Theorem rdgeq2 6340

Description: Equality theorem for the recursive definition generator. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 9-May-2015.)

Assertion

Ref	Expression
rdgeq2	⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))

Proof of Theorem rdgeq2

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

Step	Hyp	Ref	Expression
1		uneq1 3269	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))) = (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))
2	1	mpteq2dv 4073	. . 3 ⊢ (𝐴 = 𝐵 → (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) = (𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
3		recseq 6274	. . 3 ⊢ ((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) = (𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) → recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))))
4	2, 3	syl 14	. 2 ⊢ (𝐴 = 𝐵 → recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))))
5		df-irdg 6338	. 2 ⊢ rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
6		df-irdg 6338	. 2 ⊢ rec(𝐹, 𝐵) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
7	4, 5, 6	3eqtr4g 2224	1 ⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))

Syntax hints:   → wi 4   = wceq 1343  Vcvv 2726   ∪ cun 3114  ∪ ciun 3866   ↦ cmpt 4043  dom cdm 4604  ‘cfv 5188  recscrecs 6272  reccrdg 6337

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ral 2449  df-rex 2450  df-v 2728  df-un 3120  df-uni 3790  df-br 3983  df-opab 4044  df-mpt 4045  df-iota 5153  df-fv 5196  df-recs 6273  df-irdg 6338

This theorem is referenced by:  rdg0g  6356  oav  6422