Theorem frrlem5 8322

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 8313	. 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 4882	. 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 2796	1 ⊢ 𝐹 = ∪ 𝐵

This proof depends on syntax axioms:   ∧ wa 401   ∧ w3a 1103   = wceq 1569  ∃wex 1811  {cab 2741  ∀wral 3080   ⊆ wss 3900  ∪ cuni 4870   ↾ cres 5657  Predcpred 6310   Fn wfn 6542  ‘cfv 6547  (class class class)co 7430  frecscfrecs 8312

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-v 3456  df-ss 3901  df-uni 4871  df-frecs 8313

This theorem is used by:  frrlem6  8323  frrlem7  8324  frrlem8  8325  frrlem9  8326  frrlem10  8327  frrlem14  8331