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Theorem rdgeq1 5981
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 5197 . . . . . 6 (𝐹 = 𝐺 → (𝐹‘(𝑔𝑥)) = (𝐺‘(𝑔𝑥)))
21iuneq2d 3703 . . . . 5 (𝐹 = 𝐺 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)) = 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))
32uneq2d 3126 . . . 4 (𝐹 = 𝐺 → (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))) = (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))
43mpteq2dv 3869 . . 3 (𝐹 = 𝐺 → (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) = (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
5 recseq 5944 . . 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 5980 . 2 rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))))
8 df-irdg 5980 . 2 rec(𝐺, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
96, 7, 83eqtr4g 2138 1 (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1284  Vcvv 2601  cun 2971   ciun 3678  cmpt 3839  dom cdm 4363  cfv 4922  recscrecs 5942  reccrdg 5979
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063
This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rex 2354  df-v 2603  df-un 2977  df-in 2979  df-ss 2986  df-uni 3602  df-iun 3680  df-br 3786  df-opab 3840  df-mpt 3841  df-iota 4887  df-fv 4930  df-recs 5943  df-irdg 5980
This theorem is referenced by:  omv  6058  oeiv  6059
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