Theorem rdgeq2 6331

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 3264	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))) = (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥))))
2	1	mpteq2dv 4067	. . 3 ⊢ (𝐴 = 𝐵 → (𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))) = (𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
3		recseq 6265	. . 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 6329	. 2 ⊢ rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
6		df-irdg 6329	. 2 ⊢ rec(𝐹, 𝐵) = recs((𝑔 ∈ V ↦ (𝐵 ∪ ∪ 𝑥 ∈ dom 𝑔(𝐹‘(𝑔‘𝑥)))))
7	4, 5, 6	3eqtr4g 2222	1 ⊢ (𝐴 = 𝐵 → rec(𝐹, 𝐴) = rec(𝐹, 𝐵))

Syntax hints:   → wi 4   = wceq 1342  Vcvv 2721   ∪ cun 3109  ∪ ciun 3860   ↦ cmpt 4037  dom cdm 4598  ‘cfv 5182  recscrecs 6263  reccrdg 6328

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 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-10 1492  ax-11 1493  ax-i12 1494  ax-bndl 1496  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522  ax-ext 2146

This theorem depends on definitions:  df-bi 116  df-tru 1345  df-nf 1448  df-sb 1750  df-clab 2151  df-cleq 2157  df-clel 2160  df-nfc 2295  df-ral 2447  df-rex 2448  df-v 2723  df-un 3115  df-uni 3784  df-br 3977  df-opab 4038  df-mpt 4039  df-iota 5147  df-fv 5190  df-recs 6264  df-irdg 6329

This theorem is referenced by:  rdg0g  6347  oav  6413