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Theorem rdgeq1 6350
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 5495 . . . . . 6 (𝐹 = 𝐺 → (𝐹‘(𝑔𝑥)) = (𝐺‘(𝑔𝑥)))
21iuneq2d 3898 . . . . 5 (𝐹 = 𝐺 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)) = 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))
32uneq2d 3281 . . . 4 (𝐹 = 𝐺 → (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))) = (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))
43mpteq2dv 4080 . . 3 (𝐹 = 𝐺 → (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) = (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
5 recseq 6285 . . 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 6349 . 2 rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))))
8 df-irdg 6349 . 2 rec(𝐺, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
96, 7, 83eqtr4g 2228 1 (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1348  Vcvv 2730  cun 3119   ciun 3873  cmpt 4050  dom cdm 4611  cfv 5198  recscrecs 6283  reccrdg 6348
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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-v 2732  df-un 3125  df-in 3127  df-ss 3134  df-uni 3797  df-iun 3875  df-br 3990  df-opab 4051  df-mpt 4052  df-iota 5160  df-fv 5206  df-recs 6284  df-irdg 6349
This theorem is referenced by:  omv  6434  oeiv  6435
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