Theorem frrlem5 8308

Description: Lemma for well-founded recursion. State the well-founded recursion generator in terms of the acceptable functions. (Contributed by Scott Fenton, 27-Aug-2022.)

Hypotheses

Ref	Expression
frrlem5.1	⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
frrlem5.2	⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)

Assertion

Ref	Expression
frrlem5	⊢ 𝐹 = ∪ 𝐵

Distinct variable groups:   𝐴,𝑓,𝑥,𝑦   𝑓,𝐺,𝑥,𝑦   𝑅,𝑓,𝑥,𝑦

Allowed substitution hints:   𝐵(𝑥, 𝑦, 𝑓)   𝐹(𝑥, 𝑦, 𝑓)

Proof of Theorem frrlem5

Step	Hyp	Ref	Expression
1		df-frecs 8299	. 2 ⊢ frecs(𝑅, 𝐴, 𝐺) = ∪ {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
2		frrlem5.2	. 2 ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)
3		frrlem5.1	. . 3 ⊢ 𝐵 = {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
4	3	unieqi 4879	. 2 ⊢ ∪ 𝐵 = ∪ {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))}
5	1, 2, 4	3eqtr4i 2794	1 ⊢ 𝐹 = ∪ 𝐵

This proof depends on syntax axioms:   ∧ wa 401   ∧ w3a 1103   = wceq 1570  ∃wex 1812  {cab 2739  ∀wral 3077   ⊆ wss 3899  ∪ cuni 4867   ↾ cres 5653  Predcpred 6303   Fn wfn 6533  ‘cfv 6538  (class class class)co 7420  frecscfrecs 8298

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-ss 3916  df-uni 4868  df-frecs 8299

This theorem is used by:  frrlem6  8309  frrlem7  8310  frrlem8  8311  frrlem9  8312  frrlem10  8313  frrlem14  8317