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Theorem rdgeq1 6580
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 5647 . . . . . 6 (𝐹 = 𝐺 → (𝐹‘(𝑔𝑥)) = (𝐺‘(𝑔𝑥)))
21iuneq2d 4000 . . . . 5 (𝐹 = 𝐺 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)) = 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))
32uneq2d 3363 . . . 4 (𝐹 = 𝐺 → (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))) = (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))
43mpteq2dv 4185 . . 3 (𝐹 = 𝐺 → (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) = (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
5 recseq 6515 . . 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 6579 . 2 rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))))
8 df-irdg 6579 . 2 rec(𝐺, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
96, 7, 83eqtr4g 2289 1 (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1398  Vcvv 2803  cun 3199   ciun 3975  cmpt 4155  dom cdm 4731  cfv 5333  recscrecs 6513  reccrdg 6578
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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213
This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-ral 2516  df-rex 2517  df-v 2805  df-un 3205  df-in 3207  df-ss 3214  df-uni 3899  df-iun 3977  df-br 4094  df-opab 4156  df-mpt 4157  df-iota 5293  df-fv 5341  df-recs 6514  df-irdg 6579
This theorem is referenced by:  omv  6666  oeiv  6667
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