Theorem freceq2 6220

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 108	. . . . . . . . 9 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → 𝐴 = 𝐵)
2	1	eleq2d 2169	. . . . . . . 8 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	2	anbi2d 455	. . . . . . 7 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → ((dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴) ↔ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵)))
4	3	orbi2d 745	. . . . . 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 2217	. . . . 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 3958	. . . 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 6133	. . . 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 4754	. 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 6218	. 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 6218	. 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 2157	1 ⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))

Syntax hints:   → wi 4   ∧ wa 103   ∨ wo 670   = wceq 1299   ∈ wcel 1448  {cab 2086  ∃wrex 2376  Vcvv 2641  ∅c0 3310   ↦ cmpt 3929  suc csuc 4225  ωcom 4442  dom cdm 4477   ↾ cres 4479  ‘cfv 5059  recscrecs 6131  freccfrec 6217

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 671  ax-5 1391  ax-7 1392  ax-gen 1393  ax-ie1 1437  ax-ie2 1438  ax-8 1450  ax-10 1451  ax-11 1452  ax-i12 1453  ax-bndl 1454  ax-4 1455  ax-17 1474  ax-i9 1478  ax-ial 1482  ax-i5r 1483  ax-ext 2082

This theorem depends on definitions:  df-bi 116  df-tru 1302  df-nf 1405  df-sb 1704  df-clab 2087  df-cleq 2093  df-clel 2096  df-nfc 2229  df-ral 2380  df-rex 2381  df-v 2643  df-in 3027  df-uni 3684  df-br 3876  df-opab 3930  df-mpt 3931  df-res 4489  df-iota 5024  df-fv 5067  df-recs 6132  df-frec 6218

This theorem is referenced by:  seqeq1  10062  seqeq3  10064