Theorem freceq2 6664

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 109	. . . . . . . . 9 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → 𝐴 = 𝐵)
2	1	eleq2d 2308	. . . . . . . 8 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	2	anbi2d 468	. . . . . . 7 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → ((dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴) ↔ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵)))
4	3	orbi2d 802	. . . . . 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 2358	. . . . 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 4221	. . . 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 6577	. . . 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 5062	. 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 6662	. 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 6662	. 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 2296	1 ⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ∨ wo 720   = wceq 1402   ∈ wcel 2209  {cab 2224  ∃wrex 2529  Vcvv 2821  ∅c0 3520   ↦ cmpt 4192  suc csuc 4510  ωcom 4737  dom cdm 4774   ↾ cres 4776  ‘cfv 5377  recscrecs 6575  freccfrec 6661

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-in 3226  df-uni 3936  df-br 4131  df-opab 4193  df-mpt 4194  df-res 4786  df-iota 5337  df-fv 5385  df-recs 6576  df-frec 6662

This theorem is used by:  seqeq1  10887  seqeq3  10889