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Theorem recsfval 6408
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 6398 . 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 3862 . 2 𝐴 = {𝑓 ∣ ∃𝑥 ∈ On (𝑓 Fn 𝑥 ∧ ∀𝑦𝑥 (𝑓𝑦) = (𝐹‘(𝑓𝑦)))}
41, 3eqtr4i 2230 1 recs(𝐹) = 𝐴
Colors of variables: wff set class
Syntax hints:  wa 104   = wceq 1373  {cab 2192  wral 2485  wrex 2486   cuni 3852  Oncon0 4414  cres 4681   Fn wfn 5271  cfv 5276  recscrecs 6397
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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-rex 2491  df-uni 3853  df-recs 6398
This theorem is referenced by:  tfrlem6  6409  tfrlem7  6410  tfrlem8  6411  tfrlem9  6412  tfrlemibfn  6421  tfrlemiubacc  6423  tfrlemi14d  6426  tfrexlem  6427
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