Theorem frrrel 8309

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 2765	. 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 8294	1 ⊢ Rel 𝐹

This proof depends on syntax axioms:   ∧ wa 401   ∧ w3a 1103   = wceq 1570  ∃wex 1812  {cab 2743  ∀wral 3081   ⊆ wss 3906   ↾ cres 5665  Rel wrel 5668  Predcpred 6305   Fn wfn 6535  ‘cfv 6540  (class class class)co 7419  frecscfrecs 8283

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 2148  ax-9 2156  ax-11 2195  ax-12 2216  ax-ext 2737

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 2744  df-cleq 2757  df-clel 2840  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-uni 4875  df-iun 4960  df-br 5112  df-opab 5176  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-rn 5674  df-res 5675  df-ima 5676  df-pred 6306  df-iota 6496  df-fun 6542  df-fn 6543  df-fv 6548  df-ov 7422  df-frecs 8284

This theorem is used by:  wfrrel  8323