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Theorem rdgeq1 6515
Description: Equality theorem for the recursive definition generator. (Contributed by NM, 9-Apr-1995.) (Revised by Mario Carneiro, 9-May-2015.)
Assertion
Ref Expression
rdgeq1 (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))

Proof of Theorem rdgeq1
Dummy variables 𝑥 𝑔 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 fveq1 5625 . . . . . 6 (𝐹 = 𝐺 → (𝐹‘(𝑔𝑥)) = (𝐺‘(𝑔𝑥)))
21iuneq2d 3989 . . . . 5 (𝐹 = 𝐺 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)) = 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))
32uneq2d 3358 . . . 4 (𝐹 = 𝐺 → (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))) = (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))
43mpteq2dv 4174 . . 3 (𝐹 = 𝐺 → (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) = (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
5 recseq 6450 . . 3 ((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) = (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))) → recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))))) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))))
64, 5syl 14 . 2 (𝐹 = 𝐺 → recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))))) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))))
7 df-irdg 6514 . 2 rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))))
8 df-irdg 6514 . 2 rec(𝐺, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
96, 7, 83eqtr4g 2287 1 (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1395  Vcvv 2799  cun 3195   ciun 3964  cmpt 4144  dom cdm 4718  cfv 5317  recscrecs 6448  reccrdg 6513
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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211
This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ral 2513  df-rex 2514  df-v 2801  df-un 3201  df-in 3203  df-ss 3210  df-uni 3888  df-iun 3966  df-br 4083  df-opab 4145  df-mpt 4146  df-iota 5277  df-fv 5325  df-recs 6449  df-irdg 6514
This theorem is referenced by:  omv  6599  oeiv  6600
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