𝑦𝑥 Pred ( 𝑅 , 𝐴 , 𝑦 ) ⊆ 𝑥 ) ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝑦 𝐺 ( 𝑓 ↾ Pred ( 𝑅 , 𝐴 , 𝑦 ) ) ) ) } = { 𝑓 ∣ ∃ 𝑥 ( 𝑓 Fn 𝑥 ∧ ( 𝑥𝐴 ∧ ∀ 𝑦𝑥 Pred ( 𝑅 , 𝐴 , 𝑦 ) ⊆ 𝑥 ) ∧ ∀ 𝑦𝑥 ( 𝑓𝑦 ) = ( 𝑦 𝐺 ( 𝑓 ↾ Pred ( 𝑅 , 𝐴 , 𝑦 ) ) ) ) }
3 2 1 frrlem7 dom 𝐹𝐴