Theorem rdgeq2 6537

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 3354	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))) = (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))
2	1	mpteq2dv 4180	. . 3 ⊢ (𝐴 = 𝐵 → (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) = (𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
3		recseq 6471	. . 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 6535	. 2 ⊢ rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
6		df-irdg 6535	. 2 ⊢ rec(𝐹, 𝐵) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
7	4, 5, 6	3eqtr4g 2289	1 ⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))

Syntax hints:   → wi 4   = wceq 1397  Vcvv 2802   ∪ cun 3198  ∪ ciun 3970   ↦ cmpt 4150  dom cdm 4725  ‘cfv 5326  recscrecs 6469  reccrdg 6534

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-ral 2515  df-rex 2516  df-v 2804  df-un 3204  df-uni 3894  df-br 4089  df-opab 4151  df-mpt 4152  df-iota 5286  df-fv 5334  df-recs 6470  df-irdg 6535

This theorem is referenced by:  rdg0g  6553  oav  6621