Theorem rdgeq2 6425

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 3306	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))) = (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))
2	1	mpteq2dv 4120	. . 3 ⊢ (𝐴 = 𝐵 → (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) = (𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
3		recseq 6359	. . 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 6423	. 2 ⊢ rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
6		df-irdg 6423	. 2 ⊢ rec(𝐹, 𝐵) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
7	4, 5, 6	3eqtr4g 2251	1 ⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))

Syntax hints:   → wi 4   = wceq 1364  Vcvv 2760   ∪ cun 3151  ∪ ciun 3912   ↦ cmpt 4090  dom cdm 4659  ‘cfv 5254  recscrecs 6357  reccrdg 6422

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-rex 2478  df-v 2762  df-un 3157  df-uni 3836  df-br 4030  df-opab 4091  df-mpt 4092  df-iota 5215  df-fv 5262  df-recs 6358  df-irdg 6423

This theorem is referenced by:  rdg0g  6441  oav  6507