Theorem freceq2 6502

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 2277	. . . . . . . 8 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵))
3	2	anbi2d 464	. . . . . . 7 ⊢ ((𝐴 = 𝐵 ∧ 𝑔 ∈ V) → ((dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐴) ↔ (dom 𝑔 = ∅ ∧ 𝑥 ∈ 𝐵)))
4	3	orbi2d 792	. . . . . 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 2325	. . . . 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 4150	. . . 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 6415	. . . 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 4977	. 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 6500	. 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 6500	. 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 2265	1 ⊢ (𝐴 = 𝐵 → frec(𝐹, 𝐴) = frec(𝐹, 𝐵))

Syntax hints:   → wi 4   ∧ wa 104   ∨ wo 710   = wceq 1373   ∈ wcel 2178  {cab 2193  ∃wrex 2487  Vcvv 2776  ∅c0 3468   ↦ cmpt 4121  suc csuc 4430  ωcom 4656  dom cdm 4693   ↾ cres 4695  ‘cfv 5290  recscrecs 6413  freccfrec 6499

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2189

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2194  df-cleq 2200  df-clel 2203  df-nfc 2339  df-ral 2491  df-rex 2492  df-v 2778  df-in 3180  df-uni 3865  df-br 4060  df-opab 4122  df-mpt 4123  df-res 4705  df-iota 5251  df-fv 5298  df-recs 6414  df-frec 6500

This theorem is referenced by:  seqeq1  10632  seqeq3  10634