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Theorem recsfval 6118
Description: Lemma for transfinite recursion. The definition recs is the union of all acceptable functions. (Contributed by Mario Carneiro, 9-May-2015.)
Hypothesis
Ref Expression
tfrlem.1 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
Assertion
Ref Expression
recsfval recs(𝐹) = 𝐴
Distinct variable group:   𝑥,𝑓,𝑦,𝐹
Allowed substitution hints:   𝐴(𝑥,𝑦,𝑓)

Proof of Theorem recsfval
StepHypRef Expression
1 df-recs 6108 . 2 recs(𝐹) = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
2 tfrlem.1 . . 3 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
32unieqi 3685 . 2 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
41, 3eqtr4i 2118 1 recs(𝐹) = 𝐴
Colors of variables: wff set class
Syntax hints:  wa 103   = wceq 1296  {cab 2081  wral 2370  wrex 2371   cuni 3675  Oncon0 4214  cres 4469   Fn wfn 5044  cfv 5049  recscrecs 6107
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 668  ax-5 1388  ax-7 1389  ax-gen 1390  ax-ie1 1434  ax-ie2 1435  ax-8 1447  ax-10 1448  ax-11 1449  ax-i12 1450  ax-bndl 1451  ax-4 1452  ax-17 1471  ax-i9 1475  ax-ial 1479  ax-i5r 1480  ax-ext 2077
This theorem depends on definitions:  df-bi 116  df-tru 1299  df-nf 1402  df-sb 1700  df-clab 2082  df-cleq 2088  df-clel 2091  df-nfc 2224  df-rex 2376  df-uni 3676  df-recs 6108
This theorem is referenced by:  tfrlem6  6119  tfrlem7  6120  tfrlem8  6121  tfrlem9  6122  tfrlemibfn  6131  tfrlemiubacc  6133  tfrlemi14d  6136  tfrexlem  6137
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