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Theorem rdgeq1 6374
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 5516 . . . . . 6 (𝐹 = 𝐺 → (𝐹‘(𝑔𝑥)) = (𝐺‘(𝑔𝑥)))
21iuneq2d 3913 . . . . 5 (𝐹 = 𝐺 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)) = 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))
32uneq2d 3291 . . . 4 (𝐹 = 𝐺 → (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥))) = (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥))))
43mpteq2dv 4096 . . 3 (𝐹 = 𝐺 → (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))) = (𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
5 recseq 6309 . . 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 6373 . 2 rec(𝐹, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐹‘(𝑔𝑥)))))
8 df-irdg 6373 . 2 rec(𝐺, 𝐴) = recs((𝑔 ∈ V ↦ (𝐴 𝑥 ∈ dom 𝑔(𝐺‘(𝑔𝑥)))))
96, 7, 83eqtr4g 2235 1 (𝐹 = 𝐺 → rec(𝐹, 𝐴) = rec(𝐺, 𝐴))
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1353  Vcvv 2739  cun 3129   ciun 3888  cmpt 4066  dom cdm 4628  cfv 5218  recscrecs 6307  reccrdg 6372
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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159
This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2741  df-un 3135  df-in 3137  df-ss 3144  df-uni 3812  df-iun 3890  df-br 4006  df-opab 4067  df-mpt 4068  df-iota 5180  df-fv 5226  df-recs 6308  df-irdg 6373
This theorem is referenced by:  omv  6458  oeiv  6459
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