Theorem rdgeq2 6262

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 3218	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))) = (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))
2	1	mpteq2dv 4014	. . 3 ⊢ (𝐴 = 𝐵 → (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) = (𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
3		recseq 6196	. . 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 6260	. 2 ⊢ rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
6		df-irdg 6260	. 2 ⊢ rec(𝐹, 𝐵) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
7	4, 5, 6	3eqtr4g 2195	1 ⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))

Syntax hints:   → wi 4   = wceq 1331  Vcvv 2681   ∪ cun 3064  ∪ ciun 3808   ↦ cmpt 3984  dom cdm 4534  ‘cfv 5118  recscrecs 6194  reccrdg 6259

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 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2119

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2124  df-cleq 2130  df-clel 2133  df-nfc 2268  df-ral 2419  df-rex 2420  df-v 2683  df-un 3070  df-uni 3732  df-br 3925  df-opab 3985  df-mpt 3986  df-iota 5083  df-fv 5126  df-recs 6195  df-irdg 6260

This theorem is referenced by:  rdg0g  6278  oav  6343