Theorem freceq2 6140

Description: Equality theorem for the finite recursive definition generator. (Contributed by Jim Kingdon, 30-May-2020.)

Assertion

Ref	Expression
freceq2	⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))

Proof of Theorem freceq2

Dummy variables 𝑥 𝑔 𝑚 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		simpl 107	. . . . . . . . 9 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → 𝐴 = 𝐵)
2	1	eleq2d 2157	. . . . . . . 8 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	2	anbi2d 452	. . . . . . 7 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → ((dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴) ↔ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵)))
4	3	orbi2d 739	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → ((∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴)) ↔ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))))
5	4	abbidv 2205	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))} = {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))})
6	5	mpteq2dva 3920	. . . 4 ⊢ (𝐴 = 𝐵 → (𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))}) = (𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))}))
7		recseq 6053	. . . 4 ⊢ ((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))}) = (𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))}) → recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))})) = recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))})))
8	6, 7	syl 14	. . 3 ⊢ (𝐴 = 𝐵 → recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))})) = recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))})))
9	8	reseq1d 4700	. 2 ⊢ (𝐴 = 𝐵 → (recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))})) ↾ ω) = (recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))})) ↾ ω))
10		df-frec 6138	. 2 ⊢ frec(𝐹, 𝐴) = (recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴))})) ↾ ω)
11		df-frec 6138	. 2 ⊢ frec(𝐹, 𝐵) = (recs((𝑔 ∈ V ↦ {𝑥 ∣ (∃𝑚 ∈ ω (dom 𝑔 = suc 𝑚 ∧ 𝑥 ∈ (𝐹‘(𝑔‘𝑚))) ∨ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵))})) ↾ ω)
12	9, 10, 11	3eqtr4g 2145	1 ⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))

Syntax hints:   → wi 4   ∧ wa 102   ∨ wo 664   = wceq 1289   ∈ wcel 1438  {cab 2074  ∃wrex 2360  Vcvv 2619  ∅c0 3284   ↦ cmpt 3891  suc csuc 4183  ωcom 4395  dom cdm 4428   ↾ cres 4430  ‘cfv 5002  recscrecs 6051  freccfrec 6137

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070

This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-ral 2364  df-rex 2365  df-v 2621  df-in 3003  df-uni 3649  df-br 3838  df-opab 3892  df-mpt 3893  df-res 4440  df-iota 4967  df-fv 5010  df-recs 6052  df-frec 6138

This theorem is referenced by:  iseqeq1  9823  iseqeq3  9825