Theorem frrrel 8306

Description: Show without using the axiom of replacement that the well-founded recursion generator gives a relation. (Contributed by Scott Fenton, 18-Nov-2024.)

Hypothesis

Ref	Expression
frrrel.1	⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)

Assertion

Ref	Expression
frrrel	⊢ Rel 𝐹

Proof of Theorem frrrel

Dummy variables 𝑓 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqid 2760	. 2 ⊢ {𝑓 ∣ ∃𝑥(𝑓 Fn 𝑥 ∧ (𝑥 ⊆ 𝐴 ∧ ∀𝑦 ∈ 𝑥 Pred(𝑅, 𝐴, 𝑦) ⊆ 𝑥) ∧ ∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑦𝐺(𝑓 ↾ Pred(𝑅, 𝐴, 𝑦))))} = {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		frrrel.1	. 2 ⊢ 𝐹 = frecs(𝑅, 𝐴, 𝐺)
3	1, 2	frrlem6 8291	1 ⊢ Rel 𝐹

This proof depends on syntax axioms:   ∧ wa 401   ∧ w3a 1103   = wceq 1570  ∃wex 1812  {cab 2738  ∀wral 3076   ⊆ wss 3899   ↾ cres 5657  Rel wrel 5660  Predcpred 6298   Fn wfn 6528  ‘cfv 6533  (class class class)co 7414  frecscfrecs 8280

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-11 2194  ax-12 2213  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-iun 4953  df-br 5104  df-opab 5168  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-pred 6299  df-iota 6489  df-fun 6535  df-fn 6536  df-fv 6541  df-ov 7417  df-frecs 8281

This theorem is used by:  wfrrel  8320